Alternate proofs for the $n$-dimensional resolution theorems

TL;DR

This paper introduces simplified, unified proofs for key resolution theorems in topology, using innovative techniques to convert cohomology problems into homology problems and providing a general method for constructing necessary maps.

Contribution

It offers new, streamlined proofs for resolution theorems, enhancing understanding and applicability in algebraic topology.

Findings

Simplified proofs for cell-like, $ ext{Z}/p$, and $ ext{Q}$-resolution theorems.

A general topological method for constructing resolution maps.

Conversion of cohomology problems into homology problems.

Abstract

We present new, unified proofs for the cell-like, $\mathbb{Z}/p$-, and $\mathbb{Q}$-resolution theorems. Our arguments employ extensions that are much simpler then those used by our predecessors. The techniques allow us to solve problems involving cohomology groups by converting them into problems about homology groups. We provide a coordinated general topological method for constructing the maps needed to witness the resolution theorems simultaneously.

Equations

$${\check{H}}^k(D;G)=\mathrm{dirlim}(H^k(D_j;G),H^k(g_j^{j+1};G)),$$

$$\mathrm{Ext}(H_{k-1}(D_s;Z),G)=0.$$

$$\lambda :\mathrm{Hom}(H_k(D_j;Z),G)\to \mathrm{Hom}(H_k(D_i;Z),G)$$

$$\mathrm{Hom}(H_k(D_j;Z),G)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$H_n(D_j;Z)\otimes Z/p\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$H_n(g_j^i;Z)(e_q)=({\alpha }_1\cdot {\overline{e}}_1,\dots ,{\alpha }_m\cdot {\overline{e}}_m)\in H_n(D_j;Z),$$

$$\mathrm{Hom}(H_n(D_j;Z),Z/p)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$F_{L,K,{\Sigma }^n}=|L|\cup \bigcup \{K_{\sigma }|(\sigma \in L)\wedge (\dim \sigma =n+1)\}.$$

$$F_{L,K,{\Sigma }^n}^{\ast }=|L^{(n)}|\cup \bigcup \{K_{\sigma }^{(n+1)}|(\sigma \in L)\wedge (\dim \sigma =n+1)\}.$$

$$F_{L,K,{\Sigma }^n}^{\ast }\subset F_{L,K,{\Sigma }^n}^{(n+1)}.\text{qed}$$

$$K_{\sigma ,+}=K_{\sigma }\cup |N(\partial \sigma ,S)|.$$

$$f|f^{-1}(|N(\partial \sigma ,S)|):f^{-1}(|N(\partial \sigma ,S)|)\to |N(\partial \sigma ,S)|\subset K_{\sigma ,+}$$

$$h=h_0\cup \bigcup \{h_{\sigma }|(\sigma \in L)\wedge (\dim \sigma =n+1)\}$$

$$h_K:(|N^{(n+1)}|,({\overline{p}}_1{)}^{-1}(|L^{(n)}|)\cap |N^{(n+1)}|)\to (F_{L,K}^{\ast },|L^{(n)}|).$$

$$h_K^{\ast }:(|N^{(n+1)}|,({\overline{p}}_1{)}^{-1}(|L^{(n)}|)\cap |N^{(n+1)}|)\to (F_{L,K}^{\ast },|L^{(n)}|)$$

$$({†}_2)p_j^k\circ p_{k,\infty }=p_{j,\infty }:I^{\infty }\to I^j.$$

$$({†}_3)p_i^j\circ p_j^k=p_i^k:I^k\to I^i.$$

$$({†}_4)\rho (p_j^k(y),y)<\frac{1}{2^j}.$$

$$({†}_5)\rho (p_{j,\infty }(x),x)\le \frac{1}{2^j}.$$

$$K\in \{K(Z,n),K(Z/p,n),K(Q,n)\})$$

$$g_j^{j+1}=g_{j,K}\circ {\phi }_{j+1}|T_{j+1}:T_{j+1}\to T_j.$$

Full text

1. Introduction

For each abelian group $G$ and compact metrizable space $X$, $\dim_{G}X$ will denote the cohomological dimension of $X$ modulo $G$. One can find definitions of and facts about the theory of cohomological dimension in [Dr] and [Ku]. However, the material in Section 6 will be sufficient for our needs. Throughout this paper, map means continuous function, and $\operatorname{{\mathbb{N}}}$ starts with 1.

We are going to provide alternate, simpler, and unified proofs of the cell-like-, $\operatorname{{\mathbb{Z}}}/p$-, and $\operatorname{{\mathbb{Q}}}$-resolution theorems in dimension $n$ ([Ed], [Wa], [Dr], [Le]):

Theorem 1.1**.**

Let $n\in\operatorname{{\mathbb{N}}}$ and $X$ be a nonempty metrizable compactum with $\dim_{\operatorname{{\mathbb{Z}}}}X\leq n$. Then there exists a metrizable compactum $Z$ with $\dim Z\leq n$ and a surjective111It should be remarked that cell-like sets are never empty–see Definition 2.1 and the remark just before it; so the term surjective here is redundant. cell-like map $\pi:Z\to X$.

Theorem 1.2**.**

Let $n\in\operatorname{{\mathbb{N}}}$, $p\in\operatorname{{\mathbb{N}}}_{\geq 2}$, and $X$ be a nonempty metrizable compactum with $\dim_{\operatorname{{\mathbb{Z}}}/p}X\leq n$. Then there exists a metrizable compactum $Z$ with $\dim Z\leq n$ and a surjective $\operatorname{{\mathbb{Z}}}/p$-acyclic map $\pi:Z\to X$.

Theorem 1.3**.**

Let $n\in\operatorname{{\mathbb{N}}}_{\geq 2}$ and $X$ be a nonempty metrizable compactum with $\dim_{\operatorname{{\mathbb{Q}}}}X\leq n$. Then there exists a metrizable compactum $Z$ with $\dim Z\leq n$ and a surjective $\operatorname{{\mathbb{Q}}}$-acyclic map $\pi:Z\to X$.

All terms with which the reader is not familiar will be defined in Section 2. Theorem 1.1 was stated in [Ed] and proved in [Wa]. Theorems 1.2 and 1.3 along with proofs can be found in [Dr] and [Le] respectively. All three of the resolution theorems are true in case $n=0$. This holds because for any abelian group $G$ and nonempty metrizable compactum $X$, $\dim_{G}X=0$ if and only if $\dim X=0$, and all singleton spaces are both cell-like and $G$-acyclic (see Section 2). So one may take $Z=X$ and $\pi=\operatorname{id}_{X}$. Theorem 1.1 with $n=1$ is true for a similar reason because, as stated in [Wa], $\dim_{\operatorname{{\mathbb{Z}}}}X=1$ if and only if $\dim X=1$. Theorem 1.3 fails to be true for $n=1$ as remarked in [Le].

One feature of all three of the previous proofs of the resolution theorems is that they rely on a version of a result due to R. Edwards (see Theorem 4.2 of [Wa]) even if somewhat hidden inside the arguments. Because of this dependence, in each case the proofs employed relatively complicated “extensions,” e.g., Edwards-Walsh resolutions as in [Wa] and [Dr] (see also [KY]), and even more intricate ones in [Le]. The extensions were associated with triangulated compact polyhedra that appeared in an inverse sequence representing the target space $X$.

Our main results yield unified proofs of Theorems 1.1, 1.2, and 1.3 that do not rely on forms of Edwards’ result. As a consequence, although we will use extensions (Sections 4, 5, 7, and 8), these will be quite mild in comparison with those that were employed by our predecessors. Furthermore, we are going to organize the previously disparate techniques by providing a series of results that involve inverse sequences of compact polyhedra (Section 2), the application of basic notions of algebraic topology that connect homology with cohomology (Sections 2 and 3), the theory of triangulated finite polyhedra (Sections 4 and 5), the connection between the theory of cohomological dimension and Eilenberg-MacLane $\operatorname{CW}$-complexes (Section 6), and point-set topology (Sections 9, 10, and 11). All spaces that require a metric will be embedded in the Hilbert cube $I^{\infty}$ and their metrics will be induced from a fixed metric $\rho$ on $I^{\infty}$ (Section 9). Section 12 contains the final steps of our proof of the resolution theorems.

2. Acyclicity

In this section we will provide the needed definitions of “acyclicity” including the concept of being “cell-like.” We will also have lemmas showing how to detect these properties in the settings that will occur later. The term continuum will refer to a compact, connected Hausdorff space. The notion of a space having the shape of a point can be found on page 45 of [MS]. It can be seen from the remarks in the first paragraph of page 45 of [MS] that whenever a space has the shape of a point, it cannot be empty and must be connected.

Definition 2.1**.**

A compact Hausdorff space is said to be cell-like if it has the shape of a point.

It follows from this that each cell-like space must be a nonempty continuum. In the original definition of the term cell-like, which was introduced by C. Lacher (consult [Ru]), it was required that such a space be metrizable. The current loosening of the requirement of metrizability is for convenience since this is essential for understanding the main result, the cell-like resolution theorem for compact Hausdorff spaces, in [MR]; it has no impact on the work herein. Other equivalent definitions of cell-likenes can be found, for example, in [Dr].

Definition 2.2**.**

Let $G$ be an abelian group and $X$ a continuum. One says that $X$ is $G$-acyclic if for all $k\in\operatorname{{\mathbb{N}}}$, ${\check{H}}^{k}(X;G)=0$.

Note that if one wants to include $k=0$ in Definition 2.2, then Čech cohomology should be reduced. All singleton spaces are cell-like and are $G$-acyclic for any abelian group $G$.

Definition 2.3**.**

Let $\pi:Z\to X$ be a map and $G$ an abelian group. It is said that $\pi$ is:

1. (1)

cell-like* if each of its fibers is cell-like;* 2. (2)

$G$-acyclic if each of its fibers is $G$-acyclic.

Some authors require that in (1) of the preceding, the map $\pi$ be proper. Since all the spaces to which we apply Definition 2.3 will be compact and metrizable, then the maps we produce will be proper anyway. So we will lose nothing by deleting the properness requirement. The maps $\pi$ that we shall design to satisfy the resolution theorems will be surjective by construction.

Lemma 2.4**.**

Let $\mathbf{X}=(X_{j},g_{j}^{j+1})$ be an inverse sequence of nonempty metrizable compacta and $X=\lim\mathbf{X}$. Then $X$ is cell-like if for each $j\in\operatorname{{\mathbb{N}}}$, there exists $i>j$ such that $g_{j}^{i}:X_{i}\to X_{j}$ is null homotopic.∎

Let $G$ be an abelian group. There are several ways to demonstrate that a given metrizable continuum $X$ is $G$-acyclic. The continuity of Čech cohomology shows that:

Lemma 2.5**.**

For any abelian group $G$ and inverse sequence $\mathbf{D}=(D_{j},g_{j}^{j+1})$ of nonempty compact polyhedra with $D=\lim\mathbf{D}$,

[TABLE]

where for each $j$, $H^{k}(g_{j}^{j+1};G):H^{k}(D_{j};G)\to H^{k}(D_{j+1};G)$ is the homomorphism induced by $g_{j}^{j+1}:D_{j+1}\to D_{j}$.∎

Lemma 2.6**.**

Let $\mathbf{D}=(D_{j},g_{j}^{j+1})$ be an inverse sequence of nonempty compact polyhedra, $D=\lim\mathbf{D}$, $k\in\operatorname{{\mathbb{N}}}$, and $G$ an abelian group. Then $\check{H}^{k}(D;G)=0$ if for each $j$, there exists $i>j$ such that $H^{k}(g_{j}^{i};G):H^{k}(D_{j};G)\to H^{k}(D_{i};G)$ is the trivial homomorphism. ∎

When we apply Lemma 2.6, there will be given $n\in\operatorname{{\mathbb{N}}}$, and we will face three cases. The first, that $k>n$, will be taken care of because it will be true in the situations we cover that $\dim D\leq n$, so $\check{H}^{k}(D;G)=0$ automatically. The second will be that $1\leq k<n$, and will be covered in Lemma 2.7. The case that $k=n$ will be managed using Lemma 2.9. We note that for all $1\leq k\leq n$, we are going to make an appeal only to homology groups. We lay the groundwork for this now.

Lemma 2.7**.**

Let $\mathbf{D}=(D_{j},g_{j}^{j+1})$ be an inverse sequence of nonempty compact polyhedra, $D=\lim\mathbf{D}$, and $k\in\operatorname{{\mathbb{N}}}$. Suppose that for all $j\in\operatorname{{\mathbb{N}}}$,

1. (1)

there exists $i>j$ such that $H_{k}(g_{j}^{i};\operatorname{{\mathbb{Z}}}):H_{k}(D_{i};\operatorname{{\mathbb{Z}}})\to H_{k}(D_{j};\operatorname{{\mathbb{Z}}})$ is the trivial homomorphism, and 2. (2)

$H_{k-1}(D_{j};\operatorname{{\mathbb{Z}}})$* is free-abelian.*

Then for any abelian group $G$, $\check{H}^{k}(D;G)=0$.

Proof.

Fix an abelian group $G$. Let $j\in\operatorname{{\mathbb{N}}}$, and choose $i>j$ as in (1) of the hypothesis. Using (2) and an application of Theorem 52.3(b), page 318 of [Mu], one has that for each $s\in\{i,j\}$

[TABLE]

This and the short exact sequence of Theorem 53.1 of [Mu] (page 320, universal coefficient theorem for cohomology) yield that the natural Kronecker homomorphisms (see page 276 of [Mu]) $\kappa_{s}:H^{k}(D_{s};G)\to\mathrm{Hom}(H_{k}(D_{s};\operatorname{{\mathbb{Z}}}),G)$, $s\in\{i,j\}$ are isomorphisms.

Let

[TABLE]

be the homomorphism induced by $H_{k}(g_{j}^{i};\operatorname{{\mathbb{Z}}}):H_{k}(D_{i};\operatorname{{\mathbb{Z}}})\to H_{k}(D_{j};\operatorname{{\mathbb{Z}}})$. Then we have a commutative diagram of homomorphisms,

[TABLE]

Let $u\in H^{k}(D_{j};G)$. We claim that $H^{k}(g_{j}^{i};G)(u)=0$. Since $\kappa_{i}$ is an isomorphism, then the preceding is true if and only if $\kappa_{i}\circ H^{k}(g_{j}^{i};G)(u)=0$. By the diagram, this is true if and only if $\lambda\circ\kappa_{j}(u)=0$. But $\kappa_{j}$ is also an isomorphism, so we only have to show that $\lambda$ is the trivial homomorphism. Let $f\in\mathrm{Hom}(H_{k}(D_{j};\operatorname{{\mathbb{Z}}}),G)$. Then $f:H_{k}(D_{j};\operatorname{{\mathbb{Z}}})\to G$ is a homomorphism, and $\lambda(f)=f\circ H_{k}(g_{j}^{i};\operatorname{{\mathbb{Z}}})$. The second function in this composition is a trivial homomorphism and hence $\lambda(f)$ is the trivial element of $\mathrm{Hom}(H_{k}(D_{i};\operatorname{{\mathbb{Z}}}),G)$. Apply Lemma 2.6 to complete this proof. ∎

Corollary 2.8**.**

Let $\mathbf{D}=(D_{j},g_{j}^{j+1})$ be an inverse sequence of nonempty compact polyhedra, $D=\lim\mathbf{D}$, and $k\in\operatorname{{\mathbb{N}}}$. Suppose that for all $j\in\operatorname{{\mathbb{N}}}$,

1. (1)

there exists $i>j$ such that $H_{k}(g_{j}^{i};\operatorname{{\mathbb{Z}}}):H_{k}(D_{i};\operatorname{{\mathbb{Z}}})\to H_{k}(D_{j};\operatorname{{\mathbb{Z}}})$ is the trivial homomorphism, and 2. (2)

there exist infinitely many $i$ such that $H_{k-1}(D_{i};\operatorname{{\mathbb{Z}}})$ is free-abelian.

Then for any abelian group $G$, $\check{H}^{k}(D;G)=0$.

Lemma 2.9**.**

Let $\mathbf{D}=(D_{j},g_{j}^{j+1})$ be an inverse sequence of nonempty compact polyhedra, $D=\lim\mathbf{D}$ and $n\in\operatorname{{\mathbb{N}}}$. Suppose that for all $j\in\operatorname{{\mathbb{N}}}$,

1. (1)

there exists $i>j$ such that $H_{n}(g_{j}^{i};\operatorname{{\mathbb{Z}}}/p):H_{n}(D_{i};\operatorname{{\mathbb{Z}}}/p)\to H_{n}(D_{j};\operatorname{{\mathbb{Z}}}/p)$ is the trivial homomorphism, 2. (2)

$H_{n-1}(D_{j};\operatorname{{\mathbb{Z}}})$* is free-abelian, and* 3. (3)

$H_{n}(D_{j};\operatorname{{\mathbb{Z}}})$* is free-abelian of finite rank.*

Then $\check{H}^{n}(D;\operatorname{{\mathbb{Z}}}/p)=0$.

