Fundamental groups and path lifting for algebraic varieties

TL;DR

This paper investigates fundamental groups of algebraic varieties, focusing on how surjectivity on π₁ behaves under base change, the relationship between Zariski and Euclidean topologies, and the conditions for path lifting in morphisms.

Contribution

It provides new insights into the behavior of fundamental groups under base change, compares topological notions, and characterizes morphisms with path lifting properties.

Findings

Surjectivity on π₁ is preserved under certain base changes.

Openness in Zariski topology relates to Euclidean topology.

Criteria for morphisms to have the path lifting property.

Abstract

We study 3 basic questions about fundamental groups of algebraic varieties. For a morphism, is being surjective on $\pi_1$ preserved by base change? What is the connection between openness in the Zariski and in the Euclidean topologies? Which morphisms have the path lifting property?

Equations

$$(x_1,y),(x_1,y^{\prime })\in {\pi }_1(W)\text{mbox}and(x_2,y),(x_1,y^{\prime })\in {\pi }_2(W).$$

$$\begin{array}[]{ccc}(x^{\prime},X^{\prime})&\stackrel{{\scriptstyle h^{\prime}}}{{\longrightarrow}}&(x,X)\\ g^{\prime}\downarrow\hphantom{g^{\prime}}&&\hphantom{g}\downarrow g\\ (s^{\prime},S^{\prime})&\stackrel{{\scriptstyle h}}{{\longrightarrow}}&(s,S),\end{array}$$

$$\pi :({\mathbb{A}}_1^1\times {\mathbb{P}}^1){⨿}_{(0_1,p)=(0_2,p)}({\mathbb{A}}_2^1\times {\mathbb{P}}^1)\stackrel{\rho }{\to }{\mathbb{A}}_1^1{⨿}_{0_1=0_2}{\mathbb{A}}_2^1\stackrel{\sigma }{\to }{\mathbb{A}}^1.$$

$$s\in S\setminus h(Z\setminus U)\subset h(Z\cap U)\subset g(U)$$

$$Z{\times }_S{S}^{\prime }=W\coprod {⨿}_i(z_i^{\prime },Z_i^{\prime }).$$

$$X{\times }_S{S}^{\prime }=W\coprod {⨿}_i(x_i^{\prime },X_i^{\prime }),$$

$$X{\times }_S{S}^{\prime }={⨿}_i(x_i^{\prime },X_i^{\prime }),$$

$$X{\times }_S{S}^{\prime }=W\coprod {⨿}_i(x_i^{\prime },X_i^{\prime }),$$

$$\mathrm{Pic}(Y^{\prime })\to \mathrm{Pic}(E^{\prime })\cong \mathbb{Z}+{\mathbb{C}}^{\ast }$$

$$(s,S^{0}):=\bigl{(}xyz+x^{4}+y^{4}+z^{4}=0\bigr{)}\subset{\mathbb{A}}^{3}.\qed$$

$$X:=(x_0{y}_0+x_1{y}_1+x_2{y}_2=0)\subset {\mathbb{A}}_{\mathbf{x}}^3\times {\mathbb{P}}_{\mathbf{y}}^2$$

$$X{\times }_{S}X\subset {\mathbb{A}}_{\mathbf{x}}^3\times {\mathbb{P}}_{\mathbf{y}}^2\times {\mathbb{P}}_{\mathbf{z}}^2$$

$$U\setminus \{n\}\sim (0,1)\times L_n.$$

$$\bar{\gamma }(c-{\eta }_2)={\gamma }^{\prime }(c-{\eta }_2)\text{mbox}and\bar{\gamma }(c-{\eta }_1)={\gamma }_c^{\prime \prime }(c-{\eta }_1).$$

$$g^{-1}\bigl{(}Y_{i}\setminus Y_{i-1}\bigr{)}\to\bigl{(}Y_{i}\setminus Y_{i-1}\bigr{)}$$

