Inverse monoids associated with the complexity class NP

TL;DR

This paper explores the P versus NP problem by examining inverse monoids and their properties, aiming to understand the existence of one-way functions and the fundamental complexity class distinctions.

Contribution

It introduces a novel algebraic approach using inverse monoids to analyze P vs NP and the existence of injective one-way functions.

Findings

Inverse monoid properties relate to P vs NP status

Characterization of one-way functions through monoid structures

Provides algebraic criteria for complexity class separation

Abstract

We study the P versus NP problem through properties of functions and monoids, continuing the work of [3]. Here we consider inverse monoids whose properties and relationships determine whether P is different from NP, or whether injective one-way functions (with respect to worst-case complexity) exist.

Full text

1 Introduction

We give a few definitions before motivating the monoid approach to the P versus NP problem. Some of these notions appeared already in [3].

By function we mean a partial function $A^{*}\to A^{*}$, where $A$ is a finite alphabet (usually, $A=\{0,1\}$), and $A^{*}$ denotes the set of all finite strings over $A$. Let ${\sf Dom}(f)$ ($\subseteq A^{*}$) denote the domain of $f$, i.e., $\{x\in A^{*}:f(x)$ is defined}; and let ${\sf Im}(f)$ ($\subseteq A^{*}$) denote the image (or range) of $f$, i.e., $\{f(x):x\in{\sf Dom}(f)\}$. The length of $x\in A^{*}$ is denoted by $|x|$. The restriction of $f$ to $X\subseteq A^{*}$ is denoted by $f|_{X}$, and the identity function on $X$ is denoted by ${\sf id}_{X}$ or ${\sf id}|_{X}$.

Definition 1.1

(inverse, co-inverse, mutual inverse)*  A function $f^{\prime}$ is an inverse of a function $f$ iff $\,f\circ f^{\prime}\circ f=f$. In that case we also say that $f$ is a co-inverse of $f^{\prime}$.  If $f^{\prime}$ is both an inverse and a co-inverse of $f$, we say that $f^{\prime}$ is a mutual inverse of $f$, or that $f^{\prime}$ and $f$ are mutual inverses (of each other).111The terminology about inverses varies. In semigroup theory, $f^{\prime}$ such that $ff^{\prime}f=f$ is called a semi-inverse or a pseudo-inverse of $f$, in numerical mathematics $f^{\prime}$ is called a generalized inverse, in ring theory and in category theory it is called a weak inverse or a von Neumann inverse. In semigroup theory the term “inverse” of $f$ is only used if $f^{\prime}ff^{\prime}=f^{\prime}$ holds in addition to $ff^{\prime}f=f$. Co-inverses don’t seem to have a name in the literature. *

It is easy to see that $f^{\prime}$ is an inverse of $f$ iff for every $y\in{\sf Im}(f)$: $f^{\prime}(y)$ is defined and $f^{\prime}(y)\in f^{-1}(y)$. Hence, in particular, ${\sf Im}(f)\subseteq{\sf Dom}(f^{\prime})$. Also, $f^{\prime}$ is an inverse of $f$ iff $\,f\circ f^{\prime}|_{{\sf Im}(f)}={\sf id}_{{\sf Im}(f)}$.

An element $f$ of a semigroup $S$ is called regular (or more precisely, regular in $S$) iff there exists $f^{\prime}\in S$ such that $ff^{\prime}f=f$; in that case, $f^{\prime}$ is called an inverse of $f$ (more precisely, an inverse in $S$ of $f$), and $f$ is called a co-inverse of $f^{\prime}$ (in $S$). A semigroup $S$ is called regular iff all the elements of $S$ are regular in $S$; see e.g. [6, 5]. An inverse semigroup is, by definition, a semigroup $S$ such that every element of $S$ has exactly one mutual inverse in $S$; see e.g. [9, 6, 5].

The connection between the P versus NP problem and inverses comes from the following well known characterization: $\,{\sf P}\neq{\sf NP}\,$ iff there exists a function that is computable in polynomial time (by a deterministic Turing machine) and that is polynomially balanced, but that does not have an inverse that is computable in polynomial time; see e.g. [8] p. 33, and Section 1 of [3]. A function $f$: $A^{*}\to A^{*}$ is called polynomially balanced iff there exists a polynomial $p$ such that for all $x\in{\sf Dom}(f)$:   $|f(x)|\leq p(|x|)$ and $|x|\leq p(|f(x)|)$. Functions that are polynomial-time computable and polynomially balanced, but that have no inverse of that type, are one-way functions in the sense of worst-case complexity.222One-way functions in this sense are not necessarily useful for cryptography, but they are relevant for P vs. NP. In [3] this characterization of the P versus NP problem was reformulated in terms of regularity of semigroups, by introducing the monoid fP, defined as follows.

Definition 1.2

.**

${\sf fP}\ =\ \{f:f$* is a function $A^{*}\to A^{*}$ that is polynomially balanced, and*

computable by a deterministic polynomial-time Turing machine}.

In particular, when $f\in{\sf fP}$ then ${\sf Dom}(f)$ is in P. The set fP is closed under composition of functions, and the identity function is in fP, so fP is a monoid. The above characterization of P $\neq$ NP now becomes:

P $\neq$ NP   iff   fP is a non-regular monoid.

We will use some more definitions that apply to functions.

For any function $f:X\to Y$, the equivalence relation ${\sf mod}_{f}$ on ${\sf Dom}(f)$ is defined by: $\,x_{1}\ {\sf mod}_{f}\ x_{2}\,$ iff $\,f(x_{1})=f(x_{2})$. The set of equivalence classes of $\,{\sf mod}_{f}\,$ is $\,\{f^{-1}(y):y\in{\sf Im}(f)\}$.

A choice set for $f$ is, by definition, a subset of ${\sf Dom}(f)$ that contains exactly one element of each ${\sf mod}_{f}$-class. A choice function for $f$ is, by definition, any inverse $f^{\prime}$ of $f$ such that ${\sf Dom}(f^{\prime})={\sf Im}(f)$. A choice function $f^{\prime}$ maps each element $y\in{\sf Im}(f)$ injectively to an element of $f^{-1}(y)$, and every ${\sf mod}_{f}$-class contains exactly one element of ${\sf Im}(f^{\prime})$; so ${\sf Im}(f^{\prime})$ is a choice set. For a choice function $f^{\prime}$ we have $\,f\circ f^{\prime}=f\circ f^{\prime}|_{{\sf Im}(f)}={\sf id}_{{\sf Im}(f)}$, hence $\,ff^{\prime}f=f$; we also have $f^{\prime}ff^{\prime}=f^{\prime}$; more generally, if $f^{\prime}_{1},f^{\prime}_{2}$ are two choice functions for the same function $f$ then $\,f^{\prime}_{2}ff^{\prime}_{1}=f^{\prime}_{2}$. A choice function for $f$ is uniquely determined by $f$ and a choice set.

A representative choice function for $f$ is, by definition, any function $r$ such that ${\sf Im}(r)\subseteq{\sf Dom}(r)={\sf Dom}(f)$, ${\sf Im}(r)$ is a choice set for $f$, ${\sf mod}_{r}={\sf mod}_{f}$, and $r|_{{\sf Im}(r)}={\sf id}_{{\sf Im}(r)}$. Thus, $r$ maps each ${\sf mod}_{f}$-class $[x]_{f}$ to one and the same chosen element of $[x]_{f}$. It follows that $r$ is an idempotent; and if $r_{1},r_{2}$ are two representative choice functions for the same function $f$ then $\,r_{2}\circ r_{1}(.)=r_{2}(.)$. If $f\in{\sf fP}$ and if in addition $r\in{\sf fP}$, then $f\equiv_{\cal L}r$ (by Prop. 2.1 in [3]). So in that case $f$ is a regular element of fP, since it is $\cal L$-equivalent to an idempotent. Hence, if every function in fP had a representative choice function in fP, then fP would be regular, which would imply ${\sf P}={\sf NP}$.

There is a one-to-one correspondence between the choice functions of $f$ and the representative choice functions: If $c$ is a choice function then $\rho_{c}(.)=c\circ f(.)$ is a representative choice function; and if $r$ is a representative choice function of $f$ then $v_{r}(.)=r\circ f^{-1}(.)$ is a choice function (where $f^{-1}$ is the inverse relation of $f$, and $\circ$ is composition of relations). Moreover, if $r$ is a representative choice function of $f$ then $\rho_{v_{r}}=r$; and if $c$ is a choice function of $f$ then $v_{\rho_{c}}=c$.

For functions in general, the existence of a choice function, the existence of a representative choice function, and the existence of a choice set, are equivalent to the axiom of choice. Because of this connection we are especially interested in inverses $f^{\prime}$ of $f$ that are choice functions, i.e., that satisfy ${\sf Dom}(f^{\prime})={\sf Im}(f)$.

These three formulations of the axiom of choice can also be considered for fP: (1) “Every $f\in{\sf fP}$ has an inverse in fP.” Equivalently, “$f$ has a choice function in fP”, and also equivalently, “$f$ has a mutual inverse in fP”. (2) “Every $f\in{\sf fP}$ has a representative choice function in fP.” (3) “Every $f\in{\sf fP}$ has a choice set in P.”   It is an open problem problem whether the three statements are true.

We saw that (1) is true iff ${\sf P}={\sf NP}$. Moreover, (1) implies (2); indeed, if $f$ has an inverse $f^{\prime}\in{\sf fP}$ then $f^{\prime}f$ is a representative choice function in fP. And (2) implies (3); indeed, if $r\in{\sf fP}$ is a representative choice function of $f$, then $r$ is regular (being an idempotent), hence ${\sf Im}(r)$ belongs to P by Prop. 1.9 in [3]; and ${\sf Im}(r)$ is a choice set for $r$ and for $f$. It remains an open problem whether other implications between (1), (2), (3) hold. Nevertheless, we have the following.

Proposition 1.3

There exists an injective one-way function (for worst-case complexity) iff there exists a one-way function that has a choice set in P.

Proof. If $g\in{\sf fP}$ is injective then ${\sf Dom}(g)$ is a choice set for $g$, and ${\sf Dom}(g)\in{\sf P}$. If $g$ is also one-way, then it is therefore a one-way function with a choice set in P.

The converse follows immediately from the next Lemma.    $\Box$

Lemma 1.4

If $f\in{\sf fP}$ has a choice set $C\in{\sf P}$, but $f$ is not regular (i.e., (3) holds but (1) does not), then the restriction $f|_{C}$ is an injective one-way function.

Proof. Since $f\in{\sf fP}$ and $C\in{\sf P}$ then we have $f|_{C}\in{\sf fP}$. Moreover, if $C$ is a choice set for $f$ then $f|_{C}$ is injective. And if $C$ is a choice set, any inverse of $f|_{C}$ is also an inverse of $f$. Hence $f|_{C}$ has no inverse in fP (since $f$ is not regular).    $\Box$

The motivation for this paper is based on the following simple observation: If $f:A^{*}\to A^{*}$ is any partial function and if $f^{\prime}$ is an inverse of $f$, then the restriction $f^{\prime}|_{{\sf Im}(f)}$ is an injective inverse of $f$. Moreover, if $f\in{\sf fP}$ and $f^{\prime}\in{\sf fP}$, then ${\sf Im}(f)\in{\sf P}$ (by Prop. 1.9 in [3]); hence $f^{\prime}|_{{\sf Im}(f)}\in{\sf fP}$. Thus we have: $\,{\sf P}\neq{\sf NP}\,$ iff there exists $f\in{\sf fP}$ such that $f$ has no injective inverse in fP.

The classes P and NP are defined in terms of sets of strings (“formal languages”). To add more structure we use functions, and we characterized P and NP by properties of monoids of functions [3]. Our next step, in the present paper, is to characterize P versus NP by properties of inverse monoids, and of groups. However, whether this approach will help solve the P versus NP problem, remains to be seen.

Overview: In Section 2 we introduce the monoid invfP consisting of all injective regular functions in fP. We prove that every regular function in fP has an inverse in invfP, and that invfP is a maximal inverse submonoid of fP.

In Section 3 we show that the polynomial-time polynomially balanced injective Turing machines form a machine model for invfP, i.e., that a function $f$ belongs to invfP iff $f$ is computed by such a Turing machine. We conclude from this that invfP is finitely generated. We also consider polynomial-time polynomially balanced injective Turing machines with an NP-oracle, and we show that the set ${\sf invfP}^{\sf(NP)}$ of functions computed by such Turing machines is a finitely generated inverse monoid. To prove the latter, we show that there exist languages that are NP-complete with respect to one-one reductions in invfP. We show that every function in fP has an inverse in ${\sf invfP}^{\sf(NP)}$.

In Section 4 we show that ${\sf invfP}\neq{\sf invfP}^{\sf(NP)}$ iff ${\sf P}\neq{\sf NP}$. We introduce the set cofP of all functions in ${\sf invfP}^{\sf(NP)}$ that have an inverse in fP, i.e., that are a co-inverse of a function in fP. We show that this is a finitely generated monoid. We prove that ${\sf P}\neq{\sf NP}$ iff ${\sf invfP}\neq{\sf cofP}$, iff ${\sf cofP}\neq{\sf invfP}^{\sf(NP)}$, iff cofP is not regular.

We also introduce the monoid injfP of all injective functions in fP; $\,{\sf injfP}$ is also equal to ${\sf invfP}^{\sf(NP)}\,\cap\,{\sf fP}$. We show that the inverse monoid invfP is the set of the regular elements of injfP. Hence, injective one-way functions (for worst-case complexity) exist iff ${\sf invfP}\neq{\sf injfP}$. We do not know whether injfP is finitely generated; if it turns out that injfP is not finitely generated then ${\sf P}\neq{\sf NP}$.

In Section 5 we show that every element of fP is equivalent (with respect to inversive-reductions) to an element of fP that has an inverse in a subgroup of ${\sf invfP}^{\sf(NP)}$.

Notation for the monoids used, and their definition:

[TABLE]

The relation between these monoids is shown in Figure 1 (after Prop. 4.13).

2 Injective inverses of regular elements

For any function $f:A^{*}\to A^{*}$, we will be interested in inverses $f^{\prime}$ of $f$ that satisfy ${\sf Dom}(f^{\prime})={\sf Im}(f)$; such inverses are necessarily injective. Moreover, when ${\sf Dom}(f^{\prime})={\sf Im}(f)$ we not only have $ff^{\prime}f=f$, but also $f^{\prime}ff^{\prime}=f^{\prime}$. So $f$ and $f^{\prime}$ are mutual inverses.

An injective inverse $f^{\prime}$, as above, has exactly one inverse $f^{\prime\prime}$ that satisfies ${\sf Dom}(f^{\prime\prime})={\sf Im}(f^{\prime})$, namely $f^{\prime\prime}=f^{\prime-1}$ (the set-theoretic inverse function of $f^{\prime}$); moreover, ${\sf Im}(f^{\prime-1})={\sf Im}(f)$. And $f^{\prime-1}=f|_{{\sf Im}(f^{\prime})}$, i.e., the restriction of $f$ to the choice set ${\sf Im}(f^{\prime})$ of $f^{\prime}$.

Here are some more simple facts about inverses:  If $f^{\prime}$ is a mutual inverse of $f$ then $\,{\sf Im}(f^{\prime})\subseteq{\sf Dom}(f)$. Moreover, $f^{\prime}$ is a mutual inverse of $f\,$ iff $\,{\sf Im}(f^{\prime})={\sf Im}(f^{\prime}|_{{\sf Im}(f)})$. Note that $f^{\prime}|_{{\sf Im}(f)}$ is the choice function determined by $f^{\prime}$, so ${\sf Dom}(f^{\prime}|_{{\sf Im}(f)})={\sf Im}(f)$. Thus an inverse $f^{\prime}$ is a mutual inverse of $f$  iff  ${\sf Im}(f^{\prime})$ is the choice set determined by $f^{\prime}$ in ${\sf Dom}(f)$. As a consequence, a mutual inverse $f^{\prime}$ of $f$ is injective iff ${\sf Dom}(f^{\prime})={\sf Im}(f)$.

Proposition 2.1

If $g_{1}$ and $g_{2}$ are injective regular elements of fP, then $g_{2}\circ g_{1}$ is also injective and regular in fP.

Proof. The composite of injective functions is obviously injective. Since $g_{i}$ is regular (for $i=1,2$), it has the injective regular function $g_{i}^{-1}$ as an inverse. Indeed, ${\sf Im}(g_{i})\in{\sf P}$, and $g_{i}^{-1}=g_{i}^{\prime}|_{{\sf Im}(g_{i})}$ for any inverse $g_{i}^{\prime}\in{\sf fP}$ of $g_{i}$; hence, $g_{i}^{-1}\in{\sf fP}$. Thus, the injective function $g_{1}^{-1}\circ g_{2}^{-1}$ is a mutual inverse of $g_{2}\circ g_{1}$, and ${\sf Dom}(g_{1}^{-1}\circ g_{2}^{-1})={\sf Im}(g_{2}\circ g_{1})$.    $\Box$

Note that in general, the product of regular elements in fP need not be regular (unless ${\sf P}={\sf NP}$); but by the above proposition, the product of injective regular elements is regular.