Proof.

Let $j\in\operatorname{{\mathbb{N}}}$, and choose $i>j$ as in (1) of the hypothesis. Using (2) and an application of Theorem 54.4(c), page 328 of [Mu], one sees that the torsion product of $H_{n-1}(D_{s},\operatorname{{\mathbb{Z}}})$ with $\operatorname{{\mathbb{Z}}}/p$ equals [math] for $s\in\{i,j\}$. Hence according to Theorem 55.1 of [Mu] (universal coefficient theorem for homology), there are natural isomorphisms $\lambda_{s}:H_{n}(D_{s};\operatorname{{\mathbb{Z}}})\otimes\operatorname{{\mathbb{Z}}}/p\to H_{n}(D_{s};Z/p)$, $s\in\{i,j\}$.

[TABLE]

Since $H_{n}(g_{j}^{i};\operatorname{{\mathbb{Z}}}/p):H_{n}(D_{i};\operatorname{{\mathbb{Z}}}/p)\to H_{n}(D_{j};\operatorname{{\mathbb{Z}}}/p)$ is the trivial homomorphism by (1), it follows that $H_{n}(g_{j}^{i};\operatorname{{\mathbb{Z}}})\otimes id_{\operatorname{{\mathbb{Z}}}/p}$ from the diagram above is also the trivial homomorphism. Let us see what this means for the homomorphism $H_{n}(g_{j}^{i};\operatorname{{\mathbb{Z}}}):H_{n}(D_{i};\operatorname{{\mathbb{Z}}})\to H_{n}(D_{j};\operatorname{{\mathbb{Z}}})$. According to (3), the groups $H_{n}(D_{i};\operatorname{{\mathbb{Z}}})$ and $H_{n}(D_{j};\operatorname{{\mathbb{Z}}})$ are both (isomorphic to) finite direct sums of copies of $\operatorname{{\mathbb{Z}}}$, say $H_{n}(D_{i};\operatorname{{\mathbb{Z}}})\cong\oplus_{1}^{r}\operatorname{{\mathbb{Z}}}$ and $H_{n}(D_{j};\operatorname{{\mathbb{Z}}})\cong\oplus_{1}^{m}\operatorname{{\mathbb{Z}}}$, for $r,m\in\operatorname{{\mathbb{N}}}$. Let us take $e_{1},\ldots,e_{r}$, and $\overline{e}_{1},\ldots,\overline{e}_{m}$ to be generators of $H_{n}(D_{i};\operatorname{{\mathbb{Z}}})$ and $H_{n}(D_{j};\operatorname{{\mathbb{Z}}})$, respectively. Then, for each $q\in\{1,\ldots,r\}$, we have

[TABLE]

and since $H_{n}(g_{j}^{i};\operatorname{{\mathbb{Z}}})\otimes id_{\operatorname{{\mathbb{Z}}}/p}$ is trivial, this implies that each of $\alpha_{j}$ is an integer divisible by $p$.

Now the commutative diagram of the proof of Lemma 2.7 is valid for $G=\operatorname{{\mathbb{Z}}}/p$, $k=n$, and the homomorphisms $\kappa_{s}$, $s\in\{i,j\}$ are isomorphisms just as there because that only relies on (2).

[TABLE]

Hence we may perform the steps analogous to those in the proof of Lemma 2.7, as follows.

Let $u\in H^{n}(D_{j};\operatorname{{\mathbb{Z}}}/p)$. We claim that $H^{n}(g_{j}^{i};\operatorname{{\mathbb{Z}}}/p)(u)=0$. To show this, as in the proof of Lemma 2.7 we only need to show that $\lambda$ is the trivial homomorphism. Let $f\in\mathrm{Hom}(H_{n}(D_{j};\operatorname{{\mathbb{Z}}}),\operatorname{{\mathbb{Z}}}/p)$. Then $f:H_{n}(D_{j};\operatorname{{\mathbb{Z}}})\to\operatorname{{\mathbb{Z}}}/p$ is a homomorphism, and $\lambda(f)=f\circ H_{n}(g_{j}^{i};\operatorname{{\mathbb{Z}}})$. But the properties of $H_{n}(g_{j}^{i};\operatorname{{\mathbb{Z}}})$ guarantee that, when followed by a homomorphism landing in $\operatorname{{\mathbb{Z}}}/p$, this yields the trivial homomorphism. Hence $\lambda(f)$ is the trivial element of $\mathrm{Hom}(H_{n}(D_{i};\operatorname{{\mathbb{Z}}}),\operatorname{{\mathbb{Z}}}/p)$. Apply Lemma 2.6 to complete this proof. ∎

Corollary 2.10**.**

Let $\mathbf{D}=(D_{j},g_{j}^{j+1})$ be an inverse sequence of nonempty compact polyhedra, $D=\lim\mathbf{D}$, and $n\in\operatorname{{\mathbb{N}}}$. Suppose that for all $j\in\operatorname{{\mathbb{N}}}$,

1. (1)

there exists $i>j$ such that $H_{n}(g_{j}^{i};\operatorname{{\mathbb{Z}}}/p):H_{n}(D_{i};\operatorname{{\mathbb{Z}}}/p)\to H_{n}(D_{j};\operatorname{{\mathbb{Z}}}/p)$ is the trivial homomorphism, and there is an infinite subset $J\subset\operatorname{{\mathbb{N}}}$ such that for all $j\in J$, 2. (2)

$H_{n-1}(D_{j};\operatorname{{\mathbb{Z}}})$* is free-abelian, and* 3. (3)

$H_{n}(D_{j};\operatorname{{\mathbb{Z}}})$* is free-abelian of finite rank.*

Then $\check{H}^{n}(D;\operatorname{{\mathbb{Z}}}/p)=0$.

In [Na] (see also Theorem 30.1 of [IR]), it was proved that $\dim\leq n$ is preserved in the limit of an inverse sequence of metrizable spaces. Let us collect this fact and some others that we need about limits of inverse sequences of metrizable compacta.

Theorem 2.11**.**

Let $\mathbf{T}=(T_{j},g_{j}^{j+1})$ be an inverse sequence of metrizable compacta and $Z=\lim\mathbf{T}$. Then,

1. (1)

$Z$* is a metrizable compactum,* 2. (2)

if for all $j$, $T_{j}$ is connected, $Z$ is connected, 3. (3)

if for all $j$, $T_{j}\neq\emptyset$, $Z\neq\emptyset$, and 4. (4)

if $n\geq 0$ and for all $j$, $\dim T_{j}\leq n$, $\dim Z\leq n$.

3. Simplicial Complexes, Vertex Stars

For each finite simplicial complex $S$, we are going to write $|S|$ to denote its polyhedron with the weak topology which in this case is compact and metrizable. We shall be employing subdivisions and simplicial approximations, so let us note a couple of important facts that can be found in Chapter 3 of [Sp].

Lemma 3.1**.**

Let $S$ be a finite simplicial complex. Then,

1. (1)

if $R$ is a subdivision of $S$, then there is a simplicial approximation $\varphi:|R|\to|S|$ of $\operatorname{id}_{|S|}$, and 2. (2)

if $P$ is a compact polyhedron and $f:P\to|S|$ is a map, then there is a triangulation $T$ of $P$ such that for every subdivision $L$ of $T$, there is a simplicial approximation $g:|L|\to|S|$ of the map $f$.

Lemma 3.2**.**

Let $S$ be a finite simplicial complex such that $|S|$ is contractible. Then for each $k\in\operatorname{{\mathbb{N}}}$, the inclusion $|S^{(k)}|\hookrightarrow|S^{(k+1)}|$ is homotopic in $|S^{(k+1)}|$ to a constant map.

Proof.

Each contracting homotopy of $|S^{(k)}|\hookrightarrow|S|$ can be replaced by a $\operatorname{PL}$-homotopy $H:|S^{(k)}|\times[0,1]\to|S^{(k+1)}|$ where $H_{0}$ equals the inclusion $|S^{(k)}|\hookrightarrow|S^{(k+1)}|$ and $H_{1}$ is a constant map. ∎

Next is an elementary fact from homology theory.

Lemma 3.3**.**

Let $S$ be a finite simplicial complex, $k\in\operatorname{{\mathbb{N}}}$, and $G$ an abelian group. Then the inclusion map $|S^{(k+1)}|\hookrightarrow|S|$ induces an isomorphism of $H_{k}(|S^{(k+1)}|;G)$ onto $H_{k}(|S|;G)$.∎

Lemma 3.4**.**

Let $T$ be a finite simplicial complex and $S$ a subcomplex of $T$ such that $|S|$ is contractible. Then for each $n\in\operatorname{{\mathbb{N}}}$, $1\leq k<n$, and abelian group $G$, $H_{k}(|S^{(n)}|;G)=0$.

Proof.

Since $|S|$ is contractible, then $H_{k}(|S|;G)=0$. By Lemma 3.3, $k\in\operatorname{{\mathbb{N}}}$ implies that the inclusion $|S^{(k+1)}|\hookrightarrow|S|$ induces an isomorphism of $H_{k}(|S^{(k+1)}|;G)$ onto $H_{k}(|S|;G)$. This shows that $H_{k}(|S^{(k+1)}|;G)=0$. On the other hand, since $1\leq k<n$, Lemma 3.3 shows that the inclusion $|S^{(k+1)}|\hookrightarrow|S^{(n)}|$ induces an isomorphism of $H_{k}(|S^{(k+1)}|;G)$ onto $H_{k}(|S^{(n)}|;G)$, so the latter equals [math]. ∎

Let $T$ be a finite simplicial complex and $v\in T^{(0)}$. Then $\operatorname{st}(v,T)$ will denote the open star of $v$ in $T$ and $\overline{\operatorname{st}}(v,T)$ will denote the closed star of $v$ in $T$. Of course, $\operatorname{st}(v,T)$ is an open neighborhood of $v$ in the compact polyhedron $|T|$, $\overline{\operatorname{st}}(v,T)$ is a closed subset of the compact polyhedron $|T|$, and $\operatorname{st}(v,T)\subset\overline{\operatorname{st}}(v,T)$. Moreover, each of $\operatorname{st}(v,T)$ and $\overline{\operatorname{st}}(v,T)$ is contractible, and there is a unique subcomplex $S$ of $T$ such that $|S|=\overline{\operatorname{st}}(v,T)$. Lemma 3.4 leads to the next fact about closed stars.

Corollary 3.5**.**

Let $T$ be a finite simplicial complex, $v\in T^{(0)}$, and $S$ the unique subcomplex of $T$ that triangulates $\overline{\operatorname{st}}(v,T)$. Suppose that $T_{0}$ is a subdivision of $T$. Then,

1. (1)

there is a unique subcomplex $S_{0}$ of $T_{0}$ with $|S_{0}|=|S|$, and 2. (2)

for each $n\in\operatorname{{\mathbb{N}}}$, $1\leq k<n$, and abelian group $G$, $H_{k}(|S_{0}^{(n)}|;G)=0$.

As usual, when $G$ is an abelian group, $n\geq 0$, and $T$ is a simplicial complex, then $Z_{n}(T;G)$ will denote the set of (simplicial) $G$-cycles among the $G$-chains.222Some orientation on $T$ is assumed.

Definition 3.6**.**

Whenever $\sigma$ is an $(n+1)$-simplex of a simplicial complex $T$, then by $\partial\sigma$ we shall mean the $n$-dimensional simplicial $\operatorname{{\mathbb{Z}}}$-cycle of $T^{(n)}$, i.e., $\partial\sigma\in Z_{n}(T^{(n)};\operatorname{{\mathbb{Z}}})$.333Later we are going to use $\partial\sigma$ to denote the combinatorial boundary of $\sigma$ (its boundary as a manifold); the reader will have no difficulty distinguishing between these two uses of the same notation. Subsequently, however, when we deal with homomorphisms induced by maps, we shall also treat $\partial\sigma$ as a singular $\operatorname{{\mathbb{Z}}}$-cycle.

Lemma 3.7**.**

Let $E$ be a metrizable compactum, $p\in\operatorname{{\mathbb{N}}}_{\geq 2}$, $n\in\operatorname{{\mathbb{N}}}$, and $g\in H_{n}(E;\operatorname{{\mathbb{Z}}}/p)$. Then for each $r\in\operatorname{{\mathbb{N}}}$ having the property that $p$ divides $r$, we have $r\cdot g=0\in H_{n}(E;\operatorname{{\mathbb{Z}}}/p)$.∎

Definition 3.8**.**

Let $T$ be a finite simplicial complex, $v$ a vertex of $T$, and $n\in\operatorname{{\mathbb{N}}}$. Denote by $S$ the subcomplex of $T$ that triangulates $\overline{\operatorname{st}}(v,T)$, by $\mathcal{E}_{v,T}$ the set of $(n+1)$-simplexes of $T$ having $v$ as a vertex, and by $m_{v,T}$ the cardinality of $\mathcal{E}_{v,T}$.

Lemma 3.9**.**

Taking the notation from $\mathrm{Definition\,\,\ref{starstruck2}}$, one has that the group $H_{n}(|S^{(n)}|;\operatorname{{\mathbb{Z}}})$ is free abelian of rank $m_{v,T}$. Moreover, if $G$ is an abelian group, $m_{v,T}\geq 1$, and we list $\mathcal{E}_{v,T}=\{\sigma_{i}\,|\,1\leq i\leq m_{v,T}\}$, then:

1. (1)

for each $g\in G$ and $1\leq i\leq m_{v,T}$, $g\cdot\partial\sigma_{i}$ lies in $Z_{n}(|S^{(n)}|;G)$, 2. (2)

for each $z\in Z_{n}(|S^{(n)}|;G)$, there exists a set $\{g_{i}\,|\,1\leq i\leq m_{v,T}\}\subset G$ such that $z=\sum_{i=1}^{m_{v,T}}g_{i}\cdot\partial\sigma_{i}$, 3. (3)

if $p\in\operatorname{{\mathbb{N}}}_{\geq 2}$, $h\in Z_{n}(|S^{(n)}|;\operatorname{{\mathbb{Z}}}/p)$, and $r\in\operatorname{{\mathbb{Z}}}$ is a multiple of $p$, then $r\cdot h$ is an $n$-dimensional boundary $\operatorname{{\mathbb{Z}}}/p$-cycle, i.e., it is homologous to [math] in $Z_{n}(|S^{(n)}|;\operatorname{{\mathbb{Z}}}/p)$, and 4. (4)

if $p\in\operatorname{{\mathbb{N}}}_{\geq 2}$, $h\in H_{n}(|S^{(n)}|;\operatorname{{\mathbb{Z}}}/p)$, and $r\in\operatorname{{\mathbb{N}}}$ is a multiple of $p$, then $r\cdot h=0$. ∎

Lemma 3.10**.**

Let $T$ and $S$ be finite simplicial complexes, $n\in\operatorname{{\mathbb{N}}}$, and $f:|T|\to|S|$ a map. Suppose that $\sigma$ is an $(n+1)$-simplex of $T$.

1. (1)

If $\rho\in S^{(n)}$ and $f(\partial\sigma)\subset\rho$, then $f:\partial\sigma\to|S|$ is homotopic to a constant map, so for any abelian group $G$ and $g\in G$, $Z_{n}(f;G)(g\cdot\partial\sigma)$ is homologous to [math] in $Z_{n}(|S|;G)$. 2. (2)

Suppose there is an $(n+1)$-simplex $\rho\in S$ such that $f(\partial\sigma)\subset\partial\rho$. If $p\in\operatorname{{\mathbb{N}}}_{\geq 2}$ and $f:\partial\sigma\to\partial\rho$ has degree a multiple of $p$, then for any $g\in\operatorname{{\mathbb{Z}}}/p$, $Z_{n}(f;\operatorname{{\mathbb{Z}}}/p)(g\cdot\partial\sigma)$ is homologous to [math] in $Z_{n}(|S|;\operatorname{{\mathbb{Z}}}/p)$. If $f:\partial\sigma\to\partial\rho$ has degree [math], i.e., is null homotopic, then for any abelian group $G$ and $g\in G$, $Z_{n}(f;G)(g\cdot\partial\sigma)$ is homologous to [math] in $Z_{n}(|S|;G)$. ∎

4. Fundamental Extensions

Definition 4.1 provides the blueprint for the technique we are going to use to form the extensions that we shall need in our proof of the resolution theorems.

Definition 4.1**.**

Let $n\in\operatorname{{\mathbb{N}}}$ and $(K,\Sigma^{n})$ be a pair such that $K$ is a $\operatorname{CW}$-complex and $\Sigma^{n}\subset K$ is an embedded copy of $S^{n}$. For each finite simplicial complex $L$ and each $\sigma\in L$ with $\dim\sigma=n+1$, we attach $K$ to $|L|$ along $\partial\sigma$ via a homeomorphism $h_{\sigma}$ of $\Sigma^{n}\subset K$ to $\partial\sigma$, and we denote the attached copy $K_{\sigma}$. The $\operatorname{CW}$-complex so obtained will be denoted $F_{L,K,\Sigma^{n}}$, that is,

[TABLE]

It is of course understood that for each such $\sigma$, $K_{\sigma}\cap|L|=\partial\sigma$ and that if $\tau\in L$ with $\dim\tau=n+1$, then $K_{\sigma}\cap K_{\tau}=\partial\sigma\cap\partial\tau$.