$$\pi_{1}\bigl{(}X^{(n)}({\mathbb{C}}),x\bigr{)}\to\pi_{1}\bigl{(}X({\mathbb{C}}),x\bigr{)}\to\pi_{1}\bigl{(}Y({\mathbb{C}}),y\bigr{)}.$$

$$\gamma(t):=\left\{\begin{array}[]{l}p\circ\gamma^{+}(t)\quad\mbox{if}\quad\gamma^{+}(t)\geq 0\quad\mbox{and}\\ p\circ\gamma^{-}(-t)\quad\mbox{if}\quad\gamma^{-}(-t)\geq 0\end{array}\right.$$

$$t\mapsto\bigl{(}1-t,(1-t)\sin\bigl{(}\tfrac{1}{1-t}\bigr{)}\bigr{)}$$

Full text

The aim of these notes is to study 3 questions involving maps between the fundemantal groups of algebraic varieties.

• •

Let $X\to Y$ be a morphism of schemes that induces a surjection on the algebraic fundemantal groups. Does the same hold after a base change $X\times_{S}Z\to Y\times_{S}Z$?

• •

Let $X\to Y$ be a morphism between ${\mathbb{C}}$-schemes. When can we lift every continuous path in $Y({\mathbb{C}})$ to a path in $X({\mathbb{C}})$?

• •

Let $X\to Y$ be a morphism between ${\mathbb{C}}$-schemes. What is the connection between openness in the Zariski and in the Euclidean topologies?

An answer to the first question was used in the study of Pell surfaces [Kol19]. While this application involves only maps between algebraic curves, the curves in question are singular and non-proper, and it turns out to be not much harder to consider the general case. This is treated in Section 1.

The proof uses some basic properties of open and universally open morphisms, some of which I did not find in the literature. These are worked out in Section 2.

Various forms of path lifting are studied in Section 3. The answer is most complete for arc lifting (Definition 30) which is equivalent to openness of the morphism in the Euclidean topology and universal openness in the Zariski topology; see Theorem 31.

While the main applications are to schemes of finite type, the discussions in Sections 1–2 are formulated for arbitrary Noetherian schemes.

1. Maps between fundamental groups

Definition 1**.**

Let $X$ be a connected scheme and $x\to X$ a geometric point. The fundamental group of $X$ with base point $x$ is denoted by $\pi_{1}(X,x)$. Working with schemes, we use the algebraic fundamental group.

Let $f:X\to Y$ be a morphism of connected schemes. Fix a base point $x\to X$ and its $f$-image $y\to Y$. We get a natural group homomorphism $f_{*}:\pi_{1}(X,x)\to\pi_{1}(Y,y)$.

1. (1)

We say that $f$ is $\pi_{1}$-surjective if $f_{*}:\pi_{1}(X,x)\to\pi_{1}(Y,y)$ is surjective.

Using the correspondence between quotients of the fundamental group and finite, étale covers we get the following equivalent form.

1. (2)

$f$ is $\pi_{1}$-surjective iff for every connected, finite, étale cover $Y^{\prime}$, the fiber product $X\times_{Y}Y^{\prime}$ is also connected.

The latter formulation shows that the base point can be ignored in this definition. More generally, choosing a different $x^{\prime}\to y$, the image of $\pi_{1}(X,x^{\prime})\to\pi_{1}(Y,y)$ is a conjugate of the image of $\pi_{1}(X,x)\to\pi_{1}(Y,y)$.

One of the main aims is to show that connectedness of the fiber product $X\times_{Y}Y^{\prime}$ also holds if $Y^{\prime}\to Y$ is proper and universally open, see Theorem 2.5. We discuss open and universally open morphisms in Section 2.

Theorem 2**.**

Let $X,S$ be connected schemes and $g:X\to S$ a proper and universally open morphism. The following are equivalent.