Definition 2.2

Let invfP denote the set of injective regular elements of fP.

Corollary 2.3

The set invfP of injective regular elements of fP is an inverse monoid. Every regular element of fP has an inverse in invfP.      $\Box$

Corollary 2.3 implies that ${\sf P}={\sf NP}$ iff every element of fP has an inverse in invfP.

Another motivation for invfP will be seen in Prop. 3.11, where is is shown that there exist languages that are NP-complete with respect to one-one reductions in invfP.

A monoid $M_{1}$ is called an inverse submonoid of a monoid $M$ iff $M_{1}$ is submonoid of $M$, and $M_{1}$ is an inverse monoid by itself. So, an element of $M_{1}$ has exactly one mutual inverse in $M_{1}$, but it could have additional mutual inverses in $M$.

Proposition 2.4

The inverse submonoid ${\sf invfP}$ is a maximal inverse submonoid of fP; i.e., if $M$ is an inverse monoid such that $\,{\sf invfP}\subseteq M\subseteq{\sf fP}$, then $\,{\sf invfP}=M$.

Proof. Let $M$ be an inverse submonoid such that ${\sf invfP}\subseteq M\subseteq{\sf fP}$. For a contradiction, let us assume that ${\sf invfP}\neq M$. Then $M$ contains a non-injective element $f$, which is regular (since $M$ is inverse). Since $f$ is regular, $f$ has a choice function $f^{\prime}_{1}$; since $f$ is non-injective, $f$ has at least one other choice function $f^{\prime}_{2}$.

Moreover, since $f^{\prime}_{1}\in{\sf invfP}$, we can pick $f^{\prime}_{2}$ in such a way that $f^{\prime}_{2}\in{\sf invfP}$ too. Indeed, let $x_{1},x_{2}\in{\sf Dom}(f)$ be such that $f(x_{1})=f(x_{2})$, and $x_{1}$ is in the choice set ${\sf Im}(f_{1}^{\prime})$, and $x_{2}\not\in{\sf Im}(f_{1}^{\prime})$. Let $\tau$ be the transposition of $x_{1}$ and $x_{2}$ (and $\tau$ is the identity elsewhere). Then $f_{2}^{\prime}=f_{1}^{\prime}\circ\tau$ belongs to invfP, and is the same as $f_{1}^{\prime}$, except that $x_{2}$ has replaced $x_{1}$ in the choice set.

Since the choice functions $f_{1}^{\prime},f_{2}^{\prime}$ belong to invfP ($\subseteq M$), $f$ has two mutual inverses in $M$, which implies that $M$ is not an inverse monoid.    $\Box$

Remarks. (1) Prop. 2.4 holds, whether fP is regular or not.

(2) An inverse submonoid of fP need not consist of injective functions only, and the submonoid invfP is not the only maximal inverse submonoid of fP. E.g., fP contains non-injective idempotents, and these are contained in maximal inverse monoids that are different from invfP (since they contain non-injective elements).

Proposition 2.5

A maximal subgroup of fP, and more generally, any $\cal L$-class of fP, is either disjoint from invfP or entirely contained in invfP.

Proof. An $\cal L$-class of fP that intersects invfP contains injective elements, hence it consist entirely of injective elements (by Prop. 2.1 in [3]). Also, an $\cal L$-class of fP that intersects invfP contains regular elements, hence it consist entirely of regular elements (it is a well-known fact from semigroup theory that if a $\cal D$-class contains regular elements then it consists entirely of regular elements; see e.g. [5, 6]). Hence, this $\cal L$-class consists entirely of regular injective elements, hence it is contained in invfP.    $\Box$

On the other hand, every regular $\cal R$-class of fP intersects both invfP and ${\sf fP}-{\sf invfP}$. Indeed, for every regular element $f\in{\sf fP}$ we have $f\equiv_{\cal R}{\sf id}_{{\sf Im}(f)}\,$ (and ${\sf Im}(f)$ is in P when $f$ is regular). But a regular $\cal R$-class always contains some non-injective elements (by Prop. 2.1 in [3]).

3 Computing inverses

There is a machine model that exactly characterizes the injective inverses of the regular elements of fP, namely the polynomially balanced polynomial-time injective Turing machines. As a consequence we will prove that invfP is finitely generated.

Turing machines of the above type, with an NP-oracle added, form a model of computation for another inverse monoid, ${\sf invfP}^{\sf(NP)}$ (given in Def. 3.6), which contains some injective inverses for every element of fP.

3.1 Injective Turing machines

There is a very simple machine model for the elements of invfP, namely the polynomially balanced polynomial-time injective Turing machines. A deterministic Turing machine is called injective iff the transition table of the Turing machine describes a (finite) injective function; for details, see [1, 2].

The reverse of a Turing machine $M$ is the machine obtained by reversing every transition of $M$ (and also switching start and accept states). The reverse of a deterministic Turing machine $M$ is is not deterministic, unless $M$ is injective.

Remark. In the literature, injective Turing machines are called “reversible”, because of historic connections with the study of computation as a “reversible process” (in the sense of thermodynamics). However, calling an injective Turing machine “reversible” can be misleading, since the transitions obtained by reversing an injective Turing machine $M$ are not part of $M$ (but of a different machine).

Proposition 3.1

.**

(1)*  The reverse of a polynomially balanced polynomial-time injective Turing machine is also a polynomially balanced polynomial-time injective Turing machine. If a function $f$ is polynomially balanced and it is computed by a polynomial-time injective Turing machine, then $f^{-1}$ is also polynomially balanced and it is computed by a polynomial-time injective Turing machine. *

(2)*  There exists a polynomial-time injective Turing machine (that is not polynomially balanced) whose reverse does not have polynomial time-complexity. There exists an injective function $g$ that is computed by a polynomial-time injective Turing machine, whose inverse $g^{-1}$ is not computable by a polynomial-time Turing machine.*

Proof. (1) An injective Turing machine $M$, computing an injective function $f$, can be run in reverse. This yields a new injective Turing machine $M^{\prime}$, computing $f^{-1}$. If $M$ runs in polynomial time, with polynomial $p_{T}(.)$, and it is polynomially balanced, with polynomial $p_{B}(.)$, then $M^{\prime}$ is also polynomially balanced, with polynomial $p_{B}(.)$ too, and runs in polynomial time, with polynomial bound $p_{T}\circ p_{B}(.)$.

(2) An example of such a function is

$g:\ a^{(2^{m})}\ \longmapsto\ a^{m}$,   for all $m\in{\mathbb{N}}$,

where $a$ is a fixed letter; $\,g$ is undefined on any input that is not of the form $a^{k}$ with $k$ a power of 2.

Obviously, $g$ is injective and polynomial-time computable, but not polynomially balanced. Hence, $g^{-1}$ is not computable in polynomial time, since its output is exponentially longer than its input. However, $g$ is computable in linear time by the following injective Turing machine: We use a Turing machine with a rubber tape for the input, and an ordinary tape for the output. A rubber tape is a tape on which one can not only replace one letter by another one (as on an ordinary tape), but where one also can insert a letter or remove a letter (in one transition). A rubber tape can easily be simulated by two stacks; a stack is a special case of an ordinary tape. The machine that computes $g$ works as follows, in outline. It has a main loop (“while $\ldots$”); it has an inner loop that is executed at the beginning of the body of the main loop. For this inner loop, the machine uses two states to count the input positions modulo 2, and erases every second $a$ (which is possible on a rubber tape). When the right end of the input tape is reached, the state must correspond to an even number of letters $a$ on the input tape (otherwise, the machine rejects and has no output for this input). At this moment, one $a$ is printed on the output tape.

while the input tape is not empty: // main loop

{       in a loop, erase every second $a$ on the input tape; // inner loop

when the right end of the input tape is reached, // after the inner loop

if the number of letters $a$ read in the inner loop was even:

then add an $a$ on the output tape;

else reject;

in a loop, move the head of the input tape back to the left end; // 2nd inner loop

}

This program runs in linear time, and every step is injective.

This injective Turing machine for $g$ can be reversed, which yields an injective Turing machine for $g^{-1}$ with exponential time-complexity.    $\Box$

Proposition 3.2

Let $h$: $A^{*}\to A^{*}$ be an injective function. Then $h$ belongs to invfP  iff  $h$ is polynomially balanced, and is computable by some polynomial-time injective Turing machine.

Proof. For any $h\in{\sf invfP}$, $h$ is injective and computable in polynomial time; moreover, $h^{-1}\in{\sf invfP}$, i.e., $h^{-1}$ is also injective and computable in polynomial time. Hence, Bennett’s theorem is applicable (see [1, 2]), so $h$ can be computed by a polynomial-time injective deterministic Turing machine. Also, since both $h$ and $h^{-1}$ are computable in polynomial time, $h$ is polynomially balanced. Conversely, if $h$ is polynomially balanced and is computed by a polynomial-time injective Turing machine, then $h^{-1}$ can also be computed by such a machine, by Prop. 3.1.    $\Box$

We saw in the Introduction that $f\in{\sf fP}$ is regular iff $f$ has an inverse in invfP. Hence by Prop. 3.2, we have for all $f\in{\sf fP}$:

$f$* is regular  iff  $f$ has an inverse that is computable by a polynomially balanced polynomial-time injective Turing machine. And ${\sf P}={\sf NP}$  iff  every $f\in{\sf fP}$ has an inverse that can be computed by a polynomially balanced polynomial-time injective Turing machine.*

3.2 Evaluation functions and finite generation

Based on the transition table of a Turing machine, one can easily check whether this Turing is deterministic, and whether it is injective. Moreover, we saw in [3] and [4] that one can add a built-in polynomial time bound and balance bound into the transition table. This can be done for injective Turing machines too, without destroying injectiveness; indeed, a time bound $p(|x|)$ can be computed injectively on input $x$ (where $p$ is a stored polynomial, described by its degree and coefficients). So we can design a set of strings (programs) that describe all polynomially balanced polynomial-time injective Turing machines. And just as in [3], for every polynomial $q(n)=a\cdot(n^{k}+1)$, there exists an evaluation function ${\sf ev}_{q}$ that evaluates all injective Turing machine programs with built-in polynomial less than $q$. The details are the same as in [3], and injectiveness doesn’t change any reasoning. We call such a Turing machine description an invfP-program.

A Turing machine with program $w$ will be denoted by $M_{w}$; we denote the injective input-output function of $M_{w}$ by $\phi_{w}$.

An evaluation function maps a pair $(w,x)$, consisting of a program $w$ and an input $x$ (for $M_{w}$), to $(w,\phi_{w}(x))$, i.e., to that same program and the program-output. In order to represent a pair of strings by one string we use the prefix code $\{00,01,11\}$, and the function code(.) defined by

${\sf code}(0)=00$,   ${\sf code}(1)=01$.

The pair of strings $(w,x)$ is represented unambiguously by ${\sf code}(w)\,11\,x$, where 11 acts as a separator (since $\{00,01,11\}$ is a prefix code). The evaluation function ${\sf injEv}_{q}$ is defined by

{\sf injEv}_{q}\big{(}{\sf code}(w)\,11\,x\big{)}\ =\ ${\sf code}(w)\,11\ \phi_{w}(x)$

for any invfP-program $w$ with built-in polynomial less than $q$, and any $x\in{\sf Dom}(\phi_{w})$. The function ${\sf injEv}_{q}$ is itself injective and polynomially balanced and polynomial-time computable (with a larger polynomial than $q$ however). Moreover, ${\sf injEv}_{q}$ is regular; indeed, the unique mutual inverse of ${\sf injEv}_{q}$ is ${\sf injEv}_{q}^{-1}$, and this belongs to fP since $\phi_{w}(x)$ is injective and regular, i.e., $\phi_{w}^{-1}\in{\sf fP}$ for every invfP-program $w$. So, ${\sf injEv}_{q}\in{\sf invfP}$.

The proof of Prop. 4.5 in [3], showing that fP is finitely generated, goes through without much change. We use the relation

$(\star)$             \phi_{w}(x)\ =\ $\pi_{{}_{2\,|w^{\prime}|+2}}^{\prime}\circ{\sf contr}\circ{\sf recontr}^{2\,m}$ $\circ$ ${\sf injEv}_{q}$ $\circ$ ${\sf reexpand}^{2m}\circ{\sf expand}$ $\circ$ $\pi_{{}_{{\sf code}(w)\,11}}(x)$,

where $w$ is a program of a polynomially balanced polynomial-time injective Turing machine with built-in polynomial $<q$, and $\,w^{\prime}={\sf co}^{2m+1}\circ{\sf ex}^{2m+1}(w)$. See Section 4 of [3] for the definition of contr, recontr, reexpand, expand, $\pi_{n}^{\prime}$, $\pi_{v}$, co, and ex. The role of expand and reexpand is to increase the input length, so as to reduce complexity by a padding argument; when complexity is below $q$, ${\sf injEv}_{q}$ can be applied; after that, recontr and contr remove the padding.

The functions contr, recontr, ${\sf injEv}_{q}$, reexpand, expand, and $\pi_{{}_{{\sf code}(w)\,11}}$ are injective and regular (i.e., they belong to invfP). In the relation $(\star)$ we can replace $\pi_{{}_{2\,|w^{\prime}|+2}}^{\prime}$ by the injective regular function \,\pi_{{}_{{\sf code}(w)\,11}}^{\prime}:\ {\sf code}(w)\,11\ x\ $\longmapsto\ x$.  Moreover, $\pi_{{}_{{\sf code}(w)}}$ is generated by $\{\pi_{0},\pi_{1}\}$, and $\pi_{{}_{2\,|w^{\prime}|+2}}^{\prime}$ is generated by $\{\pi_{0}^{\prime},\pi_{1}^{\prime}\}$.

Thus, $\phi_{w}$ is generated by the finite set of functions $\,\{{\sf contr},$ ${\sf recontr},$ ${\sf injEv}_{q},$ ${\sf reexpand},$ ${\sf expand},$ $\pi_{0},$ $\pi_{1},$ $\pi_{0}^{\prime},$ $\pi_{1}^{\prime}\}$ $\subset$ ${\sf invfP}$, and we have:

Proposition 3.3

The inverse monoid invfP is finitely generated, as a monoid.        $\Box$

For every invfP-program $w$, we can easily obtain an invfP-program (let’s call it $w^{\prime}$) for $\phi_{w}^{-1}$; for this purpose we simply reverse the injective Turing machine described by $w$. Thus the function $w\mapsto w^{\prime}$ is polynomially balanced and polynomial-time computable; moreover, $w\mapsto w^{\prime}$ is injective and involutive, so the function $w\mapsto w^{\prime}$ is in invfP. Thus we proved:

Proposition 3.4

There exists a function ${\sf prog}_{\sf inv}\in{\sf invfP}$ such that ${\sf prog}_{\sf inv}(w)$ is an invfP-program for $\phi_{w}^{-1}$, for every invfP-program $w$.        $\Box$

We would like to extend the above function ${\sf prog}_{\sf inv}$ to all “regular” fP-programs, i.e., the fP-programs $w$ for which $\phi_{w}$ is regular in fP. We can do this by using universal search, a.k.a. Levin search; this is described in [10, 11], but without much detail; for a detailed exposition, see for example [12] (Theorem 7.21 in the 1993 edition). In the general regular case, ${\sf prog}_{\sf inv}(w)$ is just a program such that $\phi_{{\sf prog}_{\sf inv}(w)}$ has polynomial time-complexity, but ${\sf prog}_{\sf inv}(w)$ does not have a built-in polynomial for its time-complexity; Levin search is not able to explicitly find such a polynomial, although it exists if $\phi_{w}$ is regular.

Proposition 3.5

(inversion by Levin search).*  There exists a function ${\sf prog}_{\sf inv}\in{\sf fP}$ such that for every fP-program $w$, ${\sf prog}_{\sf inv}(w)$ is a program for a mutual inverse of $\phi_{w}$ satisfying ${\sf Dom}(\phi_{{\sf prog}_{\sf inv}(w)})$ $=$ ${\sf Im}(\phi_{w})$.*

The time-complexity of $\,\phi_{{\sf prog}_{\sf inv}(w)}\,$ is $\,\Theta(T_{\phi_{w}})+\Theta(T_{w}^{\prime})$, where $\Theta(T_{\phi_{w}})$ is (up to big-$\Theta$) the optimal time-complexity of all Turing machines for $\phi_{w}$, and $\Theta(T_{w}^{\prime})$ is (up to big-$\Theta$) the optimal time-complexity of all Turing machines for all mutual inverses of $\phi_{w}$.