For each $\operatorname{CW}$-complex $K$ and $n\geq 0$, $K^{(n)}$ will denote the $n$-skeleton of $K$.

Lemma 4.2**.**

For each $(n,L,K,\Sigma^{n})$ as in $\mathrm{Definition\,\ref{allstickons}}$, it is true that $F_{L,K,\Sigma^{n}}^{(n+1)}=|L^{(n+1)}|\cup\bigcup\{K_{\sigma}^{(n+1)}\,|\,(\sigma\in L)\wedge(\dim\sigma=n+1)\}$.∎

Definition 4.3**.**

For each $(n,L,K,\Sigma^{n})$ as in $\mathrm{Definition\,\ref{allstickons}}$, denote,

[TABLE]

Lemma 4.4**.**

For each $(n,L,K,\Sigma^{n})$ as in $\mathrm{Definition\,\ref{allstickons}}$,

[TABLE]

For each $(n,L,K,\Sigma^{n})$ as in Definition 4.1, there exists a retraction $r:F_{L,K,\Sigma^{n}}\to|L|$ such that $r(K_{\sigma}\setminus\partial\sigma)\subset\operatorname{int}\sigma$ whenever $\sigma\in L$ and $\dim\sigma=n+1$. Let us make this formal.

Definition 4.5**.**

For each $(n,L,K,\Sigma^{n})$ as in $\mathrm{Definition\,\ref{allstickons}}$, fix a retraction $r_{L,K,\Sigma^{n}}:F_{L,K,\Sigma^{n}}\to|L|$ such that $r_{L,K,\Sigma^{n}}(K_{\sigma}\setminus\partial\sigma)\subset\operatorname{int}\sigma$ whenever $\sigma\in L$ and $\dim\sigma=n+1$.

Lemma 4.6**.**

For each $(n,L,K,\Sigma^{n})$ as in $\mathrm{Definition\,\ref{allstickons}}$, the map $r_{L,K,\Sigma^{n}}:F_{L,K,\Sigma^{n}}\to|L|$ has the following properties:

1. (1)

$r_{L,K,\Sigma^{n}}(F_{L,K,\Sigma^{n}}^{*})\subset|L^{(n+1)}|$, 2. (2)

$r_{L,K,\Sigma^{n}}(x)=x$* for all $x\in|L^{(n)}|$,* 3. (3)

if $\sigma\in L$ with $\dim\sigma=n+1$, then $r_{L,K,\Sigma^{n}}(K_{\sigma})\subset\sigma$, and $r_{L,K,\Sigma^{n}}(K_{\sigma}\setminus\partial\sigma)\subset\operatorname{int}\sigma$, and 4. (4)

the restriction of $r_{L,K,\Sigma^{n}}$ to $F_{L,K,\Sigma^{n}}^{(n+1)}$ is a retraction of $F_{L,K,\Sigma^{n}}^{(n+1)}$ to its subspace $|L^{(n+1)}|$.∎

Let $n\in\operatorname{{\mathbb{N}}}$ and $L$ be a finite simplicial complex. There exists a subdivision $S$ of $L$ having the property that for each $\sigma\in L$ with $\dim\sigma=n+1$, $|N(\partial\sigma,S)|$ is a regular neighborhood of $\partial\sigma$, where of course $N(\partial\sigma,S)$ denotes the simplicial neighborhood of $\partial\sigma$ in $S$. For example, we may choose $S$ to be the second barycentric subdivision of $L$.

Lemma 4.7**.**

Let $n\in\operatorname{{\mathbb{N}}}$, $L$ a finite simplicial complex, and $S$ a subdivision of $L$ having the property that for each $\sigma\in L$ with $\dim\sigma=n+1$, $|N(\partial\sigma,S)|$ is a regular neighborhood of $\partial\sigma$. Then for each $\sigma\in L$ with $\dim\sigma=n+1$,

1. (1)

$\partial\sigma\subset|N(\partial\sigma,S)|$, 2. (2)

$\partial\sigma$* is a strong deformation retract of $|N(\partial\sigma,S)|$, and* 3. (3)

$|N(\partial\sigma,S)|$* is a compact $\operatorname{ANR}$.∎*

Definition 4.8**.**

Let $n\in\operatorname{{\mathbb{N}}}$ and $L$ be a finite simplicial complex. Let $S$ be a subdivision of $L$ having the property that for each $\sigma\in L$ with $\dim\sigma=n+1$, $|N(\partial\sigma,S)|$ is a regular neighborhood of $\partial\sigma$. We shall refer to such an $S$ as an $n$-regular subdivision of $L$.

Definition 4.9**.**

Let $(n,L,K,\Sigma^{n})$ be as in $\mathrm{Definition\,\ref{allstickons}}$. Suppose that $S$ is an $n$-regular subdivision of $L$. For each $\sigma\in L$ with $\dim\sigma=n+1$, let

[TABLE]

Taking into account Lemma 4.7(2,3) and Definition 4.9, one obtains the next fact.

Lemma 4.10**.**

Let $(n,L,K,\Sigma^{n})$ be as in $\mathrm{Definition\,\ref{allstickons}}$. Suppose that $S$ is an $n$-regular subdivision of $L$. Then for each $\sigma\in L$ with $\dim\sigma=n+1$,

1. (1)

$K_{\sigma,+}$* is an absolute neighborhood extensor for compact metrizable spaces,* 2. (2)

$K_{\sigma}$* is a strong deformation retract of $K_{\sigma,+}$, and* 3. (3)

whenever $Y$ is a metrizable compactum with $Y\tau K$, then $Y\tau K_{\sigma,+}$.∎

5. Lemma for an Extension

Lemma 5.1 is critical for our proof of Proposition 8.1. It is one of the key ingredients in our technique for avoiding complicated extensions, that is, to staying away from Edwards type lemmas as we mentioned in the Introduction.

Lemma 5.1**.**

Let $(n,L,K,\Sigma^{n})$ be as in $\mathrm{Definition\,\ref{allstickons}}$. Suppose that $E$ is a compact polyhedron, $f:E\to|L|$ a map, $S$ an $n$-regular subdivision of $L$, and that for each $\sigma\in L$ with $\dim\sigma=n+1$, the map $f|f^{-1}(|N(\partial\sigma,S)|):f^{-1}(|N(\partial\sigma,S)|)\to|N(\partial\sigma,S)|\subset K_{\sigma,+}$ extends to a map of $E$ to $K_{\sigma,+}$. Let $N$ be a triangulation of $E$ that admits a simplicial approximation $f_{1}:|N|\to|L|$ of the map $f$. Then there exists a map $h:|N^{(n+1)}|\to F_{L,K,\Sigma^{n}}^{*}$ $(\mathrm{Definition\,\ref{corepart}})$ such that,

1. (1)

$h(x)=f_{1}(x)$* for all $x\in f_{1}^{-1}(|L^{(n)}|)\cap|N^{(n+1)}|$, and* 2. (2)

for each $\sigma\in L$ with $\dim\sigma=n+1$, $h(f_{1}^{-1}(\sigma)\cap|N^{(n+1)}|)\subset K_{\sigma}$.

Proof.

Let $M$ be a subdivision of $N$ that admits a map $f_{2}:|M|\to|S|$ such that $f_{2}$ is simultaneously a simplicial approximation of $f$ and $f_{1}$. Temporarily fix $\sigma\in L$ with $\dim\sigma=n+1$. We claim that:

$({\dagger}_{1})$ $f_{2}(f_{1}^{-1}(\partial\sigma))\subset\partial\sigma$,

$({\dagger}_{2})$ $f_{1}^{-1}(\partial\sigma)\subset f_{2}^{-1}(\partial\sigma)$, and

$({\dagger}_{3})$ $f_{2}|f_{1}^{-1}(\partial\sigma):f_{1}^{-1}(\partial\sigma)\to\partial\sigma$ is homotopic to $f_{1}|f_{1}^{-1}(\partial\sigma):f_{1}^{-1}(\partial\sigma)\to\partial\sigma$.

Plainly $({\dagger}_{2})$ follows from $({\dagger}_{1})$; let us prove $({\dagger}_{1})$. Let $x\in f_{1}^{-1}(\partial\sigma)$; then $f_{1}(x)\in\partial\sigma$ which implies that there exists $\tau\in S$ such that $f_{1}(x)\in\operatorname{int}\tau\subset\tau\subset\partial\sigma$. Since $f_{2}:|M|\to|S|$ is a simplicial approximation of $f_{1}$, then $f_{2}(x)\in\tau\subset\partial\sigma$ as needed for $({\dagger}_{1})$. Since $\{f_{1}(x),f_{2}(x)\}\subset\tau\subset\partial\sigma$, then $f_{1}$ and $f_{2}$ are “straight-line” homotopic as maps to $\partial\sigma$, so $({\dagger}_{3})$ is also established.

Now let us prove some parallel facts:

$({\dagger}_{4})$ $f(f_{2}^{-1}(\partial\sigma))\subset|N(\partial\sigma,S)|$,

$({\dagger}_{5})$ $f_{2}^{-1}(\partial\sigma)\subset f^{-1}(|N(\partial\sigma,S)|$, and

$({\dagger}_{6})$ $f|f_{2}^{-1}(\partial\sigma):f_{2}^{-1}(\partial\sigma)\to|N(\partial\sigma,S)|$ is homotopic to $f_{2}|f_{2}^{-1}(\partial\sigma):f_{2}^{-1}(\partial\sigma)\to\partial\sigma\subset|N(\partial\sigma,S)|$.

Plainly $({\dagger}_{5})$ follows from $({\dagger}_{4})$. We proceed to prove $({\dagger}_{4})$. Let $x\in f_{2}^{-1}(\partial\sigma)$; there exists $\tau\in S$ with $f(x)\in\operatorname{int}\tau$. Since $f_{2}:|M|\to|S|$ is a simplicial approximation of $f$ and $\tau\in S$, then $f_{2}(x)\in\tau$. But $f_{2}(x)\in\partial\sigma\cap\tau$ which implies that $\tau\subset|N(\partial\sigma,S)|$. Hence $f(x)\in\operatorname{int}\tau\subset\tau\subset|N(\partial\sigma,S)|$, which gives us $({\dagger}_{4})$. Since $\{f(x),f_{2}(x)\}\subset\tau\subset|N(\partial\sigma,S)|$, then $f$ and $f_{2}$ are “straight-line” homotopic as maps to $|N(\partial\sigma,S)|$, so $({\dagger}_{6})$ is also established.

By hypothesis, the map

[TABLE]

extends to a map of $E$ to $K_{\sigma,+}$. This, alongside $({\dagger}_{5})$ and $({\dagger}_{6})$, shows that,

$({\dagger}_{7})$ $f_{2}|f_{2}^{-1}(\partial\sigma):f_{2}^{-1}(\partial\sigma)\to\partial\sigma\subset|N(\partial\sigma,S)|$ extends to a map of $E$ to $K_{\sigma,+}$.

Taking into consideration $({\dagger}_{7})$, $({\dagger}_{2})$, and $({\dagger}_{3})$, one concludes that:

$({\dagger}_{8})$ $f_{1}|f_{1}^{-1}(\partial\sigma):f_{1}^{-1}(\partial\sigma)\to\partial\sigma\subset|N(\partial\sigma,S)|$ extends to a map of $E$ to $K_{\sigma,+}$.

Making use of $({\dagger}_{8})$ and the strong deformation retraction guaranteed us by Lemma 4.7(2), we have,

$({\dagger}_{9})$ $f_{1}|f_{1}^{-1}(\partial\sigma):f_{1}^{-1}(\partial\sigma)\to\partial\sigma\subset|N(\partial\sigma,S)|$ extends to a map $f_{1,\sigma}:E\to K_{\sigma}$.

Having accomplished the preceding for each fixed $\sigma$, now notice that $|L^{(n+1)}|=|L^{(n)}|\cup\bigcup\{\sigma\,|\,(\sigma\in L)\wedge(\dim\sigma=n+1)\}$. Of course, $f_{1}(|N^{(n+1)}|)\subset|L^{(n+1)}|$. Hence,

$({\dagger}_{10})$ |N^{(n+1)}|=\big{(}f_{1}^{-1}(|L^{(n)}|)\cup\bigcup\{f_{1}^{-1}(\sigma)\,|\,(\sigma\in L)\wedge(\dim\sigma=n+1)\}\big{)}\cap|N^{(n+1)}|.

There exists a subcomplex $N_{0}$ of $N^{(n+1)}$ such that $|N_{0}|=f_{1}^{-1}(|L^{(n)}|)\cap|N^{(n+1)}|$. Moreover, for each $\sigma\in L$ with $\dim\sigma=n+1$, there is also a subcomplex $N_{\sigma}$ of $N^{(n+1)}$ such that $|N_{\sigma}|=f_{1}^{-1}(\sigma)\cap|N^{(n+1)}|$. Using this and $({\dagger}_{10})$ we see that,

$({\dagger}_{11})$ $|N^{(n+1)}|=|N_{0}|\cup\bigcup\{|N_{\sigma}|\,|\,(\sigma\in L)\wedge(\dim\sigma=n+1)\}$.

Taking into account $({\dagger}_{9})$, for each $\sigma\in L$ with $\dim\sigma=n+1$, we put

$({\dagger}_{12})$ h_{\sigma}=f_{1,\sigma}\big{|}|N_{\sigma}|:|N_{\sigma}|\to K_{\sigma}.

As an application of $({\dagger}_{9})$, we find that if $\sigma\in L$ and $\dim\sigma=n+1$, then

$({\dagger}_{13})$ for each $x\in f_{1}^{-1}(\partial\sigma)\cap|N^{(n+1)}|$, $f_{1,\sigma}(x)=f_{1}(x)$.

If $\tau\in L$ and $\dim\tau=n+1$, then

$({\dagger}_{14})$ $f_{1}^{-1}(\partial\sigma)\cap f_{1}^{-1}(\partial\tau)\cap|N^{(n+1)}|=f_{1}^{-1}(\partial\sigma\cap\partial\tau)\cap|N^{(n+1)}|\subset f_{1}^{-1}(|L^{(n)}|)\break\cap|N^{(n+1)}|=|N_{0}|$.

Let,

$({\dagger}_{15})$ h_{0}=f_{1}\big{|}|N_{0}|:|N_{0}|\to|L^{(n)}|.

Making use of $({\dagger}_{11})$–$({\dagger}_{15})$ and $({\dagger}_{9})$, we see that,

[TABLE]

is a map of $|N^{(n+1)}|$ to $F_{L,K,\Sigma^{n}}^{*}$ satisfying $(1)$ and $(2)$. This completes our proof. ∎

6. Eilenberg-MacLane Complexes for $\operatorname{{\mathbb{Z}}}$, $\operatorname{{\mathbb{Z}}}/p$, $\operatorname{{\mathbb{Q}}}$

When $X$ is a space, we shall write $X\tau K$ to mean that $X$ is an absolute co-extensor for $K$. This simply means that if $A$ is a closed subset of $X$, and $f:A\to K$ is a map, then there is a map $g:X\to K$ that extends $f$. We assume that the reader is familiar with the concept of an Eilenberg-MacLane $\operatorname{CW}$-complex $K$ of type $K(G,n)$; such complexes exist for each abelian group $G$, and for a given abelian group $G$ any $\operatorname{CW}$-complexes $K_{1}$, $K_{2}$ of type $K(G,n)$ are homotopy equivalent. It is well-known that if $X$ is a metrizable compactum and $K_{1}$, $K_{2}$ are Eilenberg-MacLane $\operatorname{CW}$-complexes of type $K(G,n)$, then $X\tau K_{1}$ if and only if $X\tau K_{2}$. According to Theorem 1.1 of [Dr],

Lemma 6.1**.**

For each $n\in\operatorname{{\mathbb{N}}}$, compact metrizable space $X$, and Eilenberg-MacLane $\operatorname{CW}$-complex $K$ of type $K(G,n)$, $\dim_{G}X\leq n$ if and only if $X\tau K$.

When we encounter a space $X$ which is homeomorphic to $S^{n}$, then we shall assume that there is a fixed isomorphism between $\pi_{n}(X)$ and $\operatorname{{\mathbb{Z}}}$. In this way if $Y$ is also homeomorphic to $S^{n}$ and $f:X\to Y$ is a map, then the degree of $f$, denoted $\deg(f)$, is well-defined.