1. (1)

$g$* is $\pi_{1}$-surjective.* 2. (2)

$X\times_{S}X$* is connected.* 3. (3)

$X\times_{S}\dots\times_{S}X$* ($n$ copies of $X$) is connected for some $n\geq 2$.* 4. (4)

The number of connected components of $X\times_{S}\dots\times_{S}X$ is bounded, independent of the number of factors. 5. (5)

For every connected scheme $Y$ and proper, universally open morphism $h:Y\to S$, the fiber product $X\times_{S}Y$ is connected. 6. (6)

For every connected scheme $Y$ and proper, universally open morphism $h:Y\to S$, the second projection $X\times_{S}Y\to Y$ is $\pi_{1}$-surjective.

The following example shows that in (2.5–6) we need $Y\to S$ to be universally open. It would not be enough to assume only that $Y\to S$ is finite and open.

Example 3**.**

Let $C$ be a projective, nodal rational curve over an algebraically closed field with normalization $\pi:{\mathbb{P}}^{1}\cong\bar{C}\to C$ and $c\in C$ a smooth point. Note that $\pi_{1}(C,c)\cong{\mathbb{Z}}$. Let $g_{n}:(c_{n},C_{n})\to(c,C)$ be its unique degree $n$, connected, étale cover. For $n,m\geq 1$ set $(x,X_{n,m}):=(c_{n},C_{n})\amalg_{c_{n}=c_{m}}(c_{m},C_{m})$. The maps $g_{n},g_{m}$ glue to $g_{n,m}:(x,X_{n,m})\to(c,C)$. Then

1. (1)

$g_{n,m}$ is $\pi_{1}$-surjective iff $(n,m)=1$, 2. (2)

$\bar{C}\to C$ is finite, open and surjective, yet 3. (3)

$X_{n,m}\times_{C}\bar{C}$ is the disjoint union of $\bar{C}\amalg_{c=c}\bar{C}$ and of $n+m-2$ copies of $\bar{C}$. Thus it is disconnected iff $n+m>2$.

We start the proof of Theorem 2 with a series of remarks and lemmas that establish various special cases, and then use them to settle the general case.

Remark 4** (Stein factorization).**

Let $h:X\to X^{\prime}$ be a proper morphism such that $h_{*}{\mathcal{O}}_{X}={\mathcal{O}}_{X^{\prime}}$. Then $X$ is connected if $X^{\prime}$ is connected. Since $h_{*}$ commutes with any flat base change, (1.2) shows that $h_{*}:\pi_{1}(X,x)\to\pi_{1}(X^{\prime},x^{\prime})$ is an isomorphism.

Applying this to the Stein factorization $g:X\to X^{\prime}\to S$ shows that Theorem 2 holds for proper morphisms iff it holds for finite morphisms.

Lemma 5** (2.2$\Rightarrow$ 2.5).**

Let $X,Y,S$ be connected schemes and $g:X\to S,h:Y\to S$ finite, universally open morphisms. Assume that $X\times_{S}X$ is connected. Then $X\times_{S}Y$ is also connected.

Proof. Fix a geometric point $y\to Y$ and then choose $x_{1},x_{2}\to X$ such that $g(x_{1})=g(x_{2})=h(y)$. Let $W\subset X\times_{S}X\times_{S}Y$ be the connected component that contains $(x_{1},x_{2},y)$. Since $h$ is finite and universally open, so is the projection $\pi_{12}:X\times_{S}X\times_{S}Y\to X\times_{S}X$. Thus $\pi_{12}:W\to X\times_{S}X$ is surjective by (16.3). In particular, there is a point $(x_{1},x_{1},y^{\prime})\in W$ for some $y^{\prime}\to Y$. Consider now the 2 projections $\pi_{i}:W\to X\times_{S}Y$ for $i=1,2$. Note that

[TABLE]

Since $W$ is connected, this shows that $(x_{1},y),(x_{2},y)$ are in the same connected component of $X\times_{S}Y$.