In particular:

If $\phi_{w}$ is regular then $\phi_{{\sf prog}_{\sf inv}(w)}$ has polynomial time-complexity. But the program ${\sf prog}_{\sf inv}(w)$ does not have a built-in polynomial for its time-complexity, i.e., ${\sf prog}_{\sf inv}(w)$ is not an fP-program.

If $\phi_{w}$ is not regular, ${\sf prog}_{\sf inv}(w)$ is a non-polynomial-time program. The time-complexity of $\phi_{{\sf prog}_{\sf inv}(w)}$ has nevertheless an exponential upper bound.

Proof. In our version of universal search the input has the form $(w,y)$, where $w$ is an fP-program, and $y$ is a possible output of the Turing machine $M_{w}$. Remark: For convenience we write $(w,y)$ as a pair of words, but our universal search will actually use the single word $\,{\sf code}(w)\,11\,y$.

The output of universal search is $(w,x)$ such that $x\in\phi_{w}^{-1}(y)$, if $w$ is an fP-program, and $y\in{\sf Im}(\phi_{w})$ (i.e., if $\phi_{w}^{-1}(y)\neq\varnothing$); there is no output if $w$ is not an fP-program, or if $y\not\in$ ${\sf Im}(\phi_{w})$. Thus universal search computes a mutual inverse ${\sf ev}^{\prime}$ of the general evaluation function ${\sf ev}$ for fP. Both ${\sf ev}^{\prime}$ and ${\sf ev}$ are partial recursive (but they do not belong to fP since no polynomial bound is prescribed, as for ${\sf ev}_{q}$). Since we restrict universal search to fP-programs (hence, with built-in polynomial complexity bound), the domain of ${\sf ev}$ is decidable; so universal search is an algorithm (that always halts). Indeed, by rejecting programs with no polynomial time bound, the search algorithm also rejects programs that do not halt. And for a polynomial program with known polynomial complexity, inversion has a predictable exponential-time upper-bound.

For a fixed $w$ in the input, universal search computes ${\sf ev}^{\prime}(w,\cdot)=(w,\phi_{w}^{\prime}(.))$, where $\phi_{w}^{\prime}(.)$ is a mutual inverse of $\phi_{w}$. For the details of the universal search algorithm we refer to [12] (proof of Theorem 7.21, p. 412 in the 1993 edition). From the specification of universal search and $w$, we immediately derive a program for ${\sf ev}^{\prime}(w,\cdot)$; this program is called ${\sf prog}_{\sf inv}(w)$. Since ${\sf prog}_{\sf inv}(w)$ is easily obtained from $w$, the function ${\sf prog}_{\sf inv}(.)$ belongs to fP.

The program ${\sf prog}_{\sf inv}(w)$ has minimum time-complexity $\,\Theta(T_{\phi_{w}})+\Theta(T_{w}^{\prime})$; this is proved in [12]. It follows that the time-complexity of ${\sf prog}_{\sf inv}(w)$ has an exponential upper bound (since $w$ is an fP-program). It also follows that ${\sf prog}_{\sf inv}(w)$ is a polynomial-time program if $\phi_{w}$ is regular, i.e., if $\phi_{w}$ has a polynomial-time inverse. But universal search does not find out explicitly what that polynomial is, so ${\sf prog}_{\sf inv}(w)$ cannot have a built-in polynomial for its time-complexity bound; i.e., ${\sf prog}_{\sf inv}(w)$ is not an fP-program.    $\Box$

3.3 Injective Turing machines with NP-oracle

In preparation for the next Section, we generalize polynomially balanced polynomial-time injective Turing machines by adding an oracle from NP.

Definition 3.6

By $\,{\sf invfP}^{\sf(NP)}$ we denote the set of all functions computed by polynomially balanced polynomial-time injective Turing machines, with an oracle belonging to NP.

By Prop. 3.2 we have:  ${\sf invfP}\subseteq{\sf invfP}^{\sf(NP)}$.

An injective Turing machine with an oracle is a deterministic and injective machine. Indeed, it computes deterministically and injectively inbetween oracle transitions. And in an oracle transition the only change is in the state, which goes from the query state $q_{\sf qu}$ to either $q_{\sf yes}$ or $q_{\sf no}$; the contents and head positions of the tapes (including the query tape) remain unchanged. The next transition will go to a different state than $q_{\sf yes}$ or $q_{\sf no}$, so the state $q_{\sf yes}$ (or $q_{\sf no}$) uniquely determines the previous state $q_{\sf qu}$. Thus, an oracle transition is deterministic and injective.

The reverse of an injective Turing machine with an oracle has a slightly different format than an an injective Turing machine with an oracle, as described above. Indeed, in an oracle call the following transition happens: The machine is in the query state $q_{\sf qu}$ and then it enters an answer state, either $q_{\sf yes}$ or $q_{\sf no}$, according as the word on the query tape belongs to the oracle set or not. (Before a query, a word is written on the query tape; this happens in the course of a possibly long computation. After a query, the computation continues, and during this computation the query tape content may be gradually erased or changed.)

When this sequence is reversed, an answer state occurs before the query state. From the answer state, the reverse computation goes to the query state, provided that the answer state is $q_{\sf yes}$ and the query word is in the oracle language, or if the answer state is $q_{\sf no}$ and the query word is not in the oracle language; the reverse computation rejects otherwise. We call this a reverse oracle call. Although a reverse oracle call does not have the format of an oracle call, it can easily be simulated by an oracle call and a few more transitions. Hence we have:

Proposition 3.7

Let $M$ be a polynomially balanced polynomial-time injective Turing machine with NP-oracle. Then $M$ computes a polynomially balanced injective function.

The machine $M^{\prime}$, obtained by running $M$ in reverse, is equivalent to a polynomial-time injective Turing machines with NP-oracle, and computes $f^{-1}$ (where $f$ is the injective function computed by $M$).        $\Box$

Conversely, every injective oracle Turing machine $M$ can be simulated by a reverse injective Turing machine with the same oracle. Indeed, let $M^{\prime}$ be the reverse of $M$, and let $M^{\prime}_{1}$ be an ordinary injective oracle machine that simulates $M^{\prime}$. Finally, let $(M^{\prime}_{1})^{\prime}$ be the reverse of $M^{\prime}_{1}$. Then $(M^{\prime}_{1})^{\prime}$ is a reverse injective Turing machine, with the same oracle as $M$, simulating $M$.

So, there is no intrinsic difference between injective oracle Turing machines and reverse injective oracle Turing machines. We could generalize injective Turing machines with oracle so as to allow oracle calls and reverse oracle calls in the same machine; but by the above discussion, it doesn’t matter much whether we use oracle calls, reverse oracle calls, or both.

Corollary 3.8

${\sf invfP}^{\sf(NP)}$* is an inverse monoid.*

Proof. The composite of two polynomially balanced injective functions, computed by polynomial-time injective Turing machine with NP-oracle, is also polynomially balanced and injective. It is computed by a polynomial-time injective Turing machine with NP-oracle, obtained by chaining the two machines. The disjoint union of two sets in NP is in NP; thus the combined machine can use a single NP-oracle language. Hence, ${\sf invfP}^{\sf(NP)}$ is a monoid. By Prop. 3.7, ${\sf invfP}^{\sf(NP)}$ is an inverse monoid.    $\Box$

Remark. Not all functions in ${\sf fP}^{\sf(NP)}$ have an inverse in ${\sf fP}^{\sf(NP)}$ (unless the polynomial hierarchy PH collapses, see Sect. 6 of [3]). Here, ${\sf fP}^{\sf(NP)}$ denotes the set of polynomially balanced functions computed by deterministic polynomial-time Turing machines with NP-oracle. But all functions in ${\sf invfP}^{\sf(NP)}$ have an inverse in ${\sf invfP}^{\sf(NP)}$ (by Cor. 3.8).

Proposition 3.9

Suppose $f:A^{*}\to A^{*}$ is an injective and polynomially balanced function such that both $f$ and $f^{-1}$ can be computed by deterministic polynomial-time Turing machines with NP-oracle. Then $f$ can be computed by an injective polynomial-time Turing machine with NP-oracle.

Proof. Bennett’s proof in [1, 2] applies to Turing machines with NP-oracle. An injective Turing machine for $f$ (with NP-oracle) is obtained by first using the Turing machine for $f$ (on input $x$), but with a history tape added (which makes the machine injective). The input $x$ is still present. Once $f(x)$ has been computed, a copy of it is made on the output tape. The history tape is then used to run the previous computation in reverse, thus erasing the history tape and the work-tape copy of $f(x)$. The input $x$ is still on the input tape at this moment, and a copy of $f(x)$ is on the output tape. See Lemma 1 of [2] (except that now the machine also has an NP-oracle).

To erase $x$ injectively, the Turing machine for $f^{-1}$ is used on input $f(x)$ (copied from the output tape). A history tape is added to make the computation injective; $x$ is (re-)computed (while $f(x)$ is kept on the output tape). Once $x$ has been (re-)computed, the history tape is used to run the previous computation of the $f^{-1}$-machine in reverse, thus erasing the history tape, as well as $x$. A copy of $f(x)$ is kept as the output tape. See Theorem 2(b) of [2] (except that now the machine also has an NP-oracle).    $\Box$

We will prove next that $\,{\sf invfP}^{\sf(NP)}$ is finitely generated. For this we want to replace all NP-oracles by one NP-complete set. Moreover, we want the reduction functions to be in invfP.

Recall that for sets $L_{1},L_{2}\subseteq A^{*}$, a many-one reduction from $L_{1}$ to $L_{2}$ is a polynomial-time computable function $f$: $A^{*}\to A^{*}$ such that $L_{1}=f^{-1}(L_{2})$; equivalently, $(\forall x\in A^{*})[\,x\in L_{1}\Leftrightarrow f(x)\in L_{2}\,]$. A one-one reduction is a many-one reduction $f$ that is injective.

Definition 3.10

For sets $L_{1},L_{2}\subseteq A^{*}$, an invfP-reduction from $L_{1}$ to $L_{2}$ is any element $f\in{\sf invfP}$ such that $L_{1}=f^{-1}(L_{2})$.

This definition generalizes the usual many-one reductions in the sense that the elements of invfP are partial functions (with domain in P). Note that if $L_{1}=f^{-1}(L_{2})$ then $L_{1}\subseteq{\sf Dom}(f)$.

By the definition of invfP, if $f\in{\sf invfP}$ then $f^{-1}\in{\sf invfP}$, and ${\sf Im}(f)\in{\sf P}$ (by Prop. 1.9 in [3], since $f$ is regular). If $f$ reduces $L_{1}$ to $L_{2}$ then $f^{-1}$ reduces $L_{2}\cap{\sf Im}(f)$ to $L_{1}$. So, unless $L_{2}\subseteq{\sf Im}(f)$, $f^{-1}$ does not reduce $L_{2}$ to $L_{1}$, and invfP-reducibility is not a symmetric relation.

As we saw in Prop. 3.2, an invfP-reduction can be computed by a polynomially balanced polynomial-time injective Turing machine.

Proposition 3.11

There exists languages that are NP-complete with respect to invfP-reductions.

Proof. An example is the “universal NP-complete language” of Hartmanis [7], defined as follows (slightly reformulated):

L_{\rm univ}^{\sf NP}\ =\ $\{{\sf code}(w)\ 11\,x\,11\,0^{|w|\cdot p_{w}(|x|)}:$

$w$ is a nondeterministic polynomial-time Turing machine program,

$p_{w}$ is the built-in polynomial of $w$, and $x\in L_{w}\}$.

In [7] only many-one reductions were considered, but invfP-reductions can be used too. Indeed, for any language $L_{v}\in{\sf NP}$ accepted by a nondeterministic polynomial-time Turing machine with program $v$, we define the function $g_{v}$ for all $x\in A^{*}$ by

$x\ \longmapsto$ $\ {\sf code}(v)\ 11\,x\,11\,0^{|v|\cdot p_{v}(|x|)}$.

Then $g_{v}$ belongs to invfP, since a fixed program $v$ is chosen for the language $L_{v}$. And $g_{v}$ is a invfP-reduction reduction from $L_{v}$ to $L_{\rm univ}^{\sf NP}$. So, $L_{\rm univ}^{\sf NP}$ is complete with respect to invfP-reductions.    $\Box$

Proposition 3.12

The inverse monoid $\,{\sf invfP}^{\sf(NP)}$ is finitely generated, as a monoid.

Proof. By Prop. 3.11 there exist sets that are NP-complete with respect to invfP-reductions. Hence, for all polynomially balanced polynomial-time injective Turing machines with NP-oracle, we can use a fixed set $N$ as the oracle, where $N$ is NP-complete with respect to invfP-reductions. This changes the time-complexity of each function in ${\sf invfP}^{\sf(NP)}$ by a polynomial amount.

Now the proof of Prop. 3.3 goes through if we replace invfP-programs by ${\sf invfP}^{\sf(NP)}$-programs. The latter programs use the three additional states $q_{\sf qu}$, $q_{\sf yes}$, and $q_{\sf no}$, and one instruction for implementing a call to the oracle $N$. We replace ${\sf injEv}_{q}$ by ${\sf injEv}_{q}^{(N)}$, where the latter is computed by an injective Turing machine that is similar to the one for ${\sf injEv}_{q}$, but with oracle calls to $N$ added.    $\Box$

In the next section we will consider inverses and co-inverses that are computed by polynomially balanced polynomial-time injective Turing machines with NP-oracle.

4 Inverses and co-inverses of any element of fP

If P $\neq$ NP then there exist non-regular functions $f\in{\sf fP}$. In that case we can nevertheless consider the injective inverses and co-inverses of $f$ in ${\sf invfP}^{\sf(NP)}$, as we shall see in Prop. 4.9. First we derive some general results about injective inverses and co-inverses.

4.1 Inverses, co-inverses, sub-inverses

Here we consider arbitrary functions $f,f^{\prime}:A^{*}\to A^{*}$, unless more precise conditions are stated. A subfunction of a function $f$ is, by definition, any function $g$ such that $g\subseteq f$; equivalently, ${\sf Dom}(g)\subseteq{\sf Dom}(f)$ and for all $x\in{\sf Dom}(g)$: $g(x)=f(x)$.

Lemma 4.1

If $\,ff^{\prime}f=f\,$ then $\,{\sf Im}(f)\subseteq{\sf Dom}(f^{\prime})$.

Proof. If $y=f(x)\in{\sf Im}(f)$ then $ff^{\prime}(y)=ff^{\prime}f(x)=f(x)=y$, so $ff^{\prime}(y)$ is defined, hence $f^{\prime}(y)$ is defined, so $y\in{\sf Dom}(f^{\prime})$.    $\Box$

Lemma 4.2

.**

(1)*  If $\,f^{\prime}ff^{\prime}=f^{\prime}\,$ then $\,{\sf Im}(f^{\prime})\subseteq{\sf Dom}(f)$.*

(2)*  If in addition $f^{\prime}$ is injective, then $\,{\sf Dom}(f^{\prime})\subseteq{\sf Im}(f)$. This does not hold in general when $f^{\prime}$ is not injective.*

Proof. (1) This is equivalent to Lemma 4.1, up to notation.

(2) If $z\in{\sf Dom}(f^{\prime})$ then $f^{\prime}(z)=f^{\prime}ff^{\prime}(z)$ is defined, hence $ff^{\prime}(z)$ is defined and $ff^{\prime}(z)\in{\sf Im}(f)$. When $f^{\prime}$ is injective we can consider its set-theoretic inverse $f^{\prime-1}$, and we have $f^{\prime-1}f^{\prime}={\sf id}_{{\sf Dom}(f^{\prime})}$. Applying $f^{\prime-1}$ on the left to $f^{\prime}(z)=f^{\prime}ff^{\prime}(z)$ yields ${\sf id}_{{\sf Dom}(f^{\prime})}\circ f^{\prime}f(z)=z$ (if $z\in{\sf Dom}(f^{\prime})$). Hence ${\sf id}_{{\sf Dom}(f^{\prime})}\circ ff^{\prime}(z)$ is defined, and ${\sf id}_{{\sf Dom}(f^{\prime})}\circ ff^{\prime}(z)=ff^{\prime}(z)=z$, so $z\in{\sf Im}(f)$.