In what follows, we will use standard constructions of the $\operatorname{CW}$-complexes $K(\operatorname{{\mathbb{Z}}},n)$, $K(\operatorname{{\mathbb{Z}}}/p,n)$, and $K(\operatorname{{\mathbb{Q}}},n)$ for $n\in\operatorname{{\mathbb{N}}}$ and $p\geq 2$. For $K(\operatorname{{\mathbb{Z}}},n)$ and $K(\operatorname{{\mathbb{Q}}},n)$ we shall require that $n\geq 2$, but for $K(\operatorname{{\mathbb{Z}}}/p,n)$, $n\geq 1$ will be permitted. These restrictions on $n$ will typically be implicit in the sequel. Each of $K(\operatorname{{\mathbb{Z}}},n)$ and $K(\operatorname{{\mathbb{Z}}}/p,n)$ respectively contains a unique “canonical” copy $\Sigma_{\operatorname{{\mathbb{Z}}}}=K(\operatorname{{\mathbb{Z}}},n)^{(n)}$ and $\Sigma_{\operatorname{{\mathbb{Z}}}/p}=K(\operatorname{{\mathbb{Z}}}/p,n)^{(n)}$ of $S^{n}$. In this setting, $K(\operatorname{{\mathbb{Z}}},n)^{(n+1)}=K(\operatorname{{\mathbb{Z}}},n)^{(n)}=\Sigma_{\operatorname{{\mathbb{Z}}}}$; moreover $K(\operatorname{{\mathbb{Z}}}/p,n)^{(n+1)}$ is a standard Moore space of type $(\operatorname{{\mathbb{Z}}}/p,n)$, i.e., an $(n+1)$-cell $B^{n+1}$ attached to $\Sigma_{\operatorname{{\mathbb{Z}}}/p}$ via a map of $\partial(B^{n+1})$ of degree $p$. We shall denote $\kappa_{\operatorname{{\mathbb{Z}}}}:\Sigma_{\operatorname{{\mathbb{Z}}}}\hookrightarrow K(\operatorname{{\mathbb{Z}}},n)$ and $\kappa_{\operatorname{{\mathbb{Z}}}/p}:\Sigma_{\operatorname{{\mathbb{Z}}}/p}\hookrightarrow K(\operatorname{{\mathbb{Z}}}/p,n)$.

$(*)$ In case $f:S^{n}\to\Sigma_{\operatorname{{\mathbb{Z}}}/p}$ is a map and $\pi_{n}(\kappa_{\operatorname{{\mathbb{Z}}}/p})\circ\pi_{n}(f):\pi_{n}(S^{n})\to\pi_{n}(K(\operatorname{{\mathbb{Z}}}/p,n))$ is the trivial map, then $\deg(f)$ is a multiple of $p$.

The situation with $K(\operatorname{{\mathbb{Q}}},n)$ is different because $K(\operatorname{{\mathbb{Q}}},n)^{(n)}$ is a countable wedge, say $\bigvee\{S_{i}^{n}\,|\,i\in\operatorname{{\mathbb{N}}}\}$ of copies $S_{i}^{n}$ of $S^{n}$. So there is not a canonically unique choice of $S^{n}$ as we had above for the groups $\operatorname{{\mathbb{Z}}}$ and $\operatorname{{\mathbb{Z}}}/p$.

Definition 6.2**.**

Select a fixed isomorphism $\psi:\pi_{n}(K(\operatorname{{\mathbb{Q}}},n))\to\operatorname{{\mathbb{Q}}}$. We may do this so that for some $i\in\operatorname{{\mathbb{N}}}$, if we define $\Sigma_{\operatorname{{\mathbb{Q}}}}=S_{i}^{n}$, then the following is true. Let $\kappa_{\operatorname{{\mathbb{Q}}}}:\Sigma_{\operatorname{{\mathbb{Q}}}}\hookrightarrow K(\operatorname{{\mathbb{Q}}},n)$. Then the composition $\psi\circ\pi_{n}(\kappa_{\operatorname{{\mathbb{Q}}}}):\pi_{n}(\Sigma_{\operatorname{{\mathbb{Q}}}})\to\operatorname{{\mathbb{Q}}}$ sends $\pi_{n}(\Sigma_{\operatorname{{\mathbb{Q}}}})$ isomorphically onto the standard copy of $\operatorname{{\mathbb{Z}}}$ in $\operatorname{{\mathbb{Q}}}$.

Lemma 6.3**.**

Let $n\in\operatorname{{\mathbb{N}}}$ and $\mu$ be an $(n+1)$-simplex.

1. (1)

If $f:\partial\mu\to\Sigma_{\operatorname{{\mathbb{Z}}}}$ is a map such that $\kappa_{\operatorname{{\mathbb{Z}}}}\circ f:\partial\mu\to K(\operatorname{{\mathbb{Z}}},n)$ extends to a map $f^{*}:\mu\to K(\operatorname{{\mathbb{Z}}},n)$, then $f$ is homotopic to a constant map. 2. (2)

If $f:\partial\mu\to\Sigma_{\operatorname{{\mathbb{Z}}}/p}$ is a map such that $\kappa_{\operatorname{{\mathbb{Z}}}/p}\circ f:\partial\mu\to K(\operatorname{{\mathbb{Z}}}/p,n)$ extends to a map $f^{*}:\mu\to K(\operatorname{{\mathbb{Z}}}/p,n)^{(n+1)}$, then $\deg(f)$ is a multiple of $p$. 3. (3)

If $f:\partial\mu\to\Sigma_{\operatorname{{\mathbb{Q}}}}$ is a map such that $\kappa_{\operatorname{{\mathbb{Q}}}}\circ f:\partial\mu\to K(\operatorname{{\mathbb{Q}}},n)$ extends to a map $f^{*}:\mu\to K(\operatorname{{\mathbb{Q}}},n)$, then $\deg(f)=0$.

Proof.

Note that $\pi_{n}(\kappa_{\operatorname{{\mathbb{Z}}}}):\pi_{n}(\Sigma_{\operatorname{{\mathbb{Z}}}})\to\pi_{n}(K(\operatorname{{\mathbb{Z}}},n))$ is an isomorphism. Assume that $f$ is not homotopic to a constant map. It follows that there exists $k\in\operatorname{{\mathbb{N}}}$ such that $\deg(f)\in\{k,-k\}$. Select a generator $g$ of $\pi_{n}(\partial\mu)$. Then for some generator $\overline{g}$ of $\pi_{n}(\Sigma_{\operatorname{{\mathbb{Z}}}})$, $\pi_{n}(f)(g)=k\cdot\overline{g}$ is a nontrivial element of $\pi_{n}(\Sigma_{\operatorname{{\mathbb{Z}}}})$. Hence $\pi_{n}(\kappa_{\operatorname{{\mathbb{Z}}}})(k\cdot\overline{g})=\pi_{n}(\kappa_{\operatorname{{\mathbb{Z}}}})(\pi_{n}(f)(g))=\pi_{n}(\kappa_{\operatorname{{\mathbb{Z}}}}\circ f)(g)$ is a nontrivial element of $\pi_{n}(K(\operatorname{{\mathbb{Z}}},n))$. But by the assumption in (1), $\kappa_{\operatorname{{\mathbb{Z}}}}\circ f:\partial\mu\to K(\operatorname{{\mathbb{Z}}},n)$ extends to the map $f^{*}:\mu\to K(\operatorname{{\mathbb{Z}}},n)$ showing that $\pi_{n}(\kappa_{\operatorname{{\mathbb{Z}}}}\circ f)(g)=\pi_{n}(\kappa_{\operatorname{{\mathbb{Z}}}})\circ\pi_{n}(f)(g)$ is the trivial element of $\pi_{n}(K(\operatorname{{\mathbb{Z}}},n))$. Since $\pi_{n}(\kappa_{\operatorname{{\mathbb{Z}}}})$ is an isomorphism, it has to be true that $\pi_{n}(f)(g)$ is the trivial element of $\pi_{n}(K(\operatorname{{\mathbb{Z}}},n))$, a contradiction. We have demonstrated (1).

Next we prove (2). Note that $K(\operatorname{{\mathbb{Z}}}/p,n)^{(n+1)}$ is a Moore space of type $(\operatorname{{\mathbb{Z}}}/p,n)$. We have that $\kappa_{\operatorname{{\mathbb{Z}}}/p}\circ f:\partial\mu\to K(\operatorname{{\mathbb{Z}}}/p,n)^{(n+1)}$ extends to a map $f^{*}:\mu\to K(\operatorname{{\mathbb{Z}}}/p,n)^{(n+1)}$. This shows that $\pi_{n}(\kappa_{\operatorname{{\mathbb{Z}}}/p})\circ\pi_{n}(f)=\pi_{n}(\kappa_{\operatorname{{\mathbb{Z}}}/p}\circ f):\pi_{n}(\partial\mu)\to\pi_{n}(K(\operatorname{{\mathbb{Z}}}/p,n)^{(n+1)})$ is the trivial map of $\pi_{n}(\partial\mu)$. It follows from $(*)$ that $\deg(f)$ is a multiple of $p$.

Let us prove (3). Suppose that $\deg(f)\neq 0$. Consider the case that $\deg(f)\in\{1,-1\}$. Then $\pi_{n}(f):\pi_{n}(\partial\mu)\to\pi_{n}(\Sigma_{\operatorname{{\mathbb{Q}}}})$ is an isomorphism. This and Definition 6.2, show that

$({\dagger}_{1})$ $\psi\circ\pi_{n}(\kappa_{\operatorname{{\mathbb{Q}}}})\circ\pi_{n}(f)=\psi\circ\pi_{n}(\kappa_{\operatorname{{\mathbb{Q}}}}\circ f)$ sends $\pi_{n}(\partial\mu)$ isomorphically onto $\operatorname{{\mathbb{Z}}}\subset\operatorname{{\mathbb{Q}}}$.

But the assumption that $\kappa_{\operatorname{{\mathbb{Q}}}}\circ f$ extends to a map of $\mu$ into $K(\operatorname{{\mathbb{Q}}},n)$ shows that the homomorphism $\pi_{n}(\kappa_{\operatorname{{\mathbb{Q}}}}\circ f)$ is trivial, which in turn implies that $\psi\circ\pi_{n}(\kappa_{\operatorname{{\mathbb{Q}}}}\circ f)$ sends $\pi_{n}(\partial\mu)$ trivially into $\operatorname{{\mathbb{Q}}}$, a contradiction.

We consider the case that $k\geq 2$ and $\deg(f)\in\{k,-k\}$. Then

$({\dagger}_{2})$ there is a generator $\overline{g}$ of $\pi_{n}(\Sigma_{\operatorname{{\mathbb{Q}}}})$ such that $\pi_{n}(f)(g)=k\cdot\overline{g}$ which is a nontrivial element of $\pi_{n}(\Sigma_{\operatorname{{\mathbb{Q}}}})$.

Hence, $\psi\circ\pi_{n}(\kappa_{\operatorname{{\mathbb{Q}}}})(k\cdot\overline{g})=\psi\circ\pi_{n}(\kappa_{\operatorname{{\mathbb{Q}}}})(\pi_{n}(f)(g))=\psi\circ\pi_{n}(\kappa_{\operatorname{{\mathbb{Q}}}})\circ\pi_{n}(f)(g)=\psi\circ\pi_{n}(\kappa_{\operatorname{{\mathbb{Q}}}}\circ f)(g)$ is a nontrivial element of $\operatorname{{\mathbb{Z}}}\subset\operatorname{{\mathbb{Q}}}$. But the assumption in (3) implies that $\kappa_{\operatorname{{\mathbb{Q}}}}\circ f$ is homotopic to a constant map, so $\psi\circ\pi_{n}(\kappa_{\operatorname{{\mathbb{Q}}}}\circ f)(g)$ is the trivial element of $\operatorname{{\mathbb{Z}}}\subset\operatorname{{\mathbb{Q}}}$, a contradiction. ∎

7. Simple Extensions

In [Wa], [Dr] (see also [RT]), and [Le], the proofs of Theorems 1.1–1.3 made use of increasingly more complex extensions. One of our goals, as stated in the Introduction, is to provide proofs of these theorems that employ much simpler extensions. The groundwork for producing them was laid in Sections 4 and 6. Now we shall provide the precise definitions for the extensions we are going to use.

When $n\in\operatorname{{\mathbb{N}}}$ is given, $G\in\{\operatorname{{\mathbb{Z}}},\operatorname{{\mathbb{Z}}}/p,\operatorname{{\mathbb{Q}}}\}$, and $L$ is a finite simplicial complex, then the $\operatorname{CW}$-complex $F_{L,K,\Sigma^{n}}$ of Definition 4.1 will be constructed with $\Sigma^{n}=\Sigma_{G}$, the particular canonical copy of $S^{n}$ in $K$ (see Definition 6.2 in case $G=\operatorname{{\mathbb{Q}}}$), and we shall conserve notation by simply denoting $F_{L,K,\Sigma^{n}}=F_{L,K}$. For each $\sigma\in L$ with $\dim\sigma=n+1$, the inclusion $\kappa_{\sigma}:\partial\sigma\hookrightarrow K_{\sigma}$ will correspond to the inclusion $\kappa_{G}:\Sigma_{G}\hookrightarrow K$, $G\in\{\operatorname{{\mathbb{Z}}},\operatorname{{\mathbb{Z}}}/p,\operatorname{{\mathbb{Q}}}\}$.

Lemma 6.3 can now be applied to arrive at the following.

Lemma 7.1**.**

Let $L$ be a finite simplicial complex, $\sigma\in L$ with $\dim\sigma=n+1$, and $\mu$ an $(n+1)$-simplex.

1. (1)

If $K=K(\operatorname{{\mathbb{Z}}},n)$ and $f:\partial\mu\to\partial\sigma$ is a map such that $f=\kappa_{\sigma}\circ f:\partial\mu\to K_{\sigma}$ extends to a map $f^{*}:\mu\to K_{\sigma}$, then $\deg(f)=0$. 2. (2)

If $K=K(\operatorname{{\mathbb{Z}}}/p,n)$ and $f:\partial\mu\to\partial\sigma$ is a map such that $f=\kappa_{\sigma}\circ f:\partial\mu\to K_{\sigma}$ extends to a map $f^{*}:\mu\to K_{\sigma}^{(n+1)}$, then $\deg(f)$ is a multiple of $p$. 3. (3)

If $K=K(\operatorname{{\mathbb{Q}}},n)$ and $f:\partial\mu\to\partial\sigma$ is a map such that $f=\kappa_{\sigma}\circ f:\partial\mu\to K_{\sigma}$ extends to a map $f^{*}:\mu\to K_{\sigma}$, then $\deg(f)=0$.

8. Proposition for Control of the $(n+1)$-skeleton

Proposition 8.1**.**

Let $n\in\operatorname{{\mathbb{N}}}$, $K\in\{K(\operatorname{{\mathbb{Z}}},n),K(\operatorname{{\mathbb{Z}}}/p,n),K(\operatorname{{\mathbb{Q}}},n)\}$, $E$ a compact polyhedron, $L$ a finite simplicial complex, $\overline{p}:E\to|L|$ a map, and $S$ an $n$-regular subdivision of $L$. Suppose that for each $\sigma\in L$ with $\dim\sigma=n+1$, the map $\overline{p}|(\overline{p})^{-1}(|N(\partial\sigma,S)|):(\overline{p})^{-1}(|N(\partial\sigma,S)|)\to|N(\partial\sigma,S)|\subset K_{\sigma,+}$ extends to a map of $E$ to $K_{\sigma,+}$. Let $N$ be a triangulation of $E$ that admits a simplicial approximation $\overline{p}_{1}:|N|\to|L|$ of the map $\overline{p}$, $L_{0}$ a subdivision of $N$, and $\varphi:|L_{0}|\to|N|$ a simplicial approximation to $\operatorname{id}:|L_{0}|\to|N|$. Then there exists a map $g_{K}:|N^{(n+1)}|\to|L^{(n+1)}|$ such that:

1. (1)

$g_{K}$* is an $L$-modification of both \overline{p}_{1}\big{|}|N^{(n+1)}|:|N^{(n+1)}|\to|L^{(n+1)}|\subset|L| and \overline{p}\big{|}|N^{(n+1)}|:|N^{(n+1)}|\to|L|;* 2. (2)

$g_{K}(|N^{(n+1)}|)\subset|L^{(n+1)}|$, $g_{K}(|N^{(n)}|)\subset|L^{(n)}|$, $g_{K}\circ\varphi(|L_{0}^{(n+1)}|)\subset|L^{(n+1)}|$, and $g_{K}\circ\varphi(|L_{0}^{(n)}|)\subset|L^{(n)}|$; 3. (3)

for each $x\in|L_{0}^{(n+1)}|$, there exists $\sigma\in L$ with $\{\overline{p}(x),g_{K}\circ\varphi(x)\}\subset\sigma$; 4. (4)

for each $\nu\in L_{0}^{(n+1)}$, $\overline{p}_{1}(\varphi(\nu))\in L^{(n+1)}$, and $g_{K}(\varphi(\nu))\subset\overline{p}_{1}(\varphi(\nu))$; 5. (5)

$g_{K}(x)=\overline{p}_{1}(x)$* for all $x\in(\overline{p}_{1})^{-1}(|L^{(n)}|)\cap|N^{(n+1)}|$;* 6. (6)

in case $K=K(\operatorname{{\mathbb{Z}}},n)$ and $\tau\in N^{(n+1)}$, then $g_{K}(\tau)\subset|L^{(n)}|$ so $g_{K}(|N^{(n+1)}|)\subset|L^{(n)}|$; the preceding implies that $g_{K}\circ\varphi(|L_{0}^{(n+1)}|)\subset g_{K}(|N^{(n+1)}|)\subset|L^{(n)}|$.