Let $V_{j}\subset X\times_{S}Y$ be the connected components. Projection to $Y$ is universally open, so the projections $V_{j}\to Y$ are surjective by (16.3). All preimages of $y\in Y$ are in the same connected component by the above argument, hence $X\times_{S}Y$ has only 1 connected component. ∎

Using (1.2) this implies the following.

Corollary 6** (2.2 $\Leftrightarrow$ 2.1).**

Let $X,S$ be connected schemes and $g:X\to S$ a finite, universally open morphism. Assume that $X\times_{S}X$ is connected. Then $\pi_{1}(X)\to\pi_{1}(S)$ is surjective. ∎

Lemma 7**.**

Let $g:X\to S$ be a universally open morphism of finite type. Assume that the diagonal $\Delta_{X/S}$ is a connected component of $X\times_{S}X$. Then $g$ uniquely factors as $g:X\to S^{\prime}\to S$ where $X\to S^{\prime}$ is a universal homeomorphism and $S^{\prime}\to S$ is étale.

Proof. Such a factorization is unique, so it is enough to construct it étale locally on $S$. By Corollary 21, after an étale base change we may assume that $g$ is of the form $g:\amalg_{i}(x_{i},X_{i})\to(s,S)$ where each restriction $g_{i}:(x_{i},X_{i})\to(s,S)$ is finite, local and $k(x_{i})/k(s)$ is purely inseparable. Thus $X_{i}\times_{S}X_{i}$ is connected, since all of its irreducible components contain the 1-pointed scheme $x_{i}\times_{s}x_{i}$. Thus $\Delta_{X_{i}/S}=X_{i}\times_{S}X_{i}$, so $k(x^{\prime}_{i})/k(g(x^{\prime}_{i}))$ is also purely inseparable for every $x^{\prime}_{i}\in X_{i}$. Hence $g_{i}$ is a universal homeomorphism by [Gro71, I.3.7–8]. Thus the factorization is $g:\amalg_{i}(x_{i},X_{i})\to\amalg_{i}(s,S)\to(s,S)$. ∎

Lemma 8**.**

Let $X$ be a connected scheme and $g:X\to S$ a finite, universally open morphism. Then $g$ uniquely factors as $g:X\to S^{\prime}\to S$ where $S^{\prime}\to S$ is finite, étale and $X\times_{S^{\prime}}X$ is connected.

Proof. Let $\Delta_{X/S}^{\rm conn}$ denote the connected component of $\Delta_{X/S}$ in $X\times_{S}X$. It is a finite equivalence relation on $X$ and the geometric quotient $S_{1}:=X/\Delta_{X/S}^{\rm conn}$ exists by [Kol12, Lem.17]. The natural map $X\times_{S_{1}}X\to X\times_{S}X$ is a universal homeomorphism onto $\Delta_{X/S}^{\rm conn}$, thus $X\times_{S_{1}}X$ is connected. We apply Lemma 7 to $S_{1}\to S$ to get $S_{1}\to S^{\prime}\to S$ where $S_{1}\to S^{\prime}$ is a universal homeomorphism and $S^{\prime}\to S$ is étale. ∎

Combining Lemma 8 and Corollary 6 we get the finite case of following. For proper morphisms we also use Remark 4.

Corollary 9**.**

Let $g:X\to S$ be a proper, universally open morphism of connected schemes. Let $x\to X$ be a geometric point and $s\to S$ its image. Then $\operatorname{im}[\pi_{1}(X,x)\to\pi_{1}(S,s)]$ has finite index in $\pi_{1}(S,s)$. ∎

The normalization of the nodal plane cubic shows that the conclusion does not hold if $g$ is only assumed finite and open.

10****Proof of Theorem 2.

As we noted in Remark 4, it is enough to prove the special case when $X\to S$ and $Y\to S$ are both finite.