When $f^{\prime}$ is not injective, the result might not hold. Consider for example $f=\{(a,b)\}$ and $f^{\prime}=\{(a,a),(b,a)\}$. Then $f^{\prime}ff^{\prime}=f^{\prime}\,$ (and $ff^{\prime}f=f$ as well, so $f$ and $f^{\prime}$ are mutual inverses); but ${\sf Dom}(f^{\prime})=\{a,b\}\,\not\subseteq\,\{b\}={\sf Im}(f)$.    $\Box$

Corollary 4.3

If $f,f^{\prime}$ are mutual inverses and $f^{\prime}$ is injective then ${\sf Dom}(f^{\prime})={\sf Im}(f)$. This does not hold in general when $f^{\prime}$ is not injective.        $\Box$

Lemma 4.4

If $\,f^{\prime}ff^{\prime}=f^{\prime}\,$ and $f^{\prime}$ is injective, then  $f^{\prime-1}\ =\ f|_{{\sf Im}(f^{\prime})}$;  equivalently,

f^{\prime}\ =\ \big{(}f|_{{\sf Im}(f^{\prime})}\big{)}^{-1}* $\ =\ f^{-1}\ \cap\ {\sf Dom}(f^{\prime})\times{\sf Im}(f^{\prime})$.*

Hence, $\,f^{\prime-1}\,\subseteq\,f$.

Proof. Applying $f^{\prime-1}$ on the left and the right to $f^{\prime}=f^{\prime}ff^{\prime}$ yields $\,{\sf id}_{{\sf Dom}(f^{\prime})}\circ f\circ{\sf id}_{{\sf Im}(f^{\prime})}$ $=f^{\prime-1}$. This is equivalent to $\,f|_{{\sf Im}(f^{\prime})\,\cap\,f^{-1}({\sf Dom}(f^{\prime}))}=f^{\prime-1}$. Hence, $f^{\prime-1}\subseteq f$, so $\,f^{\prime}=(f^{\prime-1})^{-1}\subseteq f^{-1}$, where $f^{-1}=\{(y,x):y=f(x)\}$. Thus, ${\sf Im}(f^{\prime})=f^{\prime}({\sf Dom}(f^{\prime}))\subseteq f^{-1}({\sf Dom}(f^{\prime}))$, hence $\,{\sf Im}(f^{\prime})\,\cap\,f^{-1}({\sf Dom}(f^{\prime}))={\sf Im}(f^{\prime})$; so $\,f|_{{\sf Im}(f^{\prime})\,\cap\,f^{-1}({\sf Dom}(f^{\prime}))}=$ $f|_{{\sf Im}(f^{\prime})}=f^{\prime-1}$.

The equality $\,f^{\prime}\,=\,f^{-1}\,\cap\,{\sf Dom}(f^{\prime})\times{\sf Im}(f^{\prime})$ is just a logical reformulation of $f^{\prime}=(f|_{{\sf Im}(f^{\prime})})^{-1}$.    $\Box$

Corollary 4.5

If $f^{\prime}$ is an injective co-inverse of $f$, then $f^{\prime}$ is a mutual inverse of $\,f|_{{\sf Im}(f^{\prime})}\,$ (which is an injective subfunction of $f$). Moreover, $f^{\prime}$ is uniquely determined by $f$ and ${\sf Im}(f^{\prime})$, as $\,f^{\prime}=(f|_{{\sf Im}(f^{\prime})})^{-1}$.

Proof. Since $\,f^{\prime}={\sf id}_{{\sf Im}(f^{\prime})}\circ f^{\prime}$, we have $\,f^{\prime}=f^{\prime}ff^{\prime}=f^{\prime}\circ f\circ{\sf id}_{{\sf Im}(f^{\prime})}\circ f^{\prime}$ $=f^{\prime}\circ f|_{{\sf Im}(f^{\prime})}\circ f^{\prime}$. Moreover, since $f|_{{\sf Im}(f^{\prime})}=f^{\prime-1}$ (by Lemma 4.4), we have $\,f|_{{\sf Im}(f^{\prime})}=f^{\prime-1}=f^{\prime-1}f^{\prime}f^{\prime-1}=$ $f|_{{\sf Im}(f^{\prime})}\circ f^{\prime}\circ f|_{{\sf Im}(f^{\prime})}$.    $\Box$

Corollary 4.6

If $\,f^{\prime}ff^{\prime}=f^{\prime}\,$ and $\,{\sf Dom}(f^{\prime})={\sf Im}(f)$, then $f^{\prime}$ is an injective mutual inverse of $f$.

Proof. When ${\sf Dom}(f^{\prime})={\sf Im}(f)$ then ${\sf Im}(f^{\prime})$ is a choice set of $f$. Indeed, for every $y\in{\sf Im}(f)$, $f^{\prime}(y)$ is defined and $f^{\prime}(y)\in f^{-1}(y)\,$ (since $f^{\prime}\subset f^{-1}$ by the second equality in Lemma 4.4). Hence $f^{\prime}$ is an inverse of $f$. Injectiveness follows from the observations made in the Introduction and at the beginning of Section 2.    $\Box$

Theorem 4.7

(anti-homomorphic property of co-inverses).**

Suppose $f_{i}^{\prime}$ is an injective co-inverse of $f_{i}$ for $i=1,2$. Then $\,f_{1}^{\prime}f_{2}^{\prime}\,$ is an injective co-inverse of $\,f_{2}f_{1}$.

Proof. We first observe that for subidentities ${\sf id}_{X}$, ${\sf id}_{Y}$, we have $\,{\sf id}_{X}\ {\sf id}_{Y}={\sf id}_{X\cap Y}$ $={\sf id}_{Y}\ {\sf id}_{X}$. Since $f_{i}^{\prime}$ is injective, $f_{i}^{\prime}f_{i}^{\prime-1}={\sf id}_{{\sf Im}(f_{i}^{\prime})}$, and $\,f_{i}^{\prime-1}f_{i}^{\prime}={\sf id}_{{\sf Dom}(f_{i}^{\prime})}$. Hence

$\,f_{1}^{\prime}f_{2}^{\prime}f_{2}^{\prime-1}f_{1}^{\prime-1}f_{1}^{\prime}f_{2}^{\prime}\,=\,$ $f_{1}^{\prime}\ {\sf id}_{{\sf Im}(f_{2}^{\prime})}\ {\sf id}_{{\sf Dom}(f_{1}^{\prime})}\ f_{2}^{\prime}\,$ $=\,f_{1}^{\prime}\ {\sf id}_{{\sf Dom}(f_{1}^{\prime})}\ {\sf id}_{{\sf Im}(f_{2}^{\prime})}\ f_{2}^{\prime}$ $\,=\,f_{1}^{\prime}f_{2}^{\prime}$.

So, f_{1}^{\prime}f_{2}^{\prime}\,=\,f_{1}^{\prime}f_{2}^{\prime}f_{2}^{\prime-1}f_{1}^{\prime-1}f_{1}^{\prime}f_{2}^{\prime}\ \subseteq\ $f_{1}^{\prime}f_{2}^{\prime}f_{2}f_{1}f_{1}^{\prime}f_{2}^{\prime}\,$ (by Lemma 4.4). Now, since $f_{1}^{\prime}f_{2}^{\prime}\,\subseteq\,f_{1}^{\prime}f_{2}^{\prime}f_{2}f_{1}f_{1}^{\prime}f_{2}^{\prime}$, we have $\,f_{1}^{\prime}f_{2}^{\prime}(x)=f_{1}^{\prime}f_{2}^{\prime}f_{2}f_{1}f_{1}^{\prime}f_{2}^{\prime}(x)\,$ when $f_{1}^{\prime}f_{2}^{\prime}(x)$ is defined. And when $f_{1}^{\prime}f_{2}^{\prime}(x)$ is not defined, $f_{1}^{\prime}f_{2}^{\prime}f_{2}f_{1}f_{1}^{\prime}f_{2}^{\prime}(x)$ is not defined either. Thus, $f_{1}^{\prime}f_{2}^{\prime}\,=\,f_{1}^{\prime}f_{2}^{\prime}f_{2}f_{1}f_{1}^{\prime}f_{2}^{\prime}$.    $\Box$

Remark. Theorem 4.7, though straightforward, is remarkable as it applies to injective co-inverses but not to injective inverses or mutual inverses.

The latter can be illustrated by the following example: Let $f_{1}=\{(0,0),(1,0)\}=f_{2}$, and let $f_{1}^{\prime}=\{(0,1)\}=f_{2}^{\prime}$, so $f_{i}$ and $f_{i}^{\prime}$ are mutual inverses $\,(i=1,2)$, and $f_{i}^{\prime}$ is injective. But $f_{1}^{\prime}\circ f_{2}^{\prime}=\theta\,$ (the empty map), which is not an inverse of $f_{2}f_{1}=\{(0,0),(1,0)\}$.

In [3] (Prop. 6.1) we showed that every element $f\in{\sf fP}$ has an inverse in ${\sf fP}^{\sf(NP)}$; one such inverse is $f_{\sf min}^{\prime}$, defined by

f_{\sf min}^{\prime}(y)\ =\ \left\{\begin{array}[]{ll}{\sf min}(f^{-1}(y))&\ \ \ \mbox{if y\in{\sf Im}(f), }\\ {\rm undefined}&\ \ \ \mbox{otherwise,}\end{array}\right.

where the min operation is taken with respect to the length-lexicographic order, $\leq_{\sf llex}$, of $\{0,1\}^{*}$. Recall that by definition, $u\leq_{\sf llex}v\,$ iff $\,|u|<|v|$ or $[\,|u|=|v|$ and $u\leq_{\sf dict}v\,]$; here, $\leq_{\sf dict}$ is the dictionary order of $\{0,1\}^{*}$ determined by $0<_{\sf dict}1$.

Proposition 4.8

For every $f\in{\sf fP}$ we have:

(1)*  $f\circ f_{\sf min}^{\prime}={\sf id}_{{\sf Im}(f)}$;  hence, $f\circ f_{\sf min}^{\prime}\circ f=f$;  also  $f_{\sf min}^{\prime}\circ f\circ f_{\sf min}^{\prime}=f_{\sf min}^{\prime}$;   (2)  $f_{\sf min}^{\prime}$ is injective;*

(3)*  ${\sf Im}(f_{\sf min}^{\prime})\in{\sf coNP}$;   (4)  ${\sf Dom}(f_{\sf min}^{\prime})={\sf Im}(f)\in{\sf NP}$;   (5)  $f_{\sf min}^{\prime-1}\in{\sf fP}^{\sf(NP)}$.*

Proof.  (1) is obvious from the definition of $f_{\sf min}^{\prime}$.

(2) Injectiveness of $f_{\sf min}^{\prime}$ follows from the fact that the sets $f^{-1}(y)$ (for $y\in{\sf Im}(f)$) are two-by-two disjoint.

(3) We have: $\,x\in{\sf Im}(f_{\sf min}^{\prime})$  iff  $x\in{\sf Dom}(f)$ and $(\forall z)[\,z<_{\sf llex}x$ $\Rightarrow f(z)\neq f(x)\,]\,$ (the latter sentence expresses that $x$ is the $\leq_{\sf llex}$-minimum element in its ${\sf mod}_{f}$-class). Since ${\sf Dom}(f)\in{\sf P}$, since the relation $\,\{(z,x):z<_{\sf llex}x\Rightarrow f(z)\neq f(x)\}\,$ is in P, and since this relation is universally quantified (by $(\forall z,z<_{\sf llex}x)$), it follows that ${\sf Im}(f_{\sf min}^{\prime})\in{\sf coNP}$.

(4) By the definition of $f_{\sf min}^{\prime}$ we have ${\sf Dom}(f_{\sf min}^{\prime})={\sf Im}(f)$. And ${\sf Im}(f)\in{\sf NP}$ for every $f\in{\sf fP}$.

(5) As a consequence of (2), $f_{\sf min}^{\prime-1}\in{\sf fP}^{\sf(NP)}$. Indeed, $f_{\sf min}^{\prime-1}=f|_{{\sf Im}(f_{\sf min}^{\prime})}$ (by Lemma 4.4), and $f\in{\sf fP}$, and ${\sf Im}(f_{\sf min}^{\prime})\in{\sf coNP}$, so $f_{\sf min}^{\prime-1}$ is computed by any Turing machine for $f$ with an added oracle call to ${\sf Im}(f_{\sf min}^{\prime})$.    $\Box$

In the proof of Prop. 6.1 in [3] we showed that $f_{\sf min}^{\prime}\in$ ${\sf fP}^{\sf(NP)}$. Next we strengthen this by showing that the oracle Turing machine that computes $f_{\sf min}^{\prime}$ is an injective Turing machine with an NP-oracle. In Def. 3.6 we introduced the set ${\sf invfP}^{\sf(NP)}$ of polynomially balanced functions computed by polynomial-time injective Turing machines with an NP-oracle.

Proposition 4.9

Every $f\in{\sf fP}$ has an inverse $f^{\prime}$ in $\,{\sf invfP}^{\sf(NP)}\,$ (hence $f^{\prime-1}\in{\sf invfP}^{\sf(NP)}$), with the additional property that $\,{\sf Im}(f^{\prime})\in{\sf coNP}$.

Proof.  Let $f^{\prime}$ be $f_{\sf min}^{\prime}$, which is injective as we just saw, hence $f^{\prime-1}$ is an injective function. Both $f^{\prime}$ and $f^{\prime-1}$ are in ${\sf fP}^{\sf(NP)}$; this holds for $f^{\prime}$ by Prop. 6.1 in [3], and for $f^{\prime-1}$ by Prop. 4.8. Hence by Prop. 3.9, $f^{\prime}\in{\sf invfP}^{\sf(NP)}$.

Since $f^{\prime}$ is computed by an injective Turing machine (with oracle), $f^{\prime-1}$ can be computed by the same machine run backwards; so $f^{\prime-1}\in{\sf invfP}^{\sf(NP)}$.

We proved in Prop. 4.9 that ${\sf Im}(f_{\sf min}^{\prime})\in{\sf coNP}$.    $\Box$

Remark. As an alternative proof of Prop. 4.9, observe that the algorithm for computing $f_{\sf min}^{\prime}(y)$ in the proof of Prop. 6.1 in [3] actually describes an injective Turing machine; it makes oracle calls to the set $\,\{(z,u):z\in f(uA^{*})\}\,$ (which is in NP for any fixed $f\in{\sf fP}$).

Based on the anti-homomorphic property of co-inverses (Theorem 4.7) we define:

Definition 4.10

The monoid of co-inverses of fP is

${\sf cofP}\ =$* $\ \{f^{\prime}\in{\sf invfP}^{\sf(NP)}:\,$ $f^{\prime}$ is a co-inverse of some element of ${\sf fP}\}$.*

Equivalently, an element $f^{\prime}$ of ${\sf invfP}^{\sf(NP)}$ belongs to ${\sf cofP}$  iff  $f^{\prime}$ has an inverse in fP.

By Theorem 4.7, ${\sf cofP}$ is indeed a monoid.

Also, $f_{\sf min}^{\prime}\in{\sf cofP}$ for every $f\in{\sf fP}$, since $f$ is a mutual inverse of $f_{\sf min}^{\prime}$ (Prop. 4.8). Hence, for every $f\in{\sf fP}$, $\,{\sf cofP}$ contains a mutual inverse of $f$.

One could also consider the monoid of inverses of fP, namely $\,\{g\in{\sf invfP}^{\sf(NP)}:\,g$ is an inverse of some element of ${\sf fP}\}$. However, this is just all of ${\sf invfP}^{\sf(NP)}$, since every element of ${\sf invfP}^{\sf(NP)}$ is an inverse of the empty map $\theta\in{\sf fP}$. So, being an inverse of an element in fP is a trivial notion; but having an inverse in fP (i.e., being a co-inverse of an element in fP) is non-trivial. The question whether ${\sf cofP}$ is equal to ${\sf invfP}^{\sf(NP)}$ has an interesting “answer”, which is presented in Prop. 4.20 below.

Proposition 4.11

${\sf cofP}\ \cap\ {\sf fP}\ =\ {\sf invfP}$.

Proof. $[\supseteq]$  If $f^{\prime}\in{\sf invfP}$ then $f^{\prime-1}\in{\sf invfP}\subseteq{\sf fP}$. Moreover, $f^{\prime}\,f^{\prime-1}\,f^{\prime}=f^{\prime}$, so $f^{\prime}$ is a co-inverse of an element in fP. Hence $f^{\prime}\in{\sf cofP}$.

$[\subseteq]$  If $f^{\prime}\in{\sf cofP}\cap{\sf fP}$ then there exists $f\in{\sf fP}$ such that $f^{\prime}ff^{\prime}=f^{\prime}$, hence $f^{\prime}$ is a regular element of fP. Since we assume $f^{\prime}\in{\sf cofP}$, it follows that $f^{\prime}$ is injective. So, $f^{\prime}$ is an injective regular element of fP, i.e., $f^{\prime}\in{\sf invfP}$.    $\Box$

On the other hand, ${\sf invfP}^{\sf(NP)}\ \cap\ {\sf fP}$ contains all injective one-way functions. In fact, ${\sf invfP}^{\sf(NP)}\,\cap\,{\sf fP}\ =\ {\sf invfP}$  iff  injective one-way functions do not exist. The latter is the case iff $\,{\sf P}={\sf UP}\,$ (see e.g. section 12.1 in [13]). But cofP contains no one-way function, since every element of cofP has an inverse in fP.