Now let $v\in N^{(0)}$, $N_{v}$, $L_{0,v}$ respectively be the subcomplexes of $N$, $L_{0}$ with $|N_{v}|=|L_{0,v}|=\overline{\operatorname{st}}(v,N)$, and $\widetilde{L}$ be a subcomplex of $L$ with $g_{K}(|N_{v}|)\subset\widetilde{L}$. Then, 7. (7)

both $g_{K}(|N_{v}^{(n)}|)\subset|\widetilde{L}^{(n)}|$ and $g_{K}(\varphi(|L_{0,v}^{(n)}|))\subset|\widetilde{L}^{(n)}|$, 8. (8)

In case $K=K(\operatorname{{\mathbb{Z}}},n)$, then g_{K}\circ\varphi\big{|}|L_{0,v}^{(n)}|:|L_{0,v}^{(n)}|\to|\widetilde{L}^{(n)}| is null homotopic; 9. (9)

in case $K=K(\operatorname{{\mathbb{Z}}}/p,n)$, then the induced homomorphism H_{n}(g_{K}\circ\varphi\big{|}|L_{0,v}^{(n)}|;\operatorname{{\mathbb{Z}}}/p):H_{n}(|L_{0,v}^{(n)}|;\operatorname{{\mathbb{Z}}}/p)\to H_{n}(|\widetilde{L}^{(n)}|;\operatorname{{\mathbb{Z}}}/p) is trivial, and 10. (10)

in case $K=K(\operatorname{{\mathbb{Q}}},n)$, then the induced homomorphism H_{n}(g_{K}\circ\varphi\big{|}|L_{0,v}^{(n)}|;\operatorname{{\mathbb{Z}}}):H_{n}(|L_{0,v}^{(n)}|;\operatorname{{\mathbb{Z}}})\to H_{n}(|\widetilde{L}^{(n)}|;\operatorname{{\mathbb{Z}}}) is trivial.

Proof.

We shall treat the pair \big{(}|N^{(n+1)}|,(\overline{p}_{1})^{-1}(|L^{(n)}|)\cap|N^{(n+1)}|\big{)} as a relative $\operatorname{CW}$-complex (see 7.6.4, p. 401 of [Sp] for the notion of a relative $\operatorname{CW}$-complex). We want to apply Lemma 5.1 to $(n,L,K,\Sigma^{n})$ with $K$ as given and $\Sigma^{n}$ the canonical copy of $S^{n}$ in $K$ (see Section 7). The map $f$ of Lemma 5.1 is replaced by our $\overline{p}:E\to|L|$; the simplicial approximation $f_{1}$ is the current $\overline{p}_{1}$. Let $h_{K}:|N^{(n+1)}|\to F_{L,K}^{*}$ be a map as granted to us by Lemma 5.1. From Lemma 5.1(1), it is seen that $h_{K}$ is a pair map,

[TABLE]

Of course, $(F_{L,K}^{*},|L^{(n)}|)$ is also a relative $\operatorname{CW}$-complex. Part $(1)$ of Lemma 5.1 shows in addition that since $h_{K}$ equals the simplicial map $\overline{p}_{1}$ on $(\overline{p}_{1})^{-1}(|L^{(n)}|)\break\cap|N^{(n+1)}|$, then

$({\dagger}_{1})$ $h_{K}$ is simplicial from $(\overline{p}_{1})^{-1}(|L^{(n)}|)\cap|N^{(n+1)}|$ to $|L^{(n)}|$.

Using the cellular approximation theorem (7.17, p. 404 of [Sp]), one finds a relative cellular approximation

[TABLE]

of the map $h_{K}$. This is done so that

$({\dagger}_{2})$ $h_{K}^{*}=h_{K}=\overline{p}_{1}$ on $(\overline{p}_{1})^{-1}(|L^{(n)}|)\cap|N^{(n+1)}|$.

The cellular approximation theorem requires that if $x\in|N^{(n+1)}|$ and $h_{K}(x)$ lies in a subcomplex $M$ of $F_{L,K}^{*}$, then $h_{K}^{*}(x)\in M^{(n+1)}$. It then follows that

$({\dagger}_{3})$ if $\sigma\in L$ with $\dim\sigma=n+1$, $x\in|N^{(n+1)}|$, and $h_{K}(x)\in K_{\sigma}$, then $h_{K}^{*}(x)\in K_{\sigma}^{(n+1)}$.

Using Lemma 4.6(1,2), we may treat

$({\dagger}_{4})$ $r_{L,K}:(F_{L,K}^{*},|L^{(n)}|)\to(|L^{(n+1)}|,|L^{(n)}|)$ as a pair map where $r_{L,K}(x)=x$ for all $x\in|L^{(n)}|$.

Now we define

$({\dagger}_{5})$ $g_{K}:|N^{(n+1)}|\to|L^{(n+1)}|$ by $g_{K}(x)=r_{L,K}\circ h_{K}^{*}(x)$.

From this and $({\dagger}_{4})$, it follows that,

$({\dagger}_{6})$ if $x\in|N^{(n+1)}|$ and $h_{K}^{*}(x)\in|L^{(n)}|$, then $g_{K}(x)=h_{K}^{*}(x)$.

To check (1), let $x\in|N^{(n+1)}|$ and let $\tau\in L$ be the unique simplex with $\overline{p}(x)\in\operatorname{int}\tau$. Since $\overline{p}_{1}$ is a simplicial approximation of $\overline{p}$, then $\overline{p}_{1}(x)\in\tau$. But $x\in|N^{(n+1)}|$ and $\overline{p}_{1}$ is simplicial, so there exists a unique $\sigma\in L^{(n+1)}$ with $\sigma\subset\tau$ and $\overline{p}_{1}(x)\in\operatorname{int}\sigma$.

If $x\in(\overline{p}_{1})^{-1}(|L^{(n)}|)\cap|N^{(n+1)}|$, then by $({\dagger}_{2})$, $h_{K}^{*}(x)=h_{K}(x)=\overline{p}_{1}(x)\in|L^{(n)}|$. Using $({\dagger}_{6})$, one sees that $g_{K}(x)=h_{K}^{*}(x)=\overline{p}_{1}(x)\in\sigma\subset\tau$. So $g_{K}$ is an $L$-modification of $\overline{p}_{1}$ with respect to the domain $(\overline{p}_{1})^{-1}(|L^{(n)}|)\cap|N^{(n+1)}|$.

The alternate case is that $x\notin(\overline{p}_{1})^{-1}(|L^{(n)}|)$ and hence $\overline{p}_{1}(x)\in|L^{(n+1)}|\setminus|L^{(n)}|$. So $\dim\sigma=n+1$. By Lemma 5.1(2), $h_{K}((\overline{p}_{1})^{-1}(\sigma)\cap|N^{(n+1)}|)\subset K_{\sigma}$. But $x\in(\overline{p}_{1})^{-1}(\sigma)\cap|N^{(n+1)}|$, so $h_{K}(x)\in K_{\sigma}$. This puts $h_{K}^{*}(x)\in K_{\sigma}$. Applying Lemma 4.6(2,3), one sees that $r_{L,K}(K_{\sigma})\subset\sigma$. Therefore $g_{K}(x)=r_{L,K}\circ h_{K}^{*}(x)\in\sigma\subset\tau$. So $g_{K}$ is an $L$-modification of $\overline{p}_{1}$ with respect to the complementary part of the domain.

Putting the two cases together, one sees that $g_{K}$ is an $L$-modification of \overline{p}_{1}\big{|}|N^{(n+1)}|:|N^{(n+1)}|\to|L^{(n+1)}|. So the first part of (1) is true. To show that $g_{K}$ is an $L$-modification of \overline{p}\big{|}|N^{(n+1)}|:|N^{(n+1)}|\to|L|, let $x\in|N^{(n+1)}|$ and $\sigma\in L$ be the unique simplex with $\overline{p}(x)\in\operatorname{int}\sigma$. Then because $\overline{p}_{1}:|N|\to|L|$ is a simplicial approximation of $\overline{p}$, one has that $\overline{p}_{1}(x)\in\sigma$. By the first part of (1), $g_{K}(x)\in\sigma$, as required to complete our proof of (1).

Statement (2) follows from (1) and the fact that \overline{p}_{1}\big{|}|N^{(n+1)}|:|N^{(n+1)}|\to|L^{(n+1)}| is simplicial, and hence both $g_{K}(|N^{(n+1)}|)\subset|L^{(n+1)}|$ and $g_{K}(|N^{(n)}|)\break\subset|L^{(n)}|$ hold.

Now we prove (3). Let $x\in|L_{0}^{(n+1)}|\subset|N|$, $\sigma\in L$ be the unique simplex with $\overline{p}(x)\in\operatorname{int}\sigma$, $\mu\in L_{0}^{(n+1)}$ the unique simplex with $x\in\operatorname{int}\mu$, and $\lambda\in N$ the unique simplex with $x\in\operatorname{int}\lambda$. Then $\varphi(\mu)\in N^{(n+1)}$ and $\varphi(x)\in\varphi(\mu)\subset\lambda$. Now, $\overline{p}_{1}:|N|\to|L|$ is a simplicial approximation of $\overline{p}$, so $\overline{p}_{1}(x)\in\sigma$. It follows that $\overline{p}_{1}(\lambda)\subset\sigma$, so $\overline{p}_{1}\circ\varphi(x)\in\sigma$. But, $g_{K}$ is an $L$-modification of $\overline{p}_{1}$, so $g_{K}\circ\varphi(x)\in\sigma$. Thus $\{g_{K}\circ\varphi(x),\overline{p}(x)\}\subset\sigma$.

The first part of (4) is true because both $\overline{p}_{1}$ and $\varphi$ are simplicial. The second part comes from the first part and (1). One obtains (5) from $({\dagger}_{2})$, $({\dagger}_{4})$, and $({\dagger}_{5})$.

In order to prove properties from (6) onward, let us notice some additional facts. Since $\overline{p}_{1}:|N|\to|L|$ is simplicial, then for each $\tau\in N^{(n+1)}$, $\overline{p}_{1}(\tau)\in L^{(n+1)}$. Let us use this in proving that,

$({\dagger}_{7})$ if $\tau\in N^{(n+1)}$ and $\dim\tau=n+1=\dim\overline{p}_{1}(\tau)$, then $h_{K}^{*}(\tau)\subset K_{\overline{p}_{1}(\tau)}^{(n+1)}$, and $h_{K}^{*}(\partial\tau)\subset K_{\overline{p}_{1}(\tau)}^{(n)}=\partial(\overline{p}_{1}(\tau))$.

Now $\tau\subset(\overline{p}_{1})^{-1}(\overline{p}_{1}(\tau))\cap|N^{(n+1)}|$. By Lemma 5.1(2) with $\overline{p}_{1}(\tau)$ in place of $\sigma$, $h_{K}((\overline{p}_{1})^{-1}(\overline{p}_{1}(\tau))\cap|N^{(n+1)}|)\subset K_{\overline{p}_{1}(\tau)}$. By $({\dagger}_{3})$, $h_{K}^{*}((\overline{p}_{1})^{-1}(\overline{p}_{1}(\tau))\cap|N^{(n+1)}|)\subset K_{\overline{p}_{1}(\tau)}^{(n+1)}$. Hence the first part of $({\dagger}_{7})$ holds true. The second part is true because $h_{K}^{*}$ is cellular. Next we show that,

$({\dagger}_{8})$ if $\tau\in N^{(n+1)}$ and $\dim\tau=n+1=\dim\overline{p}_{1}(\tau)$, then $g_{K}(\tau)\subset\overline{p}_{1}(\tau)$, $g_{K}(\partial\tau)\subset\partial(\overline{p}_{1}(\tau))$, and $g_{K}|\partial\tau=h_{K}^{*}|\partial\tau$.

By the first part of $({\dagger}_{7})$, $h_{K}^{*}(\tau)\subset K_{\overline{p}_{1}(\tau)}^{(n+1)}$, and by $({\dagger}_{5})$, $x\in\tau$ implies that $g_{K}(x)=r_{L,K}\circ h_{k}^{*}(x)$. Apply this with Lemma 4.6(3) to obtain the first part of $({\dagger}_{8})$. The second and third parts follow respectively from the second part of $({\dagger}_{7})$ and $({\dagger}_{6})$.

If we put together $({\dagger}_{7})$ and $({\dagger}_{8})$, we arrive at the next fact.

$({\dagger}_{9})$ if $\tau\in N^{(n+1)}$ and $\dim\tau=n+1=\dim(\overline{p}_{1}(\tau))$, then $g_{K}|\partial\tau:\partial\tau\to\partial(\overline{p}_{1}(\tau))$ extends to a map of $\tau$ to $K_{\overline{p}_{1}(\tau)}$.

To prove (6), consider first the case that $\overline{p}_{1}(\tau)\subset|L^{(n)}|$, i.e., $\overline{p}_{1}(\tau)\in L^{(n)}$. Then use (5) to see that $g_{K}(\tau)=\overline{p}_{1}(\tau)\subset|L^{(n)}|$. In the complementary case, $\overline{p}_{1}(\tau)\in L^{(n+1)}$ and $\dim\overline{p}_{1}(\tau)=n+1$. Note that since $K=K(\operatorname{{\mathbb{Z}}},n)$, then $K_{\overline{p}_{1}(\tau)}^{(n+1)}=K_{\overline{p}_{1}(\tau)}^{(n)}=\partial(\overline{p}_{1}(\tau))$. This and $({\dagger}_{7})$ show that $h_{K}^{*}(\tau)\subset\partial(\overline{p}_{1}(\tau))\subset|L^{(n)}|$. Now employ $({\dagger}_{6})$ to see that $g_{K}(\tau)=h_{K}^{*}(\tau)\subset\partial(\overline{p}_{1}(\tau))\subset|L^{(n)}|$. The first part of (7) follows from the second part of (2); its second part uses the first part and the fact that $\varphi$ is a simplicial approximation of the identity. As to (8), use Lemma 3.2 to see that there is a homotopy $H:|L_{0,v}^{(n)}|\times[0,1]\to|L_{0,v}^{(n+1)}|$ between the inclusion $\iota:|L_{0,v}^{(n)}|\hookrightarrow|L_{0,v}^{(n+1)}|$ and a constant map. Now apply (6) to the homotopy $g_{K}\circ\varphi\circ H:|L_{0,v}^{(n)}|\times[0,1]\to|\widetilde{L}^{(n)}|$ (with appropriate domain restrictions) to get a null homotopy as desired in (8).

In proving (9) and (10) we shall use the following fact from Lemma 3.9.

$({\dagger}_{10})$ There exists $m\in\operatorname{{\mathbb{N}}}$ and a set $\{\sigma_{i}\,|\,1\leq i\leq m\}$ of $(n+1)$-simplexes of $N_{v}$, such that for any abelian group $G$ and each $z\in Z_{n}(|N_{v}^{(n)}|;G)$, there is a set $\{g_{i}\,|\,1\leq i\leq m\}\subset G$ with $z=\sum_{i=1}^{m}g_{i}\cdot\partial\sigma_{i}$.

Statement (9) will be true if we can prove that

$({\dagger}_{11})$ $H_{n}(g_{K};\operatorname{{\mathbb{Z}}}/p):H_{n}(|N_{v}^{(n)}|;\operatorname{{\mathbb{Z}}}/p)\to H_{n}(\widetilde{L}^{(n)};\operatorname{{\mathbb{Z}}}/p)$ is trivial.