We already proved that 2.1 $\Leftrightarrow$ 2.2 and that they imply 2.5. Setting $X=Y$ shows that 2.5 $\Rightarrow$ 2.2 and we get 2.3 by induction on the number of factors.

We use the shorthand $X^{n}_{S}:=X\times_{S}\dots\times_{S}X$ for the fiber product with $n$ factors. The coordinate projection $\pi_{12}:X^{n}_{S}\to X\times_{S}X$ is surjective, thus 2.3 $\Rightarrow$ 2.2.

Since every connected component of $X^{n}_{S}$ dominates $S$, if $X^{n}_{S}$ has at least $2$ connected components then $X^{mn}_{S}$ has at least $2^{m}$ connected components. Thus (2.3) $\Leftrightarrow$ (2.4).

If $g$ is not $\pi_{1}$-surjective then by Lemma 8 it factors as $X\to S^{\prime}\to S$ where $S^{\prime}\to S$ is finite, étale and of degree $\geq 2$. Setting $Y:=S^{\prime}$ shows that (2.6) $\Rightarrow$ (2.1).

Conversely, assume (2.1) and fix $h:Y\to S$. We already know that $X\times_{S}Y$ is connected. If $X\times_{S}Y\to Y$ is not $\pi_{1}$-surjective then by (1.2) there is a nontrivial, finite étale cover $Y^{\prime}\to Y$ such that $(X\times_{S}Y)\times_{Y}Y^{\prime}$ is not connected. Applying (2.5) to $Y^{\prime}\to S$ we get that $X\times_{S}Y^{\prime}\cong X\times_{S}Y\times_{Y}Y^{\prime}$ is connected, a contradiction. Thus (2.1) $\Rightarrow$ (2.6). ∎

Next we consider 3 variants of $\pi_{1}$-surjectivity.

11****Topological fundamental group.

Let $X$ be a connected ${\mathbb{C}}$-scheme of finite type and $x\in X$ a point. We then have the topological fundamental group \pi^{\rm top}\bigl{(}X({\mathbb{C}}),x\bigr{)}. A morphism $f:X\to Y$ between connected ${\mathbb{C}}$-schemes of finite type is $\pi^{\rm top}$-surjective if $f_{*}:\pi^{\rm top}(X)\to\pi^{\rm top}(Y)$ is surjective.

Although the natural map $\pi^{\rm top}(X)\to\pi_{1}(X)$ can have infinite kernel, as a consequence of Lemma 39 we see that if $f$ is proper and universally open then the index of \operatorname{im}\bigl{[}\pi^{\rm top}(X)\to\pi^{\rm top}(Y)\bigr{]} in $\pi^{\rm top}(Y)$ equals the index of \operatorname{im}\bigl{[}\pi(X)\to\pi(Y)\bigr{]} in $\pi(Y)$. In particular, $f$ is $\pi^{\rm top}$-surjective iff it is $\pi$-surjective. Thus Theorem 2 holds for the topological fundamental group as well.

12****First homology group.

Let $X$ be a connected scheme and $x\to X$ a geometric point. The first homology group, denoted by $H_{1}(X)$, is defined as the abelianization of $\pi_{1}(X,x)$; it is independent of the base point. Since we start with the algebraic fundamental group, $H_{1}(X)$ is a $\hat{\mathbb{Z}}$ module, where $\hat{\mathbb{Z}}\cong\oplus_{p}{\mathbb{Z}}_{p}$ is the profinite completion of ${\mathbb{Z}}$. If $k={\mathbb{C}}$ then $H_{1}(X)$ is the profinite completion of H_{1}\bigl{(}X({\mathbb{C}}),{\mathbb{Z}}\bigr{)}.

We say that $f$ is $H_{1}$-surjective if $f_{*}:H_{1}(X)\to H_{1}(Y)$ is surjective. Note that $\pi_{1}$-surjective implies $H_{1}$-surjective.