Definition 4.12

Let injfP denote the monoid of all injective functions in fP.

The set of regular elements of injfP is exactly the submonoid invfP (by the definition of invfP), hence the set of injective one-way functions (for worst-case complexity, see Prop. 1.3) is $\,{\sf injfP}-{\sf invfP}$. We saw that invfP and ${\sf invfP}^{\sf(NP)}$ are regular monoids (in fact, inverse monoids). On the other hand, injfP is regular iff injective one-way functions do not exist. Thus we have:

*injective one-way functions exist  iff *   ${\sf invfP}\neq{\sf injfP}\neq{\sf invfP}^{\sf(NP)}$.

Since invfP consists of the regular elements of injfP, we also have:

*injective one-way functions exist  iff *   ${\sf invfP}\neq{\sf injfP}$.

Since ${\sf invfP}^{\sf(NP)}$ is regular we have,

*if injective one-way functions exist, then *  ${\sf injfP}\neq{\sf invfP}^{\sf(NP)}$.

We will also see (Theorem 4.18) that one-way functions exist  iff  ${\sf invfP}\neq{\sf invfP}^{\sf(NP)}$.

We saw that cofP contains no one-way functions (see the observations after Prop. 4.11), so we have:  $\,{\sf injfP}\ \cap\ {\sf cofP}\ =\ {\sf invfP}$.  Moreover:

Proposition 4.13

${\sf invfP}^{\sf(NP)}\,\cap\,{\sf fP}\ =\ {\sf injfP}$.

Proof. $[\subseteq]$  Every $f\in{\sf invfP}^{\sf(NP)}$ is injective, so the inclusion $\subseteq$ holds.

$[\supseteq]$  If $f\in{\sf injfP}$ then $f_{\sf min}^{\prime}=f^{-1}\in{\sf invfP}^{\sf(NP)}$, by Prop. 4.9. It follows that $(f^{-1})^{-1}=f\in$ ${\sf invfP}^{\sf(NP)}$, by Prop. 3.7. Hence $f\in{\sf invfP}^{\sf(NP)}\cap{\sf injfP}$ $\subseteq$ ${\sf invfP}^{\sf(NP)}\cap{\sf fP}$.    $\Box$

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${\sf invfP}^{\sf(NP)}$

cofP

invfP

injfP

fP

Figure 1:  Relations between $\,{\sf fP}$, $\,{\sf invfP}^{\sf(NP)}$, $\,{\sf cofP}$, $\,{\sf invfP}$, and $\,{\sf injfP}$.

There is an interesting connection between co-inverses and sub-inverses of $f$.

Definition 4.14

A function $g^{\prime}$ is a sub-inverse of a function $f$ iff there exists $g\subseteq f$ such that $gg^{\prime}g=g$ and $g^{\prime}gg^{\prime}=g^{\prime}$. I.e., the sub-inverses of $f$ are the mutual inverses of the subfunctions of $f$.

Proposition 4.15

For any function $f:A^{*}\to A^{*}$, the injective co-inverses of $f$ are the same as the injective sub-inverses of $f$.

Proof. $[\Rightarrow]$ Let $f^{\prime}$ be an injective co-inverse of $f$; then ${\sf Dom}(f^{\prime})\subseteq{\sf Im}(f)$ and ${\sf Im}(f^{\prime})\subseteq{\sf Dom}(f)$, by Lemmas 4.1 and 4.2. Let us consider the following restriction of $f$:  $g=f|_{{\sf Im}(f^{\prime})}$. We claim that $f^{\prime}$ is a mutual inverse of $g$, hence $f^{\prime}$ is a sub-inverse of $f$. Recall that $f^{\prime}=(f|_{{\sf Im}(f^{\prime})})^{-1}$, by Lemma 4.4. It follows from this that $\,gf^{\prime}g=f|_{{\sf Im}(f^{\prime})}\circ(f|_{{\sf Im}(f^{\prime})})^{-1}$ $\circ$ $f|_{{\sf Im}(f^{\prime})}=f|_{{\sf Im}(f^{\prime})}=g$; similarly it follows that $\,f^{\prime}gf^{\prime}=(f|_{{\sf Im}(f^{\prime})})^{-1}$ $\circ$ $f|_{{\sf Im}(f^{\prime})}$ $\circ$ $(f|_{{\sf Im}(f^{\prime})})^{-1}$ $=(f|_{{\sf Im}(f^{\prime})})^{-1}=f^{\prime}$.

So, $f^{\prime}$ is an injective sub-inverses of $f$.

$[\Leftarrow]$ Let $g\subset f$ be any restriction of $f$, and let $g^{\prime}$ be any injective mutual inverse of $g$; in other words, $g^{\prime}$ is an injective sub-inverse of $f$. We claim that $g^{\prime}$ is a co-inverse of $f$.

By Corollary 4.3, and Lemma 4.4, we have: ${\sf Dom}(f^{\prime})={\sf Im}(f)$, and $g^{\prime}=(g|_{{\sf Im}(g^{\prime})})^{-1}$. Since $g\subset f$ the latter implies $g^{\prime}=(f|_{{\sf Im}(f^{\prime})})^{-1}$. Hence, $g^{\prime}fg^{\prime}=(f|_{{\sf Im}(f^{\prime})})^{-1}\circ f\circ$ $(f|_{{\sf Im}(f^{\prime})})^{-1}=(f|_{{\sf Im}(f^{\prime})})^{-1}=g^{\prime}$. So, $g^{\prime}$ is a co-inverse of $f$.    $\Box$

Note that in part $[\Leftarrow]$ of the above proof, $g$ need not be injective, but $g|_{{\sf Im}(g^{\prime})}=f|_{{\sf Im}(g^{\prime})}$ is injective. Thus we have:

If $g^{\prime}$ is an injective sub-inverse of $f$ and it is a mutual inverse of a subfunction of $f$, then $g^{\prime}$ is also a mutual inverse of an injective subfunction of $f$.

The following gives a relation between two notions of sub-inverse. We defined a sub-inverse of $f$ to be a mutual inverse of a subfunction of $f$ (Def. 4.14). We could, instead, have defined a sub-inverse of $f$ to be a subfunction of a mutual inverse of $f$.

Proposition 4.16

.**

(1)*  If $g^{\prime}$ is a subfunction of a mutual inverse of $f$, then $g^{\prime}$ is a sub-inverse of $f$. (For this fact, $g^{\prime}$ need not be injective, and $f$ need not be in fP.)*

(2)*  If $g^{\prime}\in{\sf invfP}^{\sf(NP)}$ is a sub-inverse of $f\in{\sf fP}$, then $g^{\prime}$ is a subfunction of some mutual inverse $f^{\prime}\in{\sf invfP}^{\sf(NP)}$ of $f$.*

Proof. (1) Let $g^{\prime}\subseteq f^{\prime}$ with $ff^{\prime}f=f$ and $f^{\prime}ff^{\prime}=f^{\prime}$. Let $g=fg^{\prime}f$; then $g\subseteq ff^{\prime}f=f$. And for all $x\in{\sf Dom}(g)$ we have $gg^{\prime}g(x)=ff^{\prime}f(x)=f(x)=g(x)$ (the latter since $x\in{\sf Dom}(g)$). So $gg^{\prime}g=g$. Moreover, for all $y\in{\sf Dom}(g^{\prime})$ we have $g^{\prime}gg^{\prime}(y)=f^{\prime}ff^{\prime}(y)$, since $g^{\prime}\subseteq f^{\prime}$ and $g\subseteq f$; and $f^{\prime}ff^{\prime}(y)=y$. So, $g^{\prime}gg^{\prime}=g^{\prime}$.

(2) Since $g^{\prime}\in{\sf invfP}^{\sf(NP)}$, $g^{\prime}$ is injective, hence it is an injective co-inverse of $f$ (by Prop. 4.15). Hence, $g^{\prime}=(f|_{{\sf Im}(g^{\prime})})^{-1}$, and $g^{\prime}$ is a mutual inverse of $g=f|_{{\sf Im}(g^{\prime})}\subseteq f$. We can extend $g^{\prime}$ to the following mutual inverse of $f$: For all $y\in{\sf Im}(f)$ let

f^{\prime}(y)\ =\ \left\{\begin{array}[]{ll}g^{\prime}(y)&\ \ \ \mbox{if ,y\in{\sf Dom}(g^{\prime}),}\\ f_{\sf min}^{\prime}(y)&\ \ \ \mbox{otherwise.}\end{array}\right.

It follows that $f^{\prime}$ is injective. Indeed, both $g^{\prime}$ and $f_{\sf min}^{\prime}$ are injective. Moreover, $g^{\prime}(y_{1})\neq f_{\sf min}^{\prime}(y_{2})$ for all $y_{1}\in{\sf Dom}(g^{\prime})$ and all $y_{2}\in{\sf Im}(f)-{\sf Dom}(g^{\prime})$, since if we had equality then $fg^{\prime}(y_{1})=ff_{\sf min}^{\prime}(y_{2})$; but $fg^{\prime}(y_{1})=y_{1}$ since $g^{\prime}=(f|_{{\sf Im}(g^{\prime})})^{-1}$, and $ff_{\sf min}^{\prime}(y_{2})=y_{2}$; so we would have $y_{1}=y_{2}$, but $y_{1}\in{\sf Dom}(g^{\prime})$ and $y_{2}\not\in{\sf Dom}(g^{\prime})$.

Also, $f^{\prime}$ is clearly a mutual inverse of $f$.

Since $g^{\prime}$ is computed by a injective Turing machine with NP-oracle, and likewise $f_{\sf min}^{\prime}$, there is an injective Turing machine with NP-oracle for $f^{\prime}$; the Turing machine for $g^{\prime}$, being deterministic, can check whether $g^{\prime}(y)$ is defined, i.e., whether $y\in{\sf Dom}(g^{\prime})$. Hence, $f^{\prime}\in{\sf invfP}^{\sf(NP)}$.    $\Box$

4.2 Connections with NP

The next theorem motivates the study of the monoid ${\sf cofP}$ (and of invfP) in the context of the P versus NP problem. We prove an easy Lemma first.

Lemma 4.17

Let $f,h$ be functions $A^{*}\to A^{*}$ and let $Z\subset A^{*}$. If $\ f\circ h|_{Z}\circ f=f$, then $fhf=f$.

Proof. For $x\not\in{\sf Dom}(f)$, $fhf(x)$ and $f(x)$ are both undefined. For $x\in{\sf Dom}(f)$, $\,h|_{Z}(f(x))$ is defined, otherwise $f\circ h|_{Z}\circ f(x)$ would be undefined, and thus would not be equal to $f(x)$. Hence we have $f(x)\in Z$, so ${\sf Im}(f)\subseteq Z$. It follows that $h|_{Z}(f(x))=h(f(x))$, for all $x\in{\sf Dom}(f)$. Hence $h|_{Z}\circ f=h\circ f$, thus $f\circ h|_{Z}\circ f=f\,h\,f=f$.    $\Box$

Theorem 4.18

The following are equivalent:

${\sf P}={\sf NP}$;

${\sf cofP}={\sf invfP}$;

${\sf cofP}\subset{\sf fP}$;

${\sf cofP}\,$* is regular;*

${\sf invfP}^{\sf(NP)}\ =\ {\sf invfP}$.

Proof. We use the following logical fact: If $X\Rightarrow Y_{i}$ and $\overline{X}\Rightarrow\overline{Y}_{i}$, then $X\Leftrightarrow Y_{i}$ (for $i=1,2,\ldots\,$). Here, “${\sf P}={\sf NP}$” plays the role of $X$.

$[\Rightarrow]$  We assume ${\sf P}={\sf NP}$. We obviously have ${\sf invfP}\subseteq{\sf cofP}$. If ${\sf P}={\sf NP}$ then ${\sf fP}^{\sf(NP)}={\sf fP}$, hence ${\sf cofP}\subseteq{\sf fP}$. Also, fP is regular when ${\sf P}={\sf NP}$. Since all elements of ${\sf cofP}$ are injective, all elements of ${\sf cofP}$ are injective regular elements of fP, i.e., ${\sf cofP}\subseteq{\sf invfP}$. In conclusion, ${\sf cofP}={\sf invfP}$.

If ${\sf cofP}={\sf invfP}$ then ${\sf cofP}\subset{\sf fP}$. The inclusion is strict because fP has some non-injective elements.

And if ${\sf cofP}={\sf invfP}$ then, since ${\sf invfP}$ is regular, ${\sf cofP}$ is regular.

$[\Leftarrow]$  If ${\sf P}\neq{\sf NP}$ then there exists $f\in{\sf fP}$ such that $f$ is not regular. For $f$ there exists a mutual inverse $f^{\prime}\in{\sf invfP}^{\sf(NP)}$ such that ${\sf Dom}(f^{\prime})={\sf Im}(f)$; hence, $f^{\prime}\in{\sf cofP}$. Moreover, ${\sf Dom}(f^{\prime})={\sf Im}(f)\in{\sf NP}$; and $f^{\prime}$ can be chosen so that in addition, ${\sf Im}(f^{\prime})\in{\sf coNP}$ (e.g., by choosing $f^{\prime}=f^{\prime}_{\sf min}$ as in Prop. 4.9). Hence, $f^{\prime-1}\in{\sf invfP}^{\sf(NP)}$ (since $f^{\prime-1}\,=\,$ $f|_{{\sf Im}(f^{\prime})}$ and ${\sf Im}(f^{\prime})\in{\sf coNP}$). However, $f^{\prime}\not\in{\sf fP}$, since $f$ is not regular in fP.

Let us assume by contradiction that the monoid ${\sf cofP}$ is regular. Then $f^{\prime}$ (as above) is a regular element of ${\sf cofP}$, hence (by Lemma 4.9), $f^{\prime-1}\in{\sf cofP}$. This implies (by the definition of ${\sf cofP}$) that $\,f^{\prime-1}\,h\,f^{\prime-1}=f^{\prime-1}$ for some $h\in{\sf fP}$. Composing on the left and on the right with $f^{\prime}$ yields  ${\sf id}_{{\sf Im}(f^{\prime})}\ h\ {\sf id}_{{\sf Dom}(f^{\prime})}=f^{\prime}$; equivalently, $h|_{{\sf Dom}(f^{\prime})\,\cap\,h^{-1}({\sf Im}(f^{\prime}))}=f^{\prime}$. Hence, $ff^{\prime}f=$ $f\ h|_{{\sf Dom}(f^{\prime})\,\cap\,h^{-1}({\sf Im}(f^{\prime}))}\ f$ $=$ $f$. It follows (by Lemma 4.17) that $h$ is an inverse of $f$, since the restriction $h|_{{\sf Dom}(f^{\prime})\,\cap\,h^{-1}({\sf Im}(f^{\prime}))}$ is an inverse of $f$. Hence (since $f$ is not regular), $h\not\in{\sf fP}$; but this contradicts the previous assumption that $h\in{\sf fP}$. Thus ${\sf cofP}$ is not regular.

Since the monoid invfP is regular, we conclude that ${\sf cofP}\neq{\sf invfP}$.

And ${\sf cofP}\neq{\sf invfP}$ implies ${\sf cofP}\not\subset{\sf fP}$; indeed, if we had ${\sf cofP}\subset{\sf fP}$ then we would have ${\sf cofP}={\sf cofP}\,\cap\,{\sf fP}=$ ${\sf invfP}$.

Finally, if ${\sf invfP}^{\sf(NP)}={\sf invfP}$ then ${\sf cofP}={\sf invfP}$, hence by what we proved above, ${\sf P}={\sf NP}$. Conversely, if ${\sf P}={\sf NP}$ then every polynomial-time injective Turing machine with NP-oracle is equivalent to a polynomial-time deterministic Turing machine (without oracle). Thus if $f\in{\sf invfP}^{\sf(NP)}$ (hence by Prop. 3.7, $f^{-1}\in{\sf invfP}^{\sf(NP)}$), then $f,f^{-1}\in{\sf fP}$. By Bennett’s results [1, 2] (especially Lemma 1 and Theorem 2 (b) in [2]), this implies $f\in{\sf invfP}$.    $\Box$

Although every element of ${\sf cofP}$ has an inverse in fP (by the definition of ${\sf cofP}$), not every element of ${\sf cofP}$ has an inverse in ${\sf cofP}$, unless ${\sf P}={\sf NP}$; the latter follows from the fact that ${\sf cofP}$ is regular iff ${\sf P}={\sf NP}$.

By definition, ${\sf cofP}\subseteq{\sf invfP}^{\sf(NP)}$. We will now address the question whether this inclusion is strict; it turns out, interestingly, that this is equivalent to the question whether ${\sf P}\neq{\sf NP}$.