We see that $({\dagger}_{11})$ is true if for each $z=\sum_{i=1}^{m}g_{i}\cdot\partial\sigma_{i}\in Z_{n}(|N_{v}^{(n)}|;\operatorname{{\mathbb{Z}}}/p)$, $Z_{n}(g_{K};\operatorname{{\mathbb{Z}}}/p)(z)$ is homologous to [math] in $Z_{n}(\widetilde{L}^{(n)};\operatorname{{\mathbb{Z}}}/p)$. Let $1\leq i\leq m$ and with $\sigma_{i}$ in place of $\tau$, let us consider the two cases that we explored in our proof of (6) above. In the first of these, we have that $\overline{p}_{1}(\sigma_{i})\in L^{(n)}$ and $g_{K}(\sigma_{i})=\overline{p}_{1}(\sigma_{i})$. It follows that $g_{K}$ maps $\partial\sigma_{i}$ trivially into $|\widetilde{L}^{(n)}|$. So, $Z_{n}(g_{K};\operatorname{{\mathbb{Z}}}/p)(g_{i}\cdot\partial\sigma_{i})$ is homologous to [math] in $Z_{n}(\widetilde{L}^{(n)};\operatorname{{\mathbb{Z}}}/p)$. In the second case, using $({\dagger}_{9})$ with $\sigma_{i}$ replacing $\tau$, one has that $g_{K}|\partial\sigma_{i}:\partial\sigma_{i}\to\partial(\overline{p}_{1}(\sigma_{i}))$ extends to a map of $\sigma_{i}$ into $K_{\overline{p}_{1}(\sigma_{i})}^{(n+1)}$. By Lemma 7.1(2), $g_{K}|\partial\sigma_{i}:\partial\sigma_{i}\to\partial(\overline{p}_{1}(\sigma_{i}))$ has degree a multiple of $p$. From this and the first part of Lemma 3.10(2), it follows that $Z_{n}(g_{K};\operatorname{{\mathbb{Z}}}/p)(g_{i}\cdot\partial\sigma_{i})$ is homologous to [math] in $Z_{n}(\partial(\overline{p}_{1}(\sigma_{i}));\operatorname{{\mathbb{Z}}}/p)$ and hence in $Z_{n}(\widetilde{L}^{(n)};\operatorname{{\mathbb{Z}}}/p)$. This gives us (9). To obtain (10), use the same argument we just made, but replace $\operatorname{{\mathbb{Z}}}/p$ with $\operatorname{{\mathbb{Z}}}$ and the first part of Lemma 3.10(2) by the second part of Lemma 3.10(2). ∎

9. Extensor Lemma

For the remainder of this paper, $I^{\infty}$ will denote the Hilbert cube, i.e., $I^{\infty}=\prod\{I_{i}\,|\,i\in\operatorname{{\mathbb{N}}}\}$ where $I_{i}=[0,1]$ for each $i$. For each $k\in\operatorname{{\mathbb{N}}}$, we use $I^{k}$ to denote $\{x\in I^{\infty}\,|\,x_{i}=0\,\,\mathrm{for}\,\,i>k\}$. Of course if $1\leq j\leq k$, then $I^{j}\subset I^{k}$, and $p_{j}^{k}:I^{k}\to I^{j}$ denotes the coordinate projection. If $P\subset I^{k}$, then $P^{[\infty]}$ will designate the set of $x$ in $I^{\infty}$ whose first $k$ coordinates are the same as those of an element of $P$. Plainly if $P\subset Q\subset I^{k}$, then $P^{[\infty]}\subset Q^{[\infty]}$. One sees that if $P$ is closed, respectively open, in $I^{k}$, then $P^{[\infty]}$ is closed, respectively open, in $I^{\infty}$. Let $p_{k,\infty}:I^{\infty}\to I^{k}$ denote the coordinate projection of the Hilbert cube to its finite part, and we have that $(p_{j,\infty}|I^{k})\circ p_{k,\infty}=p_{j,\infty}:I^{\infty}\to I^{j}$. It is therefore true that

[TABLE]

If $1\leq i\leq j\leq k$, then

[TABLE]

We shall use the metric $\rho$ on $I^{\infty}$ given by $\rho(x,y)=\sum_{i=1}^{\infty}\frac{|x_{i}-y_{i}|}{2^{i}}$. With this metric on $I^{\infty}$, $\operatorname{diam}I^{\infty}=1$. Moreover, if $1\leq j\leq k$ and $y\in I^{k}$, then

[TABLE]

Similarly, for any $x\in I^{\infty}$,

[TABLE]

The main result of this section is Lemma 9.3, a type of “extensor lemma.” It provides us with a statement, see (2) of that lemma, about extending a map under the condition that a given compactum $X$ is a subspace of $I^{\infty}$. A couple of preliminaries will be useful.

Lemma 9.1**.**

Let $K$ be a $\operatorname{CW}$-complex. Then $K$ has an open cover $\mathcal{V}$ such that any two $\mathcal{V}$-close maps of any space to $K$ are homotopic.

Proof.

There exists a simplicial complex $L$ such that $|L|_{m}$, that is $|L|$ with the metric topology, is homotopy equivalent to $K$. Choose a homotopy equivalence $f:K\to|L|_{m}$. By Theorem III.11.3 (page 106) of [Hu], $|L|_{m}$ is an ANR. Theorem IV.1.1 (page 111) of [Hu] shows that there is an open cover $\mathcal{W}$ of $|L|_{m}$ having the property that any two $\mathcal{W}$-close maps of any space to $|L|_{m}$ are homotopic. The open cover needed for $K$ is $\mathcal{V}=f^{-1}(\mathcal{W})$. ∎

Lemma 9.2**.**

Let $X\subset I^{\infty}$ be compact and nonempty. Then there exist an increasing sequence $(n_{j})$ in $\operatorname{{\mathbb{N}}}$ and a sequence $(P_{j})$ of nonempty compact polyhedra $P_{j}\subset I^{n_{j}}$ such that for all $j\in\operatorname{{\mathbb{N}}}:$

1. (1)

$X\subset\operatorname{int}_{I^{\infty}}(P_{j}^{[\infty]})\subset P_{j}^{[\infty]}\subset N(X,\frac{2}{j})$, 2. (2)

$p_{n_{j}}^{n_{j+1}}(P_{j+1})\subset\operatorname{int}_{I^{n_{j}}}P_{j}$, and 3. (3)

$P_{j+1}^{[\infty]}\subset\operatorname{int}_{I^{\infty}}(P_{j}^{[\infty]})$.

Then,

$(*_{1})$* $X=\bigcap\{P_{j}^{[\infty]}\,|\,j\in\operatorname{{\mathbb{N}}}\}$.*

Suppose moreover, that $j\in\operatorname{{\mathbb{N}}}$ and $B_{j}\subset P_{j}$. Then,

$(*_{2})$* if $k\geq j$, and $B_{k}=(p_{n_{j}}^{n_{k}})^{-1}(B_{j})\cap P_{k}$, one has that $B_{k}\subset P_{k}\subset I^{n_{k}}$ and $p_{n_{j}}^{n_{k}}(B_{k})\subset B_{j}$, and*

$(*_{3})$* if we put $B_{j,\infty}=p_{n_{j},\infty}^{-1}(B_{j})\cap X$, then for any open neighborhood $S$ of $B_{j,\infty}$ in $I^{\infty}$, there exists $k\geq j$ such that for all $l\geq k$, $B_{l}\subset S$.*

If the increasing sequence $(n_{j})$ is replaced by an increasing subsequence, then all of the above still hold true.

Proof.

We shall construct the sets $P_{j}$ by recursion. Put $n_{1}=1$ and $P_{1}=I^{1}$. One sees that $P_{1}\subset P_{1}^{[\infty]}=I^{\infty}$, so $(1)$ is true in case $j=1$ since $\operatorname{diam}I^{\infty}=1$. Because we are using a recursive construction, when we come to a particular $j$, we only have to consider statements (1)–(3) “up to $j$.” Hence (2) and (3) need not be considered yet, and we will deal with $(*_{1})$–$(*_{3})$ later.

Suppose that $j\in\operatorname{{\mathbb{N}}}$, and we have found finite sequences $1=n_{1}<\dots<n_{j}$ in $\operatorname{{\mathbb{N}}}$ and compact polyhedra $P_{1},\dots,P_{j}$ such that for $1\leq s\leq j$, $P_{s}\subset I^{n_{s}}$, and $(1)$– $(3)$ are true up to $j$.

One may choose $n_{j+1}\in\operatorname{{\mathbb{N}}}$ such that $n_{j+1}>n_{j}$ and $X\subset(p_{n_{j+1},\infty}(X))^{[\infty]}\subset N(X,\frac{2}{j+1})$. There is a neighborhood $V$ of $p_{n_{j+1},\infty}(X)$ in $I^{n_{j+1}}$ such that $V^{[\infty]}\subset N(X,\frac{2}{j+1})$. Choose a compact polyhedron $P_{j+1}$ of $I^{n_{j+1}}$ so that, $p_{n_{j+1},\infty}(X)\subset\operatorname{int}_{I^{n_{j+1}}}P_{j+1}\subset P_{j+1}\subset V$. From this, $X\subset(p_{n_{j+1},\infty}(X))^{[\infty]}\subset(\operatorname{int}_{I^{\infty}}P_{j+1})^{[\infty]}\subset P_{j+1}^{[\infty]}\subset V^{[\infty]}\subset N(X,\frac{2}{j+1})$. This gives us $(1)$ for $j+1$.

Notice that $(1)$ for $j$ implies, $p_{n_{j},\infty}(X)=p_{n_{j}}^{n_{j+1}}\circ p_{n_{j+1},\infty}(X)\subset\operatorname{int}_{I^{n_{j}}}P_{j}$. Hence, $p_{n_{j+1},\infty}(X)\subset(p_{n_{j}}^{n_{j+1}})^{-1}(\operatorname{int}_{I^{n_{j}}}(P_{j}))$. Thus, making $P_{j+1}$ smaller if necessary, we may have $(1)$ and simultaneously, $P_{j+1}\subset(p_{n_{j}}^{n_{j+1}})^{-1}(\operatorname{int}_{I^{n_{j}}}(P_{j}))$. This achieves $(2)$ for $j+1$. From (2) we get (3).

One gets $(*_{1})$ readily from (1), and $(*_{2})$ is true for elementary reasons. To prove $(*_{3})$ let $S$ be an open neighborhood of $B_{j,\infty}$ in $I^{\infty}$. If the conclusion of $(*_{3})$ is not true, then there is an increasing sequence $(m_{i})$ in $\operatorname{{\mathbb{N}}}$, $m_{1}\geq j$, so that for each $i$, there exists $b_{i}\in B_{m_{i}}\setminus S\subset P_{m_{i}}$. Passing to a subsequence if necessary, we may assume that the sequence $(b_{i})$ in the compactum $I^{\infty}\setminus S$ converges in $I^{\infty}$ to $b\in I^{\infty}\setminus S$. Applying $(*_{1})$ and $(*_{2})$ along with the fact that $b_{i}\in P_{m_{i}}$, one sees that $b\in X\setminus B_{j,\infty}$, from which we deduce that $p_{n_{j},\infty}(b)\notin B_{j}$.

For each $i$, $p_{n_{j}}^{s_{i}}(b_{i})=p_{n_{j},\infty}(b_{i})\in B_{j}$, $s_{i}=n_{m_{i}}$. Hence $\{p_{n_{j},\infty}(b_{i})\,|\,i\in\operatorname{{\mathbb{N}}}\}\subset B_{j}$. Since $B_{j}$ is closed in $P_{j}$, $p_{n_{j},\infty}$ is a map, and $(b_{i})$ converges to $b$, then $p_{n_{j},\infty}(b)\in B_{j}$, a contradiction. This yields $(*_{3})$. We leave the validation of the final statement to the reader. ∎

In reading Lemma 9.3, one should consult Lemma 9.2(2) to see that whenever $j\leq l$, then $p_{n_{j}}^{n_{l}}(P_{l})\subset P_{j}$.

Lemma 9.3**.**

Let $X\subset I^{\infty}$, $(n_{j})$, $(P_{j})$ be as in $\mathrm{Lemma\,\,\ref{intersectinH}}$, and $K$ be a $\operatorname{CW}$-complex with $X\tau K$. The following are true.

1. (1)

Suppose that $j\in\operatorname{{\mathbb{N}}}$ and $B_{j}$ is a nonempty closed subset of $P_{j}$. For each $k\geq j$, let $B_{k}=(p_{n_{j}}^{n_{k}})^{-1}(B_{j})\cap P_{k}$. If $M$ is a finite subcomplex of $K$ and $f:B_{j}\to M$ is a map, then there exists $k\geq j$ such that for all $l\geq k$, there is a map $f^{*}:P_{l}\to K$ that extends the composition $f\circ p_{n_{j}}^{n_{l}}|B_{l}:B_{l}\to M$, where we treat $p_{n_{j}}^{n_{l}}|B_{l}:B_{l}\to B_{j}$. 2. (2)

Suppose that $K\in\{K(\operatorname{{\mathbb{Z}}},n),K(\operatorname{{\mathbb{Z}}}/p,n),K(\operatorname{{\mathbb{Q}}},n)\}$ and $L_{j}$ is a triangulation of $P_{j}$. Then there exists $k\geq j$ such that for all $l\geq k$, there is a triangulation $N_{l}$ of $P_{l}$, a simplicial approximation $\overline{p}_{j}:|N_{l}|\to|L_{j}|$ of $p_{n_{j}}^{n_{l}}:P_{l}\to P_{j}$, and a map $g_{K}:|N_{l}^{(n+1)}|\to|L_{j}^{(n+1)}|$ as in $\mathrm{Proposition\,\,\ref{pushandshove2}}$ where $(E,N,L,\overline{p},\overline{p}_{1})$ of that proposition is replaced by $(P_{l},N_{l},L_{j},p_{n_{j}}^{n_{l}},\overline{p}_{j})$ in the present context.

Proof.

We shall first prove (1). Observe that $B_{j,\infty}=p_{n_{j},\infty}^{-1}(B_{j})\cap X$ is a closed subset of $X$ and that $p_{n_{j},\infty}(B_{j,\infty})\subset B_{j}$. Since $X\tau K$, then the map $f\circ p_{n_{j},\infty}|B_{j,\infty}:B_{j,\infty}\to M$ extends to a map $h:X\to\widetilde{M}$ where $\widetilde{M}$ is a finite subcomplex of $K$ and $M\subset\widetilde{M}$. Since $\widetilde{M}$ is an ANR, we may additionally assume that there is an open neighborhood $U$ of $X$ in $I^{\infty}$ and that $h:U\to\widetilde{M}$.

Employing Lemma 9.1, select an open cover $\mathcal{V}$ of $\widetilde{M}$ such that for any space $Y$, any maps $g_{1}:Y\to\widetilde{M}$ and $g_{2}:Y\to\widetilde{M}$ that are $\mathcal{V}$-close are homotopic. Let $\mathcal{V}_{1}$ be an open cover of $\widetilde{M}$ that star-refines $\mathcal{V}$. Choose an open cover $\mathcal{W}$ of $B_{j}$ such that if $W\in\mathcal{W}$, then there exists $V_{W}\in\mathcal{V}_{1}$ with $f(W)\subset V_{W}$.

Select a cover $\mathcal{R}$ of $X$ by sets open in $U$ having the property that for each $R\in\mathcal{R}$, there exists $V_{R}\in\mathcal{V}_{1}$ with $h(R)\subset V_{R}$. Let $S=\bigcup\mathcal{R}\subset U$. Then $S$ is an open neighborhood of $X$ in $I^{\infty}$, and of course $B_{j,\infty}\subset X$. So by $(*_{3})$ of Lemma 9.2, we may choose $k_{0}\in\operatorname{{\mathbb{N}}}$ so that $k_{0}\geq j$ and for all $l\geq k_{0}$, $B_{l}\subset S$. Using Lemma 9.2(3) and Lemma 9.2($(*_{1})$), we may also require that for such $l$, $P_{l}\subset S$.

Put $B^{*}=B_{j,\infty}\cup\bigcup\{B_{l}\,|\,l\geq k_{0}\}\subset S$. An application of Lemma 9.2($(*_{2})$) yields that $p_{n_{j},\infty}|B^{*}:B^{*}\to B_{j}$ is a map. For each $b\in B_{j,\infty}$, select an open neighborhood $E_{b}$ of $b$ in $B^{*}$ such that $p_{n_{j},\infty}(E_{b})$ is contained in an element $W_{b}$ of $\mathcal{W}$ and that in addition, there exists $R_{b}\in\mathcal{R}$ with $E_{b}\subset R_{b}$. Let $S_{0}=\bigcup\{E_{b}\,|\,b\in B_{j,\infty}\}$. Then $S_{0}$ is an open neighborhood of $B_{j,\infty}$ in $B^{*}\subset I^{\infty}$. So there is an open subset $S_{1}$ of $I^{\infty}$ having the property that $S_{1}\cap B^{*}=S_{0}$. Plainly, $S_{1}$ is an open neighborhood of $B_{j,\infty}$ in $I^{\infty}$. An application of $(*_{3})$ of Lemma 9.2, with $S_{1}$ in place of $S$ gives us the existence of a $k\geq k_{0}$ so that for all $l\geq k$, we have $B_{l}\subset S_{1}$. For such $l$, $B_{l}\subset S_{1}\cap B^{*}=S_{0}$.

We are going to show that for all $l\geq k$, $f\circ p_{n_{j}}^{n_{l}}|B_{l}:B_{l}\to|M|\subset|\widetilde{M}|$ is homotopic to $h|B_{l}:B_{l}\to|\widetilde{M}|$. For in that case, if we define $h_{0}=h|B_{l}:B_{l}\to|\widetilde{M}|$, then of course since $P_{l}\subset S$, $h_{0}$ extends to the map $h|P_{l}:P_{l}\to|\widetilde{M}|$, and the homotopy extension theorem will complete our proof of (1).