Lemma 4.19

Let $f\in{\sf fP}$, and let $f^{\prime}\in{\sf cofP}$ be a co-inverse of $f$ such that $\,{\sf Im}(f^{\prime})\in{\sf coNP}$. Then $\,f^{\prime-1}\in{\sf fP}^{\sf(NP)}$.

For every $f\in{\sf fP}$ there exists a co-inverse $f^{\prime}\in{\sf cofP}$ of $f$ such that $\,{\sf Im}(f^{\prime})\in{\sf coNP}$ and $\,{\sf Dom}(f^{\prime})\in{\sf NP}$.

Proof. Since $f\in{\sf fP}$ and ${\sf Im}(f^{\prime})\in{\sf coNP}$, $f^{\prime-1}=f|_{{\sf Im}(f^{\prime})}$ can be computed by a Turing machine for $f$ with NP-oracle ${\sf Im}(f^{\prime})$.

We saw above (and in [3]) that for every $f\in{\sf fP}$ there exists a mutual inverse (hence a co-inverse) $f^{\prime}$ for which ${\sf Im}(f^{\prime})\in{\sf coNP}$ and ${\sf Dom}(f^{\prime})\in{\sf NP}$; e.g., let $f^{\prime}=f_{\sf min}^{\prime}$.    $\Box$

Proposition 4.20

${\sf P}\neq{\sf NP}$*  iff  $\,{\sf cofP}\neq{\sf invfP}^{\sf(NP)}$.*

In any case (i.e., whether ${\sf P}\neq{\sf NP}$ or not), ${\sf cofP}$ and ${\sf invfP}^{\sf(NP)}$ have the same set of idempotents.

Proof. $[\Leftarrow]$  If ${\sf cofP}={\sf invfP}^{\sf(NP)}$, then for every $f^{\prime}\in{\sf invfP}^{\sf(NP)}$ there exists $f\in{\sf fP}$ such that $f^{\prime}ff^{\prime}=f^{\prime}$. By Lemma 4.4, $f^{\prime-1}\subseteq f$, and by Lemma 4.19, $f^{\prime-1}\in{\sf invfP}^{\sf(NP)}$. It follows that $f^{\prime}$ has an injective inverse $f^{\prime-1}\in{\sf invfP}^{\sf(NP)}$, i.e., $f^{\prime}$ is regular in ${\sf cofP}={\sf invfP}^{\sf(NP)}$. This holds for all $f^{\prime}\in{\sf cofP}$, hence ${\sf cofP}$ is regular. Thus by Theorem 4.18, ${\sf P}={\sf NP}$.

$[\Rightarrow]$  If ${\sf P}={\sf NP}$ then $\,{\sf cofP}={\sf invfP}$, which in this case is also equal to ${\sf invfP}={\sf invfP}^{\sf(NP)}$.

If $e$ is an idempotent of ${\sf invfP}^{\sf(NP)}$ then $e$ is also a co-inverse of an element of fP; indeed, $e\,\circ\,{\sf id}_{A^{*}}\,\circ\,e=e$. Hence $e\in{\sf cofP}$.    $\Box$

Proposition 4.21

${\sf P}\neq{\sf NP}$*  iff  ${\sf cofP}$ contains regular elements that do not belong to invfP.*

Proof. $[\Leftarrow]$  If ${\sf P}={\sf NP}$ then ${\sf cofP}={\sf invfP}$ (by Theorem 4.18), so there are no (regular) elements in ${\sf cofP}$ that are not in invfP.

$[\Rightarrow]$  If ${\sf P}\neq{\sf NP}$ then we construct the following example of a regular element in ${\sf cofP}$ that does not belong to invfP. Let $L\subset A^{*}$ be a coNP-complete set. Then ${\sf id}_{L}\in{\sf invfP}^{\sf(NP)}$ and it is regular (being an idempotent). Also, ${\sf id}_{L}={\sf id}_{L}\circ{\sf id}_{A^{*}}\circ{\sf id}_{L}$, hence ${\sf id}_{L}$ is a co-inverse of an element of fP. Thus, ${\sf id}_{L}\in{\sf cofP}$. But if ${\sf P}\neq{\sf NP}$ then ${\sf id}_{L}\not\in{\sf invfP}$, since $L$ is coNP-complete.    $\Box$

We saw that invfP and ${\sf invfP}^{\sf(NP)}$ are finitely generated. Hence, if cofP were not finitely generated, this would imply that ${\sf cofP}\neq{\sf invfP}$, which would imply that ${\sf P}\neq{\sf NP}$ (by Theorem 4.18). However, we will prove next that cofP is finitely generated.

We will first show that cofP has a machine (or program) model, and that there is a corresponding evaluation function for all bounded-complexity functions in cofP. This is similar to the situation in fP, invfP, and ${\sf invfP}^{\sf(NP)}$.

A cofP*-program* is of the form $(v^{\prime},w)$, where $v^{\prime}$ is any ${\sf invfP}^{\sf(NP)}$-program (with built-in time-complexity and balance function), and $w$ is any fP-program (with built-in time-complexity and balance function). An ${\sf invfP}^{\sf(NP)}$-program describes a polynomially balanced polynomial-time injective Turing machine, with a fixed NP-oracle $N$ (where $N$ is NP-complete). The functions in fP or ${\sf invfP}^{\sf(NP)}$ specified by programs $w$ or $v^{\prime}$ are denoted by $\phi_{w}$, respectively $\psi_{v^{\prime}}$. On input $y\in A^{*}$, the program $(v^{\prime},w)$ is evaluated as follows:

(1)    $x=\psi_{v^{\prime}}(y)\,$ is computed;

(2)    $\phi_{w}(x)\,$ is computed;

(3)    if $\,\phi_{w}\circ\psi_{v^{\prime}}\circ\phi_{w}(x)=\phi_{w}(x)\,$ and $\,\psi_{v^{\prime}}\circ\phi_{w}\circ\psi_{v^{\prime}}(y)=\psi_{v^{\prime}}(y)\,$ then the output is $\psi_{v^{\prime}}(y)$;

there is no output otherwise.

The function specified by program $(v^{\prime},w)$ on input $y$ is denoted by $\Phi_{(v^{\prime},w)}$. It is easy to see that $\Phi_{(v^{\prime},w)}$ is a subinverse of $\phi_{w}$, and a subfunction of $\psi_{v^{\prime}}$. Thus, $\Phi_{(v^{\prime},w)}\in{\sf cofP}$.

Conversely, every function $h\in{\sf cofP}$ has an ${\sf invfP}^{\sf(NP)}$-program, say $v^{\prime}$; and $h$ has an inverse $\phi_{w}\in{\sf fP}$ for some fP-program $w$. Then $(v^{\prime},w)$ is a cofP-program for $h$.

Based on the cofP-programs and a polynomial bound $q$, we can construct an evaluation function ${\sf evCo}_{q}^{(N)}$ such that $\,{\sf evCo}_{q}^{(N)}({\sf code}(v^{\prime})\,11\,{\sf code}(w)\,11\ y)$ $\ =\ {\sf code}(v^{\prime})\,11\,{\sf code}(w)\,11\ \Phi_{(v^{\prime},w)}(y)$,  for $y$ and $(v^{\prime},w)$ as above. This is similar to Section 3.2.

Theorem 4.22

The monoid cofP is finitely generated.

Proof. The proof is the same as for ${\sf invfP}^{\sf(NP)}$ (Prop. 3.12), except that ${\sf evCo}_{q}$ (with programs $(v^{\prime},w)$ as above) is used instead of ${\sf injEv}_{q}^{(N)}$.    $\Box$

We saw that ${\sf P}\neq{\sf NP}$ iff cofP is not regular. The set of elements of cofP that are regular in cofP will be denoted by RegcofP. We will see that it has interesting properties.

Proposition 4.23

(1) The set RegcofP is a finitely generated inverse submonoid of cofP.

(2) If ${\sf P}\neq{\sf NP}$ then $\ {\sf invfP}\,\subsetneqq\,{\sf RegcofP}\,\subsetneqq\,$ ${\sf cofP}\,\subsetneqq\,{\sf invfP}^{\sf(NP)}$.  If ${\sf P}={\sf NP}$ then $\,{\sf invfP}={\sf invfP}^{\sf(NP)}$.

(3) An element $f\in{\sf fP}$ has an inverse in RegcofP iff $f$ is regular in fP.

Proof. (1) For any $g\in{\sf RegcofP}$, let $g^{\prime}\in{\sf RegcofP}$ be a mutual inverse of $g$. Multiplying $gg^{\prime}g=g$ on the left and the right by $g^{-1}$ we obtain: $\,{\sf id}_{{\sf Dom}(g)}\circ g^{\prime}\circ{\sf id}_{{\sf Im}(g)}=g^{-1}$. Hence $g^{-1}\in{\sf cofP}$, being a product of elements of cofP (since the idempotents ${\sf id}_{{\sf Dom}(g)}$, ${\sf id}_{{\sf Im}(g)}$ belong to cofP). Thus we proved:

For all $g\in{\sf RegcofP}:$ $\,g^{-1}\in{\sf cofP}$.

For any $g_{1},g_{2}\in{\sf RegcofP}$ we have therefore, $g_{1}^{-1},g_{2}^{-1}\in{\sf cofP}$, hence $g_{2}g_{1}$ has $g_{1}^{-1}g_{2}^{-1}\in{\sf cofP}$ as an inverse, so $g_{2}g_{1}$ is regular in cofP. This proves that RegcofP is closed under composition. Also, since all elements of RegcofP are injective, RegcofP is an inverse monoid.

The proof of finite generation of RegcofP is similar to the proof of finite generation of cofP. We construct an evaluation function for the elements of RegcofP, based on the following machine (or program) model for the elements of RegcofP. A RegcofP-program is any pair $(u,v^{\prime})$ of cofP-programs, and the function $\Phi_{(u,v^{\prime})}$ computed by this program is defined by

$\Phi_{(u,v^{\prime})}(x)\ =\ \psi_{u}(x)$   if   $\psi_{u}\,\psi_{v^{\prime}}\,\psi_{u}(x)=\psi_{u}(x)$;

and $\Phi_{(u,v^{\prime})}(x)$ is undefined otherwise. Since the relation $\psi_{u}\,\psi_{v^{\prime}}\,\psi_{u}(x)=\psi_{u}(x)$ can be checked in ${\sf P}^{\sf(NP)}$, $\Phi_{(u,v^{\prime})}$ belongs to ${\sf invfP}^{\sf(NP)}$. And since $\psi_{u}\in{\sf cofP}$ has some inverse $f\in{\sf fP}$, and $\Phi_{(u,v^{\prime})}$ is a subfunction of $\psi_{u}$, $\Phi_{(u,v^{\prime})}$ also has $f$ as an inverse; hence, $\Phi_{(u,v^{\prime})}\in{\sf cofP}$. Moreover, since $\psi_{u}\,\psi_{v^{\prime}}\,\psi_{u}(x)=\psi_{u}(x)$ for every $x\in{\sf Dom}(\Phi_{(u,v^{\prime})})$, and $\Phi_{(u,v^{\prime})}\subseteq\psi_{u}$, we have $\Phi_{(u,v^{\prime})}\,\psi_{v^{\prime}}\,\Phi_{(u,v^{\prime})}=\Phi_{(u,v^{\prime})}$. So $\Phi_{(u,v^{\prime})}$ is regular in cofP. Finally, every regular element of cofP obviously has a program of the form $(u,v^{\prime})$ as above.

The rest of the proof of finite generation is very similar to the one for cofP.

(2) The second part of (2) was already proved in Theorem 4.18. On the other hand, if ${\sf P}\neq{\sf NP}$ then cofP is not regular (by Theorem 4.18), so ${\sf RegcofP}\neq{\sf cofP}\neq{\sf invfP}^{\sf(NP)}$. Also, cofP contains regular elements that are not in invfP, by Prop. 4.21, hence ${\sf invfP}\neq{\sf RegcofP}$.

(3) If $f$ is a regular element of fP then $f$ has an inverse in invfP, by Coroll. 2.3.

In general, let $f$ be an element in fP that has an inverse $f^{\prime}\in{\sf RegcofP}$; we want to show that $f$ is regular in fP. By what we proved in (1) of the present proof, $f^{\prime-1}\in{\sf cofP}$. We have $f^{\prime-1}gf^{\prime-1}=f^{\prime-1}$ for some $g\in{\sf fP}$, since $f^{\prime-1}\in{\sf cofP}$. Multiplying this on the left and the right by $f^{\prime}$ yields $\,{\sf id}_{I}\circ g\circ{\sf id}_{D}=f^{\prime}$, where $D={\sf Dom}(f^{\prime})$ and $I={\sf Im}(f^{\prime})$. Since $f^{\prime}$ is an inverse of $f$, Lemma 4.17 implies that $g$ is an inverse of $f$. Since $g\in{\sf fP}$, it follows that $f$ is regular in fP.    $\Box$

We saw that the monoids fP, invfP, ${\sf invfP}^{\sf(NP)}$, and cofP are finitely generated. For injfP we ask similarly:

Question:  Is $\,{\sf injfP}$ finitely generated?

We do not know the answer. If we could show that injfP is not finitely generated, then this would prove that ${\sf injfP}\neq{\sf invfP}$, i.e., injective one-way functions exist, hence ${\sf P}\neq{\sf UP}$, and hence ${\sf P}\neq{\sf NP}$.

When we proved that fP, invfP, ${\sf invfP}^{\sf(NP)}$, and cofP are finitely generated, we used machine (or program) models, and evaluation functions. It seems that injfP does not have a program model, since injectiveness is a for-all property, that does not have finite witnesses in general. To illustrate the difficulty of finding a program model for injfP, here is an idea that does not work. We saw that ${\sf injfP}={\sf fP}\cap{\sf invfP}^{\sf(NP)}$, and this suggests that a function in injfP can be specified by a pair $(u,w)$, where $w$ is an fP-program (for $\phi_{w}\in{\sf fP}$), and $u$ is an ${\sf invfP}^{\sf(NP)}$-program (for $\psi_{u}\in{\sf invfP}^{\sf(NP)}$); the program $(u,w)$ specifies the injective function $\psi_{u}\cap\phi_{w}$. But $\psi_{u}\cap\phi_{w}$ ranges over all of ${\sf invfP}^{\sf(NP)}$ and does not necessarily belong to injfP (unless ${\sf injfP}=$ ${\sf invfP}^{\sf(NP)}$, which would imply that ${\sf P}={\sf NP}$). So, this approach towards proving finite generation does not work (unless one also proves that ${\sf P}={\sf NP}$). It seems that injfP is not finitely generated (but this will probably be very difficult to prove).

5 Subgroups and group-inverses

We first characterize the maximal subgroups of invfP, cofP, and ${\sf invfP}^{\sf(NP)}$. Then we consider elements of fP that have an inverse in such a subgroup (i.e., a group-inverse). Typically, elements of fP do not have a group-inverse, but we will see that all elements of fP are reduction-equivalent to elements that have group-inverses.

5.1 Maximal subgroups

We first characterize the idempotents of injfP.

Proposition 5.1

For any $f\in{\sf injfP}$ we have: $f$ is an idempotent  iff  $f={\sf id}_{Z}$ for some set $Z\subseteq A^{*}$ with $Z\in{\sf P}$. Hence injfP and invfP have the same idempotents.

Proof. If $f\in{\sf injfP}$ and $f=f\circ f$, then we compose on the left with $f^{-1}$ (which exists, since $f$ is injective, but $f^{-1}$ might not belong to injfP); this yields ${\sf id}_{D}={\sf id}_{D}\circ f$, where $D={\sf Dom}(f)$. Hence for all $x\in D={\sf Dom}(f)$ we have $x=f(x)$. Thus, $f={\sf id}_{D}$. Since $f\in{\sf fP}$, $D={\sf Dom}(f)\in{\sf P}$.

Conversely, if $Z\in{\sf P}$ then ${\sf id}_{Z}$ can be computed by a deterministic polynomial-time Turing machine. Moreover, ${\sf id}_{Z}\in{\sf invfP}$ since ${\sf id}_{Z}$ is injective, and it is regular (being an idempotent).    $\Box$

Proposition 5.2

The maximal subgroup of invfP and injfP are the same.

Let ${\sf id}_{Z}$ be an idempotent of invfP. The maximal subgroup of invfP with identity ${\sf id}_{Z}$ consists of the permutations of $Z$ that belong to invfP. In particular, the group of units of invfP consists of all permutations of $A^{*}$ that belong to invfP.

Proof. If $f$ belongs to a subgroup of injfP then $f$ is regular, hence $f\in{\sf invfP}$.