Let $x\in B_{l}$. It will be sufficient to show that $f\circ p_{n_{j}}^{n_{l}}(x)$ and $h_{0}(x)$ lie in an element of $\mathcal{V}$. There exists $b\in B_{j,\infty}$ such that $x\in E_{b}$. Now $b\in E_{b}\subset R_{b}\in\mathcal{R}$. It follows that $h(\{b,x\})=\{h(b),h(x)\}=\{h(b),h_{0}(x)\}\subset V_{1}=V_{R_{b}}\in\mathcal{V}_{1}$. One sees from the definition of $h$ and the fact that $b\in B_{j,\infty}$, that $h(b)=f\circ p_{n_{j},\infty}(b)$. So we have that $\{f\circ p_{n_{j},\infty}(b),h_{0}(x)\}\subset V_{1}\in\mathcal{V}_{1}$. We know that $p_{n_{j},\infty}(b)\in p_{n_{j},\infty}(E_{b})\subset W_{b}\in\mathcal{W}$. Thus $f\circ p_{n_{j},\infty}(b)\in f\circ p_{n_{j},\infty}(E_{b})\subset f(W_{b})\subset V_{2}=V_{W_{b}}\in\mathcal{V}_{1}$. Now $x\in E_{b}\cap B_{l}\subset E_{b}\cap P_{l}\subset E_{b}\cap I^{n_{l}}$, so $p_{n_{j},\infty}(x)=p_{n_{j}}^{n_{l}}(x)\in p_{n_{j},\infty}(E_{b})$, and we see that $f\circ p_{n_{j}}^{n_{l}}(x)\in f\circ p_{n_{j},\infty}(E_{b})\subset V_{2}$. Hence, $\{f\circ p_{n_{j},\infty}(b),f\circ p_{n_{j}}^{n_{l}}(x)\}\subset V_{2}$. Since $\mathcal{V}_{1}$ is a star-refinement of $\mathcal{V}$, $f\circ p_{n_{j},\infty}(b)\in V_{1}\cap V_{2}$, $h_{0}(x)\in V_{1}$, and $f\circ p_{n_{j}}^{n_{l}}(x)\in V_{2}$, one may find $V\in\mathcal{V}$ with $\{f\circ p_{n_{j}}^{n_{l}}(x),h_{0}(x)\}\subset V_{1}\cup V_{2}\subset V$. Our proof of (1) is complete.

Now we must prove (2). Let $S$ be an $n$-regular subdivision of $L_{j}$. Suppose that $\sigma\in L_{j}$ with $\dim\sigma=n+1$. Let $M=|N(\partial\sigma,S)|$, and treat $M$ as a finite $\operatorname{CW}$-subcomplex of $K_{\sigma,+}$. By Lemma 4.10(3), $X\tau K_{\sigma,+}$. Denote $f_{\sigma}=\operatorname{id}:M\to M$, and put $B_{j}=M$. Apply (1) to these data to get $k_{\sigma}\geq j$ as indicated there. Put $k=\max\{k_{\sigma}\,|\,(\sigma\in L_{j})\wedge(\dim\sigma=n+1)\}$. Hence for all $\sigma\in L_{j}$ with $\dim\sigma=n+1$ and $l\geq k$, there is a map $f_{\sigma}^{*}:P_{l}\to K_{\sigma,+}$ that extends the composition $f_{\sigma}\circ p_{n_{j}}^{n_{l}}|(p_{n_{j}}^{n_{l}})^{-1}(|N(\partial\sigma,S)|):(p_{n_{j}}^{n_{l}})^{-1}(|N(\partial\sigma,S)|)\to|N(\partial\sigma,S)|$.

Suppose that $l\geq k$. Let $E=P_{l}$, $\overline{p}=p_{n_{j}}^{n_{l}}|P_{l}:P_{l}\to P_{j}$, and $N_{l}$ be a triangulation of $P_{l}$ that admits a simplicial approximation $\overline{p}_{j}:|N_{l}|\to|L_{j}|$ of the map $\overline{p}$. Now apply Proposition 8.1 to get a map $g_{K}:|N_{l}^{(n+1)}|\to|L_{j}^{(n+1)}|$ as requested. ∎

10. Epsilons, Deltas, Maps, and Fibers

We rely on the notation established in the first part of Section 9 concerning the Hilbert cube $I^{\infty}$ and its metric $\rho$. Throughout this section $X\subset I^{\infty}$ will denote a nonempty compactum. Assume in addition that we are given an increasing sequence $(n_{j})$ in $\operatorname{{\mathbb{N}}}$ and a sequence $(P_{j})$ of nonempty compact polyhedra $P_{j}\subset I^{n_{j}}$ satisfying $(1)$-$(3)$ of $\mathrm{Lemma\,\,\ref{intersectinH}}$. The following technical lemma is similar to Lemma 3.1 of [AJR] (repeated in [RT] as Lemma 2.1).

Lemma 10.1**.**

Suppose that for each $j\in\operatorname{{\mathbb{N}}}$ we have selected a closed subset $T_{j}$ of $P_{j}$, $\delta_{j}>0$, $\epsilon_{j}>0$, and a map $g_{j}^{j+1}:T_{j+1}\to T_{j}$ so that:

1. (1)

if $u$, $v\in I^{\infty}$ and $\rho(u,v)\leq\epsilon_{j+1}$, then $\rho(p_{n_{j},\infty}(u),p_{n_{j},\infty}(v))<\delta_{j}$, 2. (2)

$9\cdot 2^{-n_{j}}<\epsilon_{j}$, 3. (3)

$\rho(g_{j}^{j+1}(u),p_{n_{j}}^{n_{j+1}}(u))<\delta_{j}$* for all $u\in T_{j+1}$, and* 4. (4)

$0<\delta_{j}<2^{1-n_{j}}=2\cdot 2^{-n_{j}}$.

Put $\mathbf{T}=(T_{j},g_{j}^{j+1})$, and $Z=\lim\mathbf{T}$. Then $Z$ is a metrizable compactum, and for each $z=(a_{1},a_{2},\dots)\in Z\subset\prod_{j=1}^{\infty}T_{j}$,

1. $(*_{1})$

the associated sequence $(a_{j})$ is a Cauchy sequence in $I^{\infty}$ whose limit lies in $X$, and 2. $(*_{2})$

the function $\pi:Z\to X$ given by $\displaystyle\pi(z)=\lim_{j\to\infty}(a_{j})$ is a map.

Fix $x\in X$ and for each $j\in\operatorname{{\mathbb{N}}}$, let $B_{x,j}=\overline{N}(p_{n_{j},\infty}(x),2\delta_{j})\cap T_{j},B_{x,j}^{\#}=\overline{N}(p_{n_{j},\infty}(x),\epsilon_{j})\cap T_{j}$. Then,

1. $(*_{3})$

$B_{x,j}$, $B_{x,j}^{\#}$ are closed in $T_{j}$ and $B_{x,j}\subset B_{x,j}^{\#}\subset T_{j}\subset P_{j}$, and 2. $(*_{4})$

$g_{j}^{j+1}(B_{x,j+1}^{\#})\subset B_{x,j}$.

If we let $\mathbb{T}_{x}=(B_{x,j},g_{j}^{j+1})$ and $\mathbb{T}_{x}^{\#}=(B_{x,j}^{\#},g_{j}^{j+1})$, then,

1. $(*_{5})$

$\lim\mathbb{T}_{x}=\lim\mathbb{T}_{x}^{\#}\subset Z$, 2. $(*_{6})$

$\pi^{-1}(x)=\lim\mathbb{T}_{x}$, 3. $(*_{7})$

if for each $j$ one has chosen $E_{x,j}$ with $B_{x,j}\subset E_{x,j}\subset B_{x,j}^{\#}$, then $g_{j}^{j+1}(E_{x,j+1})\subset E_{x,j}$ and with $\mathbb{E}_{x}=(E_{x,j},g_{j}^{j+1})$, $\pi^{-1}(x)=\lim\mathbb{E}_{x}=\lim\mathbb{T}_{x}$, and lastly 4. $(*_{8})$

if $B_{x,j}\neq\emptyset$ for each $j$, then $\pi^{-1}(x)\neq\emptyset$.

Proof.

Since each $T_{j}$ is a metrizable compactum, then by Theorem 2.11(1), $Z=\lim\mathbf{T}$ is a metrizable compactum. Employing $({\dagger}_{5})$ of Section 9, we have,

$({\dagger}_{1})$ for all $x\in I^{\infty}$ and $j\in\operatorname{{\mathbb{N}}}$, $\rho(p_{n_{j},\infty}(x),x)=\rho(p_{n_{j}}^{n_{j+1}}\circ p_{n_{j+1},\infty}(x),x)\leq 2^{-n_{j}}$.

The triangle inequality along with $({\dagger}_{1})$, (3), and (4) show that, independently of choice of $z=(a_{1},a_{2},\dots)\in Z$, for all $j\in\operatorname{{\mathbb{N}}}$ $\rho(a_{j},a_{j+1})=\rho(g_{j}^{j+1}(a_{j+1}),a_{j+1})\leq\rho(g_{j}^{j+1}(a_{j+1}),p_{n_{j}}^{n_{j+1}}(a_{j+1}))+\rho(p_{n_{j}}^{n_{j+1}}(a_{j+1}),a_{j+1})=\rho(g_{j}^{j+1}(a_{j+1}),p_{n_{j}}^{n_{j+1}}(a_{j+1}))+\rho(p_{n_{j},\infty}(a_{j+1}),a_{j+1})<\delta_{j}+2^{-n_{j}}<2^{1-n_{j}}+2^{-n_{j}}=2\cdot 2^{-n_{j}}+2^{-n_{j}}=3\cdot 2^{-n_{j}}<2^{2}\cdot 2^{-n_{j}}$, so

$({\dagger}_{2})$ independently of choice of $z=(a_{1},a_{2},\dots)\in Z$, for all $j\in\operatorname{{\mathbb{N}}}$ $\rho(a_{j},a_{j+1})<2^{2-n_{j}}$.

From $({\dagger}_{2})$ and the fact that $(n_{j})$ is increasing, such $(a_{j})$ is a Cauchy sequence in $I^{\infty}$. An application of Lemma 9.2(3) shows that for each $j$, $P_{j+1}^{[\infty]}\subset P_{j}^{[\infty]}$. By Lemma 9.2 $(*_{1})$, $X=\bigcap_{j=1}^{\infty}P_{j}^{[\infty]}\subset I^{\infty}$. Since $a_{j}\in T_{j}\subset P_{j}\subset P_{j}^{[\infty]}$ for each $j$, one concludes the validity of $(*_{1})$. The statement $(*_{2})$ is true since $({\dagger}_{2})$ shows that $\pi$ is the limit of the uniformly convergent sequence of maps $\pi_{j}:Z\to I^{n_{j}}\subset I^{\infty}$ where $\pi_{j}(z)=a_{j}$ whenever $z=(a_{1},a_{2},\dots)\in Z$. We prove $(*_{3})$ by noting that (4) and (2) imply that $2\delta_{j}<\epsilon_{j}$ so that $B_{x,j}\subset B_{x,j}^{\#}$.

Next let $u\in B_{x,j+1}^{\#}\subset\overline{N}(p_{n_{j+1},\infty}(x),\epsilon_{j+1})$. Thus,

$({\dagger}_{3})$ $\rho(u,p_{n_{j+1},\infty}(x))\leq\epsilon_{j+1}$, and $u\in T_{j+1}$.

By hypothesis, $g_{j}^{j+1}(u)\in T_{j}$. So it remains to prove that

$({\dagger}_{4})$ $\rho(g_{j}^{j+1}(u),p_{n_{j},\infty}(x))<2\delta_{j}$.

As a consequence of (1) and the first part of $({\dagger}_{3})$, $\rho(p_{n_{j},\infty}(u),p_{n_{j},\infty}\circ p_{n_{j+1},\infty}(x))=\rho(p_{n_{j}}^{n_{j+1}}(u),p_{n_{j},\infty}(x))<\delta_{j}$. This, the triangle inequality, and (3) show that $\rho(g_{i}^{j+1}(u),p_{n_{j},\infty}(x))\leq\rho(g_{i}^{j+1}(u),p_{n_{j},\infty}(u))+\rho(p_{n_{j},\infty}(u),\break p_{n_{j},\infty}(x))=\rho(g_{j}^{j+1}(u),p_{n_{j}}^{n_{j+1}}(u))+\rho(p_{n_{j}}^{n_{j+1}}(u),p_{n_{j},\infty}(x))<\delta_{j}+\delta_{j}=2\delta_{j}$. This validates $({\dagger}_{4})$ and hence establishes $(*_{4})$. Item $(*_{5})$ is an immediate consequence of $(*_{3})$ and $(*_{4})$.

We now want to prove $(*_{6})$, that $\pi^{-1}(x)=\lim\mathbb{T}_{x}$. If $a=(a_{1},a_{2},\dots)$ is a thread of $\mathbb{T}_{x}$, then for $j\in\operatorname{{\mathbb{N}}}$, $a_{j}\in B_{x,j}$, and hence $\rho(a_{j},p_{n_{j},\infty}(x))\leq 2\delta_{j}$. By applying this, $({\dagger}_{1})$, and (4), $\rho(a_{j},x)\leq\rho(a_{j},p_{n_{j},\infty}(x))+\rho(p_{n_{j},\infty}(x),x)\leq 2\delta_{j}+2^{-n_{j}}\leq 5\cdot 2^{-n_{j}}$. Hence, $\pi(a)=\lim(a_{j})=x$, so $\lim\mathbb{T}_{x}\subset\pi^{-1}(x)$.

Towards the opposite inclusion, suppose that a thread $(a_{1},a_{2},\dots)$ of $\mathbb{T}$ lies in $\pi^{-1}(x)$. Apply the triangle inequality, the fact that $(a_{j})$ converges to $x$, $({\dagger}_{1})$, $({\dagger}_{2})$, (4), and (2) to see that when $j>1$, $\rho(a_{j},p_{n_{j},\infty}(x))\leq\rho(a_{j},x)+\rho(x,p_{n_{j},\infty}(x))\leq\sum_{k=j}^{\infty}\rho(a_{k},a_{k+1})+2^{-n_{j}}<\sum_{k=j}^{\infty}2^{2-n_{k}}+2^{-n_{j}}\leq 2\cdot 2^{2-n_{j}}+2^{-n_{j}}=9\cdot 2^{-n_{j}}<\epsilon_{j}$. This puts $a_{j}\in B_{x,j}^{\#}$. So $(a_{1},a_{2},\dots)\in\lim\mathbb{T}_{x}^{\#}=\lim\mathbb{T}_{x}$, showing that $\pi^{-1}(x)\subset\lim\mathbb{T}_{x}$. Hence $\pi^{-1}(x)=\lim\mathbb{T}_{x}$ as we had proclaimed.

The statement $(*_{7})$ is plainly true. Employing Theorem 2.11(3) and $(*_{6})$, one sees that if $B_{x,j}\neq\emptyset$ for each $j$, then $\lim\mathbb{T}_{x}=\pi^{-1}(x)\neq\emptyset$, which establishes $(*_{8})$. ∎

Lemma 10.2**.**

Let $j\in\operatorname{{\mathbb{N}}}$, $\epsilon_{j}>0$, $N_{j}$ a triangulation of $P_{j}$ with $2\cdot\operatorname{mesh}N_{j}<\epsilon_{j}$, $\lambda_{j}$ be a Lebesgue number of the open cover $\mathcal{N}_{j}=\{\operatorname{st}(v,N_{j})\,|\,v\in N_{j}^{(0)}\}$ of $P_{j}$, $\delta_{j}>0$, and $4\cdot\delta_{j}<\lambda_{j}$. Then for all $x\in X$, there exists $v_{x,j}\in\operatorname{{\mathbb{N}}}_{j}^{(0)}$ such that $\overline{N}(p_{n_{j},\infty}(x),2\delta_{j})\subset\overline{\operatorname{st}}(v_{x,j},N_{j})\subset\overline{N}(p_{n_{j},\infty}(x),\epsilon_{j})$.

Proof.

By $(*_{1})$ of Lemma 9.2, $x\in P_{j}^{[\infty]}$, so $p_{n_{j},\infty}(x)\in P_{j}$. It follows from the fact that $4\cdot\delta_{j}<\lambda_{j}$ and the definition of $\mathcal{N}_{j}$ that there exists $v_{x,j}\in N_{j}^{(0)}$ such that $\overline{N}(p_{n_{j},\infty}(x),2\delta_{j})\subset\overline{\operatorname{st}}(v_{x,j},N_{j})$. Now, $p_{n_{j},\infty}(x)\in\overline{\operatorname{st}}(v_{x,j},N_{j})$, and $2\cdot\operatorname{mesh}(N_{j})<\epsilon_{j}$. This implies that $\overline{\operatorname{st}}(v_{x,j},N_{j})\subset\overline{N}(p_{n_{j},\infty}(x),\epsilon_{j})$. ∎

Lemma 10.3**.**

Let $n\in\operatorname{{\mathbb{N}}}$, $K\in\{K(\operatorname{{\mathbb{Z}}},n),K(\operatorname{{\mathbb{Z}}}/p,n),K(\operatorname{{\mathbb{Q}}},n)\})$, and $j\in\operatorname{{\mathbb{N}}}$. Suppose that $\epsilon_{j}>0$ and a triangulation $N_{j}$ of $P_{j}$ have been chosen so that $2\cdot\operatorname{mesh}(N_{j})<\epsilon_{j}$. Then there exist $\delta_{j}$, $L_{j}$, and $\varphi_{j}$, and for any given $m\in\operatorname{{\mathbb{N}}}$, there exist $l\in\operatorname{{\mathbb{N}}}_{\geq m}$ and $\epsilon_{l}$ such that:

1. (1)

$0<\delta_{j}<2^{1-n_{j}}=2\cdot 2^{-n_{j}}$, 2. (2)

$L_{j}$* is a subdivision of $N_{j}$ with $\operatorname{mesh}L_{j}<\delta_{j}$,* 3. (3)

$\varphi_{j}:|L_{j}|\to|N_{j}|$* is a simplicial approximation of $\operatorname{id}_{P_{j}}$,* 4. (4)

for each $x\in X$, there exists $v_{x,j}\in N_{j}^{(0)}\subset L_{j}^{(0)}$ such that we have $\overline{N}(p_{n_{j},\infty}(x),2\delta_{j})\subset\overline{\operatorname{st}}(v_{x,j},N_{j})\subset\overline{N}(p_{n_{j},\infty}(x),\epsilon_{j}),$ 5. (5)

$j<l$, 6. (6)

$9\cdot 2^{-n_{l}}<\epsilon_{l}$, 7. (7)

if $u$, $v\in I^{\infty}$ and $\rho(u,v)\leq\epsilon_{l}$, then $\rho(p_{n_{j},\infty}(u),p_{n_{j},\infty}(v))<\delta_{j}$, 8. (8)

$N_{l}$* is a triangulation of $P_{l}$ with $2\cdot\operatorname{mesh}N_{l}<\epsilon_{l}$ that admits a simplicial approximation $\overline{p}_{j}:|N_{l}|\to|L_{j}|$ of $p_{n_{j}}^{n_{l}}:P_{l}\to P_{j}$, and* 9. (9)

there is a map $g_{K}=g_{j,K}:|N_{l}^{(n+1)}|\to|L_{j}^{(n+1)}|$ as in $\mathrm{Lemma\,\,\ref{endset}(2)}$ with respect to the current $(P_{l},N_{l},L_{j},p_{n_{j}}^{n_{l}},\overline{p}_{j})$, 10. (10)

for any subdivision $L_{l}$ of $N_{l}$ and simplicial approximation $\varphi_{l}:|L_{l}|\to|N_{l}|$ of $\operatorname{id}_{P_{l}}$, $g_{j,K}(\varphi_{l}(|L_{l}^{(s)}|))\subset|L_{j}^{(s)}|$ for $s\in\{n,n+1\}$, 11. (11)

for all $x\in|L_{l}^{(n+1)}|$, there exists $\sigma\in L_{j}^{(n+1)}$ such that $\{p_{n_{j}}^{n_{l}}(x),g_{K}\circ\varphi_{l}(x)\}\subset\sigma$, and 12. (12)

for all $x\in|L_{l}^{(n+1)}|$, $\rho(p_{n_{j}}^{n_{l}}(x),g_{K}\circ\varphi_{l}(x))<\delta_{j}$.

Proof.

Let $\lambda_{j}$ be a Lebesgue number of the open cover $\mathcal{N}_{j}=\{\operatorname{st}(v,N_{j})\,|\,v\in N_{j}^{(0)}\}$ of $P_{j}$, and find $\delta_{j}>0$ so that (1) and $4\cdot\delta_{j}<\lambda_{j}$ are true. Select $L_{j}$ as in (2) and using Lemma 3.1(1), obtain $\varphi_{j}$ as needed in (3). According to Lemma 10.2, (4) holds true. Using Lemma 9.3(2), one can obtain $l$ so that (5)-(9) are in effect. Proposition 8.1(2) gives us (10), (10) and Proposition 8.1(3) give us (11); (11) and (2) lead to (12). ∎

11. Recursion

Lemma 11.1**.**

Let $X\subset I^{\infty}$ be a nonempty compactum and $n\in\operatorname{{\mathbb{N}}}$. Suppose that $(n_{j})$ is an increasing sequence in $\operatorname{{\mathbb{N}}}$, $(P_{j})$ is a sequence of nonempty compact polyhedra $P_{j}\subset I^{n_{j}}$ satisfying $(1)$-$(3)$ of $\mathrm{Lemma\,\,\ref{intersectinH}}$, and

[TABLE]

with $X\tau K$. Let $\epsilon_{1}=2^{-1}\cdot 9\cdot 2^{-n_{1}}$, and $N_{1}$ be a triangulation of $P_{1}$ with $2\cdot\operatorname{mesh}N_{1}<\epsilon_{1}$. Then there exist a function $l:\operatorname{{\mathbb{N}}}\to\operatorname{{\mathbb{N}}}$ with $l(1)=1$, a sequence $(\epsilon_{j},N_{j},\delta_{j},L_{j},\varphi_{j},\overline{p}_{j},g_{j,K})$, and for each $x\in X$, a sequence $(v_{x,j},N_{x,j},L_{x,j})$ such that for all $j\in\operatorname{{\mathbb{N}}}$,

1. (1)

$0<\delta_{j}<2^{1-n_{l(j)}}=2\cdot 2^{-n_{l(j)}}$, 2. (2)

$N_{j}$* is a triangulation of $P_{l(j)}$ and $L_{j}$ is a subdivision of $N_{j}$ with $\operatorname{mesh}L_{j}<\delta_{j}$,* 3. (3)

$\varphi_{j}:|L_{j}|\to|N_{j}|$* is a simplicial approximation to $\operatorname{id}_{P_{l(j)}}$,* 4. (4)

$v_{x,j}\in N_{j}^{(0)}\subset L_{j}^{(0)}$* and we have $\overline{N}(p_{n_{l(j)},\infty}(x),2\delta_{j})\subset\overline{\operatorname{st}}(v_{x,j},N_{j})\subset\overline{N}(p_{n_{l(j)},\infty}(x),\epsilon_{j})$,* 5. (5)

$l(j)<l(j+1)$, 6. (6)

$9\cdot 2^{-n_{l(j)}}<\epsilon_{j}$, 7. (7)

if $u$, $v\in I^{\infty}$ and $\rho(u,v)\leq\epsilon_{j+1}$, then $\rho(p_{n_{l(j)},\infty}(u),p_{n_{l(j)},\infty}(v))<\delta_{j}$, 8. (8)

$N_{j}$* is a triangulation of $P_{l(j)}$ with $2\cdot\operatorname{mesh}N_{j}<\epsilon_{j}$ that admits a simplicial approximation $\overline{p}_{j}:|N_{j+1}|\to|L_{j}|$ of $p_{n_{l(j)}}^{n_{l(j+1)}}:P_{l(j+1)}\to P_{l(j)}$,* 9. (9)

there is a map $g_{j,K}:|N_{j+1}^{(n+1)}|\to|L_{j}^{(n+1)}|$ as in $\mathrm{Lemma\,\,\ref{endset}(2)}$ with $(P_{l(j+1)},N_{j+1},L_{j},p_{n_{l(j)}}^{n_{l(j+1)}},\overline{p}_{j})$ playing the role of $(P_{l},N_{l},L_{j},p_{n_{j}}^{n_{l}},\overline{p}_{j})$ of that lemma, 10. (10)

$g_{j,K}(\varphi_{j+1}(|L_{j+1}^{(s)}|))\subset|L_{j}^{(s)}|$* for $s\in\{n,n+1\}$,* 11. (11)

if $x\in|L_{j+1}^{(n+1)}|$, there exists $\sigma\in L_{j}^{(n+1)}$ such that $\{p_{n_{l(j)}}^{n_{l(j+1)}}(x),g_{j,K}\circ\varphi_{j+1}(x)\}\subset\sigma$, and 12. (12)

if $x\in|L_{j+1}^{(n+1)}|$, $\rho(p_{n_{l(j)}}^{n_{l(j+1)}}(x),g_{j,K}\circ\varphi_{j+1}(x))<\delta_{j}$.

Proof.

Define $l(1)=1$ and $\epsilon_{1}=2^{-1}\cdot 9\cdot 2^{-n_{l(1)}}$; select a triangulation $N_{1}$ of $P_{l(1)}$ with $2\cdot\operatorname{mesh}N_{1}<\epsilon_{1}$. Then apply Lemma 10.3 recursively to obtain this result. ∎

Lemma 11.2**.**

Under the hypothesis of $\mathrm{Lemma\,\,\ref{newtechnicalrecur}}$, for each $j\in\operatorname{{\mathbb{N}}}$, let $N_{x,j}$, $L_{x,j}$ respectively be defined as the subcomplexes of $N_{j}$ and $L_{j}$ such that $\overline{\operatorname{st}}(v_{x,j},N_{j})=|N_{x,j}|=|L_{x,j}|$. Then,

1. (1)

$\overline{\operatorname{st}}(v_{x,j},N_{j})=|N_{x,j}|=|L_{x,j}|$* is a contractible metrizable continuum,* 2. (2)

for all $s\geq 0$, $\overline{\operatorname{st}}(v_{x,j},N_{j})\cap|L_{j}^{(s)}|=|L_{x,j}^{(s)}|$, 3. (3)

$v_{x,j}\in\overline{\operatorname{st}}(v_{x,j},N_{j})\cap|L_{x,j}^{(0)}|$, so $\overline{\operatorname{st}}(v_{x,j},N_{j})\cap|L_{x,j}^{(0)}|\neq\emptyset$, 4. (4)

for all $s\geq 1$, $|L_{x,j}^{(s)}|$ is a metrizable continuum and for all $s\geq 0$, $|L_{x,j}^{(s)}|$ contracts to a point in $|L_{x,j}^{(s+1)}|$, 5. (5)

for any abelian group $G$ and $1\leq s<n$, $H_{s}(|L_{x,j}^{(n)}|;G)=0$, 6. (6)

for any abelian group $G$ and $s>n$, $H_{s}(|L_{x,j}^{(n)}|;G)=0$, 7. (7)

$H_{n}((|L_{x,j}^{(n)}|;\operatorname{{\mathbb{Z}}})$* is free abelian of finite rank, and* 8. (8)

$H_{0}(|L_{x,j}^{(n)}|;\operatorname{{\mathbb{Z}}})$* is isomorphic to $\operatorname{{\mathbb{Z}}}$, so it is free abelian of finite rank.*

Proof.

As mentioned in Section 3, $\overline{\operatorname{st}}(v_{x,j},N_{j})$ is a contractible closed subset of the compact polyhedron $|N_{j}|$, so it is a contractible metrizable continuum and we get (1). It is plain that (2) is true. Surely $v_{x,j}\in\overline{\operatorname{st}}(v_{x,j},N_{j})\cap|N_{x,j}^{(0)}|\subset\overline{\operatorname{st}}(v_{x,j},N_{j})\cap|L_{x,j}^{(0)}|$, so we have (3). It follows from (1) that $|L_{x,j}^{(s)}|$ is connected, and since it is closed in $|L_{x,j}|$, it is a metrizable continuum. The rest of (4) is a result of Lemma 3.2. We get (5) from (1) and Lemma 3.4, and (6) is of course true since $\dim|L_{x,j}^{(n)}|\leq n<s$. Lemma 3.9 gives us (7). We get (8) from (4). ∎

12. Proof of Main Results

We now provide our coordinated proofs of Theorems 1.1, 1.2, and 1.3.

Proof.

Let $K\in\{K(\operatorname{{\mathbb{Z}}},n),K(\operatorname{{\mathbb{Z}}}/p,n),K(\operatorname{{\mathbb{Q}}},n)\})$, $n\in\operatorname{{\mathbb{N}}}$, and $X\subset I^{\infty}$ be a compactum with $X\tau K$. The reader should keep in mind that by Lemma 6.1, $\dim_{G}X\leq n$ is equivalent to $X\tau K(G,n)$. Use the notation and facts that were developed in Lemmas 11.1 and 11.2 as applied to the current $X$ and $n$. From Lemma 11.1(5), one sees that

$({\dagger}_{1})$ $(n_{l(j)})$ is an increasing sequence in $\operatorname{{\mathbb{N}}}$.

For each $j\in\operatorname{{\mathbb{N}}}$, let $T_{j}=|L_{j}^{(n)}|$. Then, an application of Lemma 11.1(2) to the fact that $|L_{j}|=P_{l(j)}\neq\emptyset$ shows that

$({\dagger}_{2})$ for all $j\in\operatorname{{\mathbb{N}}}$, $T_{j}$ is a nonempty compact polyhedron with $\dim T_{j}\leq n$.

Using Lemma 11.1(10), define

[TABLE]

This gives us an inverse sequence $\mathbf{T}=(T_{j},g_{j}^{j+1})$ (depending on $K$) of nonempty compact polyhedra (see $({\dagger}_{2})$). Let $Z=\lim\mathbf{T}$. The first thing to observe is that since $T_{j}$ is a nonempty compact polyhedron and $\dim T_{j}\leq n$ for each $j$, then by Theorem 2.11,

$({\dagger}_{3})$ $Z$ is a nonempty metrizable compactum, and $\dim Z\leq n$.

Items (7), (6), (12), and (1) of Lemma 11.1, respectively, give us (1)-(4) of Lemma 10.1, which then provides us with a map $\pi:Z\to X$. An internal description of the fibers of $\pi$ is now in order.

Now fix $x\in X$. Using Lemma 11.1(4) and Lemma 11.2(2), one sees that for all $j\in\operatorname{{\mathbb{N}}}$, $\overline{N}(p_{n_{l(j)},\infty}(x),2\cdot\delta_{j})\cap T_{j}=\overline{N}(p_{n_{l(j)},\infty}(x),2\cdot\delta_{j})\cap|L_{j}^{(n)}|\subset\overline{\operatorname{st}}(v_{x,j},N_{j})\cap T_{j}=\overline{\operatorname{st}}(v_{x,j},N_{j})\cap|L_{j}^{(n)}|=|L_{x,j}^{(n)}|\subset\overline{N}(p_{n_{l(j)},\infty}(x),\epsilon_{j})\cap T_{j}$. The leftmost and rightmost sets in the preceding expression correspond respectively to $B_{x,j}$ and $B_{x,j}^{\#}$ in Lemma 10.1. Looking at $(*_{7})$ of Lemma 10.1 with $E_{x,j}=|L_{x,j}^{(n)}|$, one sees that $\mathbf{E}_{x}=(|L_{x,j}^{(n)}|,g_{j}^{j+1})$ is an inverse sequence such that $\pi^{-1}(x)=\lim\mathbf{E}_{x}$. An application of Lemma 11.2(3,4) shows that each coordinate space in $\mathbf{E}_{x}$ is nonempty and each is a metrizable continuum; therefore,

$({\dagger}_{4})$ for each $x\in X$, $\pi^{-1}(x)$ is a nonempty metrizable continuum.

Proof of Theorem 1.1. This time $K=K(\operatorname{{\mathbb{Z}}},n)$. Using 11.1(9), we may apply Proposition 8.1(8) to see that each $g_{j}^{j+1}$ in $\mathbf{E}_{x}$ is null homotopic. Apply Lemma 2.4 to complete this part of the proof.

Proof of Theorems 1.2 and 1.3. In these cases we have that $K\in\{K(\operatorname{{\mathbb{Z}}}/p,n),K(\operatorname{{\mathbb{Q}}},n)\})$. By Proposition 8.1(9,10)

$({\dagger}_{5})$ if $K=K(\operatorname{{\mathbb{Z}}}/p,n)$, then the induced homomorphism $H_{n}(g_{j}^{j+1};\operatorname{{\mathbb{Z}}}/p):H_{n}(|L_{x,j+1}^{(n)}|;\operatorname{{\mathbb{Z}}}/p)\to H_{n}(|L_{x,j}^{(n)}|;\operatorname{{\mathbb{Z}}}/p)$ is trivial, and

$({\dagger}_{6})$ if $K=K(\operatorname{{\mathbb{Q}}},n)$, then the induced homomorphism $H_{n}(g_{j}^{j+1};\operatorname{{\mathbb{Z}}}):H_{n}(|L_{x,j+1}^{(n)}|;\operatorname{{\mathbb{Z}}})\to H_{n}(|L_{x,j}^{(n)}|;\operatorname{{\mathbb{Z}}})$ is trivial.

Our proof of Theorem 1.2 follows from $({\dagger}_{5})$, Lemma 11.2(5,6,7) (Lemma 11.2(8) in case $n=1$), and Lemma 2.9. Our proof of Theorem 1.3 follows from $({\dagger}_{6})$, Lemma 11.2(5,6) (remember that $n\geq 2$ in this setting), and Lemma 2.7. ∎