Obviously, the permutations in invfP with domain and image $Z$ form a subgroup of invfP. Conversely, if $f\in{\sf invfP}$ belongs to the maximal subgroup with identity ${\sf id}_{Z}$, then $f^{-1}f=ff^{-1}={\sf id}_{Z}$; so the domain and the image of $f$ are both $Z$. Since $f\in{\sf invfP}$, $f$ is injective, and $f^{-1}\in{\sf invfP}$. Hence $f$ permutes $Z$.    $\Box$

Proposition 5.3

Every maximal subgroup of invfP is also a maximal subgroup of fP. The group of units of fP is the same as the group of units of invfP.

Proof. A maximal subgroup of fP is a regular $\cal H$-class. For fP the characterization of the $\cal L$- and the $\cal R$-relation (see Prop. 2.1 in [3]) implies that every element $f$ of a maximal subgroup of fP with unit ${\sf id}_{Z}$ satisfies ${\sf Im}(f)=Z={\sf Dom}(f)$ and $f$ is injective (since $f$ has the same partition as ${\sf id}_{Z}$). Thus $f$ belongs to invfP.    $\Box$

The converse of Prop. 5.3 is of course not true; e.g., fP contains non-injective idempotents, hence subgroups that are not contained in invfP.

Let us now look at the idempotents and subgroups of ${\sf invfP}^{\sf(NP)}$ and of cofP. We saw that ${\sf invfP}^{\sf(NP)}$ and cofP have the same idempotents (Prop. 4.20). And we saw that ${\sf invfP}^{\sf(NP)}$ is an inverse monoid (Cor. 3.8); on the other hand, cofP is not regular, unless ${\sf P}={\sf NP}$ (by Theorem 4.18).

Lemma 5.4

For every $L\subset A^{*}$ we have:  $L\in{\sf P}^{\sf(NP)}$  iff  ${\sf id}_{L}\in{\sf invfP}^{\sf(NP)}$.

Proof. $[\Rightarrow]$ If $L\in{\sf P}^{\sf(NP)}$ then ${\sf id}_{L}$ can be computed by deterministic polynomial-time Turing machine with NP-oracle. Moreover, the inverse of ${\sf id}_{L}$ (which is just ${\sf id}_{L}$ itself) also has that property, hence by Prop. 3.9, ${\sf id}_{L}$ can be computed by an injective polynomial-time Turing machine with NP-oracle. Hence, by the definition of ${\sf invfP}^{\sf(NP)}$ we have ${\sf id}_{L}\in{\sf invfP}^{\sf(NP)}$.

$[\Leftarrow]$ From a polynomial-time injective Turing machine with NP-oracle, computing ${\sf id}_{L}$, one immediately obtains a deterministic polynomial-time Turing machine with NP-oracle, accepting $L$.    $\Box$

Proposition 5.5

Every idempotent of $\,{\sf invfP}^{\sf(NP)}$ (and cofP) is of the form ${\sf id}_{L}$, where $L\in{\sf P}^{\sf(NP)}$,  $L\subseteq A^{*}$. Moreover, every $L\in{\sf P}^{\sf(NP)}$ is accepted by an injective polynomial-time Turing machine with NP-oracle.

Proof. Let $f=f\circ f$ be an idempotent of ${\sf invfP}^{\sf(NP)}$. Multiplying in the left by $f^{-1}$ yields ${\sf id}_{D}={\sf id}_{D}\circ f$, where $D={\sf Dom}(f)$. Hence, $f={\sf id}_{D}$. Since $f\in$ ${\sf invfP}^{\sf(NP)}$ we have $D\in{\sf P}^{\sf(NP)}$ (by Lemma 5.4).

Conversely, if $Z\in{\sf P}^{\sf(NP)}$ then ${\sf id}_{Z}\in$ ${\sf invfP}^{\sf(NP)}$ (by Lemma 5.4). From this we obtain an injective polynomial-time Turing machine with NP-oracle, accepting $Z$.    $\Box$

Proposition 5.6

(1) The maximal subgroup of ${\sf invfP}^{\sf(NP)}$ with idempotent ${\sf id}_{L}$ (where $L\in{\sf P}^{\sf(NP)}$) consists of all the permutations of $L$ that belong to ${\sf invfP}^{\sf(NP)}$. In particular, the group of units of ${\sf invfP}^{\sf(NP)}$ consists of all permutations of $A^{*}$ that belong to ${\sf invfP}^{\sf(NP)}$.

(2) The maximal subgroup of cofP with idempotent ${\sf id}_{L}$ (where $L\in{\sf P}^{\sf(NP)}$) consists of all permutations $g$ of $L$ such that $g=f\,\cap\,(L\times L)$ and $g^{-1}=h\,\cap\,(L\times L)$ for some $f,h\in{\sf fP}$.

Hence, if $L\in{\sf P}$ then the maximal subgroup of cofP with idempotent ${\sf id}_{L}$ is a subgroup of invfP. In particular, the group of units of cofP is the same as the group of units of invfP.

Proof. (1) By injectiveness of the elements of ${\sf invfP}^{\sf(NP)}$, every element of the maximal subgroup of ${\sf invfP}^{\sf(NP)}$ with idempotent ${\sf id}_{L}$ is a permutation of $L$. Moreover, for every $g\in{\sf invfP}^{\sf(NP)}$ we have $g^{-1}\in{\sf invfP}^{\sf(NP)}$ (by Prop. 3.7); hence the permutations of $L$ in ${\sf invfP}^{\sf(NP)}$ form a subgroup of ${\sf invfP}^{\sf(NP)}$.

(2) If $g\in{\sf invfP}^{\sf(NP)}$ belongs to a maximal subgroup of cofP with idempotent ${\sf id}_{L}$, then both $g$ and $g^{-1}$ are permutations of $L$. If $g$ has an inverse $h\in{\sf fP}$ then multiplying $ghg=g$ on the left and on the right by $g^{-1}$ yields ${\sf id}_{L}\circ h\circ{\sf id}_{L}=g^{-1}$; moreover, ${\sf id}_{L}\circ h\circ{\sf id}_{L}=h\,\cap\,(L\times L)$. Similarly, for $g^{-1}\in{\sf coP}$ we obtain $f\,\cap\,(L\times L)=g$ for some $f\in{\sf fP}$.

Conversely, let $g$ be a permutation of $L\in{\sf P}^{\sf(NP)}$ such that $g=f\,\cap\,(L\times L)$ and $g^{-1}=h\,\cap\,(L\times L)$ for some $f,h\in{\sf fP}$. Then $g,g^{-1}$ belong to ${\sf invfP}^{\sf(NP)}$ since $L\in{\sf P}^{\sf(NP)}$ and $f,h\in{\sf fP}$. Moreover, $g$ and $g^{-1}$ belong to cofP; indeed, the inverses $f\,\cap\,(L\times L)$ and $h\,\cap\,(L\times L)$ can be extended to inverses $h$, respectively $f$, in fP (by Lemma 4.17). Since $g$ is a permutation of $L$, it follows now that $g$ belongs to a subgroup of cofP.    $\Box$

We saw in Prop. 4.21 that unless ${\sf P}={\sf NP}$, cofP contains regular elements that are not in invfP. The next Proposition describes the regular elements of cofP.

Proposition 5.7

Every element of cofP that is regular in cofP has the form $f\,\cap\,(K\times H)$, for some $f\in{\sf fP}$ and $K,H\in{\sf P}^{\sf(NP)}$.

Conversely, suppose $f,h\in{\sf fP}$ and $K,H\in{\sf P}^{\sf(NP)}$ are such that $f\,\cap\,(K\times H)$ and $h\,\cap\,(H\times K)$ are injective and mutual inverses. Then $f\,\cap\,(K\times H)$, $h\,\cap\,(H\times K)$ $\in$ ${\sf cofP}$, and they are regular in cofP.

Proof. If $g$ is regular in cofP then $g^{-1}\in{\sf cofP}$; hence, $g^{-1}\,f\,g^{-1}=g^{-1}$ for some $f\in{\sf fP}$. Hence by multiplying on the left and the right by $g$ we obtain: $f\,\cap\,(K\times H)=g$, where $K={\sf Im}(g)$ and $H={\sf Dom}(g)$.

For the converse, f\,\cap\,(K\times H)\ =\ $(f\,\cap\,(K\times H))\circ(h\,\cap\,(H\times K))$ $\circ$ $(f\,\cap\,(K\times H))$ \ =\ $(f\,\cap\,(K\times H))\circ h\circ(f\,\cap\,(K\times H))$; the latter holds by Lemma 4.17. Hence, $f\,\cap\,(K\times H)$ has an inverse in fP, so $f\,\cap\,(K\times H)$ belongs to cofP. Similarly, $h\,\cap\,(H\times K)$ belongs to cofP. Since they are mutual inverses, they are regular in cofP.    $\Box$

5.2 Group inverses of elements of fP

After noticing that every regular element of fP has an injective inverse, belonging to an inverse submonoid of fP, we wonder whether we can go even further: Does every regular element in fP have an inverse in a subgroup of fP? This is of course not the case; for immediate counter-examples consider the functions that are total but not surjective, or surjective but not total. Even within the subsemigroup of non-total non-surjective functions of fP there are counter-examples (due to the polynomial balance and time requirements). Nevertheless, we will find that every regular element of fP is equivalent, with respect to inversive reduction, to a regular element of fP that has an inverse in a subgroup of fP.

We will also investigate elements of fP that are possibly non-regular, but that have an inverse in a subgroup of ${\sf invfP}^{\sf(NP)}$. Again, we will find that every element of fP is equivalent, with respect to inversive reduction, to an element of fP that has an inverse in a subgroup of ${\sf invfP}^{\sf(NP)}$. In particular, there are elements of fP that are complete with respect to inversive reduction and that have an inverse in a subgroup of ${\sf invfP}^{\sf(NP)}$.

Some definitions about inversive reductions (from [3]) between functions $f_{1},f_{2}$:

$\bullet$   $f_{1}$ is simulated by $f_{2}$ (denoted by $f_{1}\preccurlyeq f_{2}$)  iff  there exist $\beta,\alpha\in{\sf fP}$ such that $f_{1}=\beta\circ f_{2}\circ\alpha$.

$\bullet$   $f_{1}$ reduces inversively to $f_{2}$ (notation, $f_{1}\leqslant_{\sf inv}f_{2}$)  iff

(1)  $f_{1}\preccurlyeq f_{2}$  and

(2)  for every inverse $f_{2}^{\prime}$ of $f_{2}$ there exists an inverse $f_{1}^{\prime}$ of $f_{1}$ such that $f_{1}^{\prime}\preccurlyeq f_{2}^{\prime}$ .

$\bullet$   $f,g\in{\sf fP}$ are equivalent via inversive reduction  iff  $f\leqslant_{\sf inv}g\,$ and $\,g\leqslant_{\sf inv}f$.

Theorem 5.8

Every function $f\in{\sf fP}$ is equivalent, via inversive reduction, to a function $f_{0}\in{\sf fP}$ such that $f_{0}$ has an inverse in a subgroup of ${\sf invfP}^{\sf(NP)}$.

In particular, $f_{0}$ has an inverse in the group of permutations of $\ 0\,{\sf Dom}(f)\,\cup\,1\,{\sf Im}(f)$, in ${\sf invfP}^{\sf(NP)}$. Moreover, $f_{0}$ has an inverse in the group of units of ${\sf invfP}^{\sf(NP)}$.

If $f$ (and hence $f_{0}$) is regular, then $f_{0}$ has an inverse in the group of permutations of $\,0\,{\sf Dom}(f)\,\cup\,1\,{\sf Im}(f)$, in invfP. Moreover, $f_{0}$ has an inverse in the group of units of invfP.

Remarks. Since $f$ and $f_{0}$ are equivalent via reduction it follows immediately that $\,f_{0}\in{\sf fP}$ iff $f\in{\sf fP}$; and $f$ is regular iff $f_{0}$ is regular. And by Theorem 4.18, if $f$ is regular then $f_{0}$ has an inverse in the inverse monoid invfP.

Proof of Theorem 5.8. (1) With $f$ we associate $f_{0}$ defined by

${\sf Dom}(f_{0})\ =\ 0\ {\sf Dom}(f)$,  and

$f_{0}(0\,x)=1\,f(x)$,  for all $x\in{\sf Dom}(f)$.

Similarly, we define $f_{1}$ by $\,f_{1}(1\,x)=0\,f(x)$.

To show that $f$ and $f_{0}$ simulate each other (and similarly for $f_{1}$), we introduce the functions $\pi_{a}$ and $\pi_{a}^{\prime}$ for each $a\in A$; they are defined for all $z\in A^{*}$ by

$\pi_{a}(z)=az$,   and   $\pi_{a}^{\prime}(az)=z$, with $\,{\sf Dom}(\pi_{a}^{\prime})=aA^{*}$.

Then we have:

$f_{0}=\pi_{1}\circ f\circ\pi_{0}^{\prime}$,   and   $f=\pi_{1}^{\prime}\circ f_{0}\circ\pi_{0}$.

Hence, $f$ and $f_{0}$ simulate each other.

Let us show that we have an inversive reduction of $f_{0}$ to $f$. If $f$ has an inverse $f^{\prime}$, let us define $f^{\prime}_{1}$ by $f^{\prime}_{1}(1\,y)=0\,f^{\prime}(y)$ for all $y\in{\sf Dom}(f^{\prime})$. Then $f^{\prime}_{1}$ is an inverse of $f_{0}$. And $f^{\prime}_{1}=\pi_{0}\circ f^{\prime}\circ\pi_{1}^{\prime}$, so $f_{1}^{\prime}$ is simulated by $f^{\prime}$.

Conversely, let us show that there is an inversive reduction of $f$ to $f_{0}$. Let $g$ be any inverse of $f_{0}$, i.e., $f_{0}gf_{0}(x)=f_{0}(0\,x)$, all $x\in{\sf Dom}(f)$. Then $h=g\,\cap\,1A^{*}\times 0A^{*}$ can be simulated by $g$ (indeed, $h(.)={\sf id}_{0A^{*}}\circ g\circ{\sf id}_{1A^{*}}(.)$), and $h$ also satisfies $f_{0}hf_{0}=f_{0}$. Moreover, $h\subset 1A^{*}\times 0A^{*}$ implies that $h=k_{1}$ for some function $k$. Then we have $fkf=f$; indeed, for all $x\in{\sf Dom}(f)$ we have $fkf(x)=\pi_{1}^{\prime}\,f_{0}\,k_{1}\,f_{0}(0x)=\pi_{1}^{\prime}\,f_{0}(0x)=f(x)$. We saw that $k$ is simulated by $k_{1}$ (indeed, $k=\pi_{0}^{\prime}\circ k_{1}\circ\pi_{1}$). Thus there exists an inverse (namely $k$) of $f$ that is simulated by $g$.

This completes the proof that $f$ is equivalent to $f_{0}$ via inversive reduction. It follows immediately that $f_{0}\in{\sf fP}\,$ iff $\,f\in{\sf fP}$, and that $f$ is regular iff $f_{0}$ is regular.

(2)  Let $f^{\prime}$ be any mutual inverse of $f$ with $f^{\prime}\in{\sf cofP}$, or with $f^{\prime}\in{\sf invfP}$ if $f$ is regular. Let us now extend $f_{1}^{\prime}$ to a group element. Recall that by definition, $f^{\prime}_{1}(1\,y)=0\,f^{\prime}(y)$ for all $y\in{\sf Dom}(f^{\prime})$, $\,{\sf Dom}(f_{1}^{\prime})=1\,{\sf Dom}(f^{\prime})=1\,{\sf Im}(f)$, and ${\sf Im}(f_{1}^{\prime})=0\,{\sf Im}(f)$. So, ${\sf Dom}(f_{1}^{\prime})\,\cap\,{\sf Im}(f_{1}^{\prime})=\varnothing$, and $f_{1}^{\prime}$ only has orbits of length 1. We extend $f_{1}^{\prime}$ to a permutation $F^{\prime}$ of $\,{\sf Dom}(f_{1}^{\prime})\cup{\sf Im}(f_{1}^{\prime})$, defined by

$F^{\prime}|_{{\sf Dom}(f_{1}^{\prime})}=f_{1}^{\prime}$    and    $F^{\prime}|_{{\sf Im}(f_{1}^{\prime})}=f_{1}^{\prime-1}$.

Hence, $F^{\prime}=f_{1}^{\prime}\cup f_{1}^{\prime-1}$. It follows that $F^{\prime}\in{\sf invfP}^{\sf(NP)}$, since both $f_{1}^{\prime}$ and $f_{1}^{\prime-1}$ belong to ${\sf invfP}^{\sf(NP)}$.

Also, $F^{\prime}$ is injective and $\,F^{\prime}\circ F^{\prime}={\sf id}_{Z}$, where $Z={\sf Dom}(f_{1}^{\prime})\cup{\sf Dom}(f_{1}^{\prime-1})$; so $F^{\prime}=F^{\prime-1}$ belongs to a two-element group. Clearly, ${\sf Dom}(F^{\prime})={\sf Im}(F^{\prime})=Z$.

Moreover, $F^{\prime}$ is an inverse of $f_{0}$. Indeed, for every $x\in{\sf Dom}(f)$ we have $\,f_{0}\,F^{\prime}\,f_{0}(0x)\,=$ $f_{0}\,F^{\prime}(1f(x))\,=\,f_{0}\,f_{1}^{\prime}(1f(x))$, since $F^{\prime}=f_{1}^{\prime}\cup f_{1}^{\prime-1}$ and $f_{1}^{\prime-1}$ is undefined on $1A^{*}$; hence, $f_{0}\,F^{\prime}\,f_{0}(0x)\,=\,f_{0}(0x)$.

Finally, let us show that if $f^{\prime}\in{\sf invfP}$ then $F^{\prime}\in{\sf invfP}$. Indeed, when $f^{\prime}\in{\sf invfP}$ then $f^{\prime-1}\in{\sf invfP}$, from which it follows (by disjointness of the domains and disjointness of the images) that $f_{1}^{\prime}\cup f_{1}^{\prime-1}\in$ invfP.

(3)  We can extend $F^{\prime}$ to a permutation of all of $A^{*}$ (belonging to the group of units of ${\sf invfP}^{\sf(NP)}$): We simply define $F^{\prime}$ as a permutation of ${\sf Dom}(f_{1}^{\prime})\cup{\sf Im}(f_{1}^{\prime})$, as above, and then extend $F^{\prime}$ to the identity function on $A^{*}-({\sf Dom}(f_{1}^{\prime})\cup{\sf Im}(f_{1}^{\prime}))$. Since $f_{1}^{\prime},\,{\sf id}_{{\sf Dom}(f_{1}^{\prime})}$, and ${\sf id}_{{\sf Im}(f_{1}^{\prime})}$ belong to ${\sf invfP}^{\sf(NP)}$, it follows that this extension of $F^{\prime}$ to all of $A^{*}$ belongs to ${\sf invfP}^{\sf(NP)}$.

Finally, if $f$ is regular then this extension of $F^{\prime}$ belongs to invfP; indeed, in that case, ${\sf Dom}(f_{1}^{\prime})$ and ${\sf Im}(f_{1}^{\prime})$ belong to P.    $\Box$

Corollary 5.9

There exists a function that is complete in fP for inversive reduction, and that has an inverse in the group of units of ${\sf invfP}^{\sf(NP)}$.

Proof. In [3] we saw examples of functions $F\in{\sf fP}$ that are complete with respect to inversive reduction. Then $F_{0}$, defined by $F_{0}(0x)=1\,F(x)$, is also complete (since $F$ and $F_{0}$ are equivalent via inversive reduction); and $F_{0}$ has an inverse in the group of units of ${\sf invfP}^{\sf(NP)}$ (as we saw in the proof of Theorem 5.8).    $\Box$

Remark. The group-inverses constructed above belong to ${\sf invfP}^{\sf(NP)}$, but not necessarily to cofP. If there were a complete function in fP that has an inverse in the group of units of cofP then ${\sf P}={\sf NP}$ (since the group of units of cofP is the same as the group of units of invfP).

6 Appendix: Simple facts about inverses (related to Section 2)

We present some additional properties of invfP that are not used in the rest of the paper.

For $f\in{\sf fP}$ we define the right fixator of $f$ by $\,{\sf RFix}(f)=\{\alpha\in{\sf fP}:f\circ\alpha=f\}$; in other words, $\alpha\in{\sf RFix}(f)$ iff $\alpha$ is a right-identity of $f$. Similarly, the left fixator of $f$ is ${\sf LFix}(f)=\{\alpha\in{\sf fP}:\alpha\circ f=f\}$.

We observe that ${\sf RFix}(f)\cap{\sf invfP}$ is an inverse monoid; i.e., $\alpha\in{\sf RFix}(f)\cap{\sf invfP}$ implies $\alpha^{-1}\in{\sf RFix}(f)\cap{\sf invfP}$. Indeed, if $\alpha\in{\sf invfP}$ satisfies $f\alpha=f$ then $f\alpha\alpha^{-1}=f\alpha^{-1}$; moreover, $f\alpha=f$ implies that ${\sf Dom}(f)\subseteq{\sf Im}(\alpha)$, hence $\alpha\alpha^{-1}={\sf id}_{{\sf Im}(\alpha)}$ acts as a right-identity on $f$; hence $f=f\alpha\alpha^{-1}$. Thus, $f=f\alpha^{-1}$.

For every $\alpha\in{\sf RFix}(f)$ we have:  $\alpha({\sf Dom}(f))\subseteq{\sf Dom}(f)$,  $\alpha|_{{\sf Dom}(f)}\in{\sf RFix}(f)$, and $\,\alpha(A^{*}-{\sf Dom}(f))\subseteq A^{*}-{\sf Dom}(f)$.

Every representative choice function of $f$ belongs to ${\sf RFix}(f)$, since any right fixator $\alpha\in{\sf RFix}(f)$ maps every ${\sf mod}_{f}$-class into itself. But the converse is not true, since a right fixator $\alpha$ does not necessarily map a ${\sf mod}_{f}$-class to a single element of the ${\sf mod}_{f}$-class (as a representative choice function does).

Proposition 6.1

Let $f_{1}^{\prime},f_{2}^{\prime}\in{\sf invfP}$ be inverses of the same element $f\in{\sf fP}$, such that ${\sf Dom}(f_{i}^{\prime})={\sf Im}(f)$ ($i=1,2$). Then:

(1)*    $f_{1}^{\prime-1}\equiv_{\cal R}f_{2}^{\prime-1}$ $\equiv_{\cal R}{\sf id}_{{\sf Im}(f)}\,$ (for the ${\cal R}$-relation of invfP),  and*

$\,{\sf id}_{{\sf Im}(f)}\equiv_{\cal R}f\,$* (for the ${\cal R}$-relation of fP); and*

$\,f_{1}^{\prime}\equiv_{\cal L}f_{2}^{\prime}\,$* (for the ${\cal L}$-relation of invfP).*

If $\,f_{1}^{\prime}\equiv_{\cal L}f_{2}^{\prime}\,$ and $\,f_{1}^{\prime}\equiv_{\cal R}f_{2}^{\prime}\,$ (for the ${\cal L}$- and ${\cal R}$-relations of fP), then $\,f_{1}^{\prime}=f_{2}^{\prime}$.

(2)*    $f_{i}^{\prime}f\in{\sf RFix}(f)$,*

$f_{2}^{\prime}f_{1}^{\prime-1}$* is a bijection from the choice set ${\sf Im}(f_{1}^{\prime})$ onto the choice set ${\sf Im}(f_{2}^{\prime})$;*

$\,f_{i}^{\prime}f_{i}^{\prime-1}={\sf id}_{{\sf Im}(f_{i}^{\prime})}$.

$\,f_{i}^{\prime-1}f_{i}^{\prime}={\sf id}_{{\sf Im}(f)}$.

(3)*    $f_{2}^{\prime}\,f\,f_{1}^{\prime}=f_{2}^{\prime}$.*

(4)*    If $\alpha\in{\sf RFix}(f)$, and if $f^{\prime}\in{\sf fP}$ is an inverse of $f\in{\sf fP}$ such that ${\sf Dom}(f^{\prime})={\sf Im}(f)$, then $\alpha\circ f^{\prime}$ is also an inverse of $f$ such that ${\sf Dom}(\alpha f^{\prime})={\sf Im}(f)$. So, the monoid ${\sf RFix}(f)$ acts on the left on the set*

$\{g^{\prime}\in{\sf invfP}:\,g^{\prime}$* is an inverse of $f$ and ${\sf Dom}(g^{\prime})={\sf Im}(f)\}$.*

Moreover, this action is transitive, and the action of $\,\{\alpha|_{{\sf Dom}(f)}:\alpha\in{\sf RFix}(f)\}\,$ is faithful.

Proof. (1) We saw that $f_{i}^{\prime-1}=f|_{{\sf Im}(f_{i}^{\prime})}$, hence $f_{i}^{\prime-1}=f\circ{\sf id}_{{\sf Im}(f_{i}^{\prime})}$, where ${\sf Im}(f_{i}^{\prime})\in{\sf P}$ (since $f_{i}^{\prime}$ is regular); hence ${\sf id}_{{\sf Im}(f_{i}^{\prime})}\in{\sf fP}$ and $f_{i}^{\prime-1}\leq_{\cal R}f$.

Also, $f_{i}^{\prime}=f_{i}^{\prime}ff_{i}^{\prime}\equiv_{\cal L}ff_{i}^{\prime}={\sf id}_{{\sf Im}(f)}$. We also have $f_{i}^{\prime-1}\equiv_{\cal R}{\sf id}_{{\sf Im}(f)}$; indeed, $f_{i}^{\prime-1}$ maps ${\sf Im}(f_{i}^{\prime})$ bijectively onto ${\sf Im}(f)$, so $f_{i}^{\prime-1}={\sf id}_{{\sf Im}(f)}\circ f_{i}^{\prime-1}\leq_{\cal R}$ ${\sf id}_{{\sf Im}(f)}$; and ${\sf id}_{{\sf Im}(f)}=f_{i}^{\prime-1}f_{i}^{\prime}$ $\leq_{\cal R}f_{i}^{\prime-1}$. Moreover, $f\equiv_{\cal R}{\sf id}_{{\sf Im}(f)}$, hence $f_{i}^{\prime-1}\equiv_{\cal R}f$.

The $\,\equiv_{\cal L}$-equivalence then follows, since in any inverse monoid, $x^{\prime}\equiv_{\cal R}y^{\prime}$ implies $x\equiv_{\cal L}y$ (where $z^{\prime}$ is the inverse of $z$ for any element $z$).

The fact that $f_{1}^{\prime}\equiv_{\cal L}f_{2}^{\prime}$ and $f_{1}^{\prime}\equiv_{\cal R}f_{2}^{\prime}$ imply $f_{1}^{\prime}=f_{2}^{\prime}$ is well-known in semigroup theory (see e.g. p. 26 in [5]).

(2) Obviously, $f\,f_{i}^{\prime}\,f=f$. The rest is straightforward.

(3) We know that $f_{1}^{\prime}$ and $f_{2}^{\prime}$ have the same domain, namely ${\sf Im}(f)$. And $f\,f_{1}^{\prime}|_{{\sf Im}(f)}={\sf id}_{{\sf Im}(f)}$, hence $f_{2}^{\prime}\,f\,f_{1}^{\prime}=f_{2}^{\prime}\,{\sf id}_{{\sf Im}(f)}=f_{2}^{\prime}$ (the latter again holds since ${\sf Dom}(f_{i}^{\prime})={\sf Im}(f)$).

(4) If $f\alpha=f$ then $f\alpha f^{\prime}f=ff^{\prime}f=f$, so $\alpha f^{\prime}$ is also an inverse of $f$. And ${\sf Dom}(\alpha f^{\prime})\subseteq{\sf Dom}(f^{\prime})={\sf Im}(f)$; since $\alpha f^{\prime}$ is an inverse of $f$ we also have ${\sf Im}(f)\subseteq{\sf Dom}(\alpha f^{\prime})$; hence ${\sf Dom}(\alpha f^{\prime})={\sf Im}(f)$. It follows that $\alpha f^{\prime}\in{\sf invfP}$, so ${\sf RFix}(f)$ acts on the given set on the left.

Transitivity follows from $(f_{2}^{\prime}f)\,f_{1}^{\prime}=f_{2}^{\prime}$ (proved in (3)), where $f_{2}^{\prime}f\in{\sf RFix}(f)$ by (2).

Proof of faithfulness: Note that ${\sf Dom}(f)\subseteq{\sf Dom}(\alpha)$ for all $\alpha\in{\sf RFix}(f)$. If $\alpha_{1}|_{{\sf Dom}(f)}\neq\alpha_{2}|_{{\sf Dom}(f)}$ then there exists $x_{0}\in{\sf Dom}(f)$ such that $\alpha_{1}(x_{0})\neq\alpha_{2}(x_{0})$. There exists an inverse $f_{0}^{\prime}\in{\sf invfP}$ of $f$ with ${\sf Dom}(f_{0}^{\prime})={\sf Im}(f)$ such that $f_{0}^{\prime}(f(x_{0}))=x_{0}$; indeed, we can start with any inverse $f_{0}^{\prime}$ with ${\sf Dom}(f_{0}^{\prime})={\sf Im}(f)$, and if $f_{0}^{\prime}(f(x_{0}))\neq x_{0}$ we can redefine $f_{0}^{\prime}$ on $f(x_{0})$; this does not change ${\sf Dom}(f_{0}^{\prime})$ (it just changes the choice set of $f_{0}^{\prime}$). Now, $\alpha_{1}f_{0}^{\prime}f(x_{0})=\alpha_{1}(x_{0})\neq\alpha_{2}(x_{0})$ $=$ $\alpha_{2}f_{0}^{\prime}f(x_{0})$, so $\alpha_{1}f_{0}^{\prime}\neq\alpha_{2}f_{0}^{\prime}$.    $\Box$

Proposition 6.2

If $f\in{\sf fP}$ is regular, and if $\,{\sf Dom}(f\circ\alpha)={\sf Dom}(f)$, then we have:

$\alpha\in{\sf RFix}(f)$*   iff   for every mutual inverse $f^{\prime}$ of $f$: $\alpha\circ f^{\prime}$ is also a mutual inverse of $f$.*

(If $f$ is not regular, the Proposition holds for inverses in ${\sf invfP}^{\sf(NP)}$.)

Proof. $[\Rightarrow]$  If $\alpha\in{\sf RFix}(f)$ and $f^{\prime}$ is a mutual inverse of $f$ then $f\,\alpha f^{\prime}\,f=ff^{\prime}f=f$; the first equality holds because $\alpha\in{\sf RFix}(f)$. And $\alpha f^{\prime}\,f\,\alpha f^{\prime}=\alpha f^{\prime}\,f\,f^{\prime}=\alpha f^{\prime}$ (again, using $f\alpha=f$). So $\alpha f^{\prime}$ is a mutual inverse of $f$.

$[\Leftarrow]$  If $\alpha\not\in{\sf RFix}(f)$ then $f\alpha(x_{0})\neq f(x_{0})$ for some $x_{0}\in A^{*}$. If $x_{0}\in{\sf Dom}(f)$, let $f^{\prime}\in{\sf fP}$ be a mutual inverse of $f$ such that $f^{\prime}f(x_{0})=x_{0}$; such a mutual inverse exists (if $f^{\prime}f(x_{0})\neq x_{0}$ we can redefine $f^{\prime}$ on the one element $f(x_{0})$). Then $f\,\alpha f^{\prime}\,f(x_{0})=f\alpha(x_{0})\neq f(x_{0})$, so $f\,\alpha f^{\prime}\,f\neq f$. So there exists a mutual inverse $f^{\prime}$ such that $\alpha f^{\prime}$ is not an inverse of $f$.    $\Box$

From the above we conclude that finding an inverse $f^{\prime}$ for $f$ with ${\sf Dom}(f^{\prime})={\sf Im}(f)$ can be broken up into two steps: (1) find a maximal injective subfunction $f^{\prime-1}$ of $f$ (being a maximal injective subfunction is equivalent to ${\sf Im}(f^{\prime-1})={\sf Im}(f)$); (2) find the inverse $f^{\prime}$ of $f^{\prime-1}$. (If $f$ is injective, step (1) is skipped.)

Proposition 6.3

When $f\in{\sf fP}$ is regular and if $\,{\sf Im}(\beta\circ f)={\sf Im}(f)$ we have:

$\beta\in{\sf LFix}(f)$*   iff   for every mutual inverse $f^{\prime}$ of $f$:  $f^{\prime}\beta$ is also a mutual inverse of $f$.*

(If $f$ is not regular, the Proposition holds for inverses in ${\sf invfP}^{\sf(NP)}$.)

Proof. $[\Rightarrow]$  If $\beta\in{\sf LFix}(f)$ and $f^{\prime}$ is a mutual inverse of $f$ then $f\,f^{\prime}\beta\,f=ff^{\prime}f=f$; and $f^{\prime}\beta\,f\,f^{\prime}\beta=f^{\prime}\,f\,f^{\prime}\beta=f^{\prime}\beta$ (using $\beta f=f$).

$[\Leftarrow]$ If $\beta\not\in{\sf LFix}(f)$ then $\beta f(x_{0})\neq f(x_{0})$ for some $x_{0}\in A^{*}$. If $\beta f(x_{0})\in{\sf Im}(f)$ then $f\,f^{\prime}\beta\,f(x_{0})=\beta\,f(x_{0})$ since $ff^{\prime}|_{{\sf Im}(f)}={\sf id}_{{\sf Im}(f)}$; and $\beta\,f(x_{0})\neq f(x_{0})$, hence $f\,f^{\prime}\beta\,f\neq f$.    $\Box$