In [Kol19] we needed to understand whether $H_{1}$-surjectivity is preserved by base change as in Theorem 2.6. The following example shows that it is not.

Let $X$ be a simply connected manifold (or variety over ${\mathbb{C}}$) on which $A_{n}$ acts freely. Assume that $n\geq 6$ is odd. Let $A_{n-1}\subset A_{n}$ be a point stabilizer and $C_{n}\subset A_{n}$ a subgroup generated by an $n$-cycle. We get a commutative diagram

[TABLE]

which is a fiber product square. Here $g$ is $H_{1}$-surjective but $g^{\prime}$ is not.

This is the reason why, although in [Kol19] the main interest is in $H_{1}$-surjectivity, we needed to understand the base change behaviour of $\pi_{1}$-surjectivity.

13****Tame fundamental group.

Let $X$ be a $k$-scheme and $\operatorname{char}k=p>0$. Let $\pi_{1}^{(p)}(X,x)$ denote the largest prime to $p$ quotient of $\pi_{1}(X,x)$, that is, the inverse limit of all quotients $\pi_{1}(X,x)\twoheadrightarrow H$ where $p\nmid|H|$. We say that $f$ is $\pi_{1}$-surjective modulo $p$ if $f_{*}:\pi_{1}^{(p)}(X)\to\pi_{1}^{(p)}(Y)$ is surjective.

The diagram (12.1) also shows that $\pi_{1}$-surjectivity modulo $p$ is not preserved by base change.

2. Open and universally open maps

We discuss properties of universally open maps that were used in the proofs of Theorems 2 and 31. We aim to treat these in their natural generality and also establish various results that are of independent interest.

Definition 14**.**

A morphism $f:X\to S$ is open at $x\in X$ if for every open $x\in U\subset X$, its image $f(U)$ contains an open neighborhood of $f(x)$. A morphism $f:X\to S$ is open (resp. open along $W\subset X$) if $f$ is open at every $x\in X$ (resp. $x\in W$).

A morphism $f:X\to S$ is universally open at $x\in X$ (resp. along $W\subset X$) if $f_{T}:X\times_{S}T\to T$ is open along $g_{X}^{-1}(x)$ (resp.  $g_{X}^{-1}(W)$) for every $g:T\to S$ where $g_{X}:X\times_{S}T\to X$ is the first projection. We say that $f$ is universally open if it is universally open at every $x\in X$. (Note that the Zariski topology on the product of 2 varieties is not the product topology, this is why open does not imply universally open.)

The following examples are good to keep in mind.

1. (1)

Let $g:X\to Y$ be a morphism of finite type ${\mathbb{C}}$-schemes. We see in Theorem 31 that $g$ is universally open iff $g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$ is open in the Euclidean topology. Thus universal openness should be viewed as the more geometric notion. 2. (2)

If $g:X\to Y$ is open then every irreducible component of $X$ dominates an irreducible component of $Y$. See Lemma 28 for a partial converse statement. 3. (3)

Let $g:C_{1}\to C_{2}$ be a quasi-finite morphism of purely 1-dimensional $k$-schemes. If $C_{2}$ is irreducible then $g$ is open since every dense, constructible subset of $C_{2}$ is open. 4. (4)

Let $p:\bar{X}\to X$ be a finite, birational morphism. Then $p$ is universally open iff it is a bijection on geometric points. 5. (5)

As a consequence we see that if $C$ is an irreducible curve with nodes then the normalization $\bar{C}\to C$ is open but not universally open. 6. (6)

Let $p:\bar{X}\to X$ be a finite, birational morphism. Assume that every irreducible component of $X$ has dimension $\geq 2$. Then $p$ is open $\Leftrightarrow$ $p$ is universally open $\Leftrightarrow$ $p$ is a bijection on geometric points.

Thus the difference between open and universally open appears mostly for 1-dimensional targets. Nonethless, many proofs involve localization and induction on the dimension, so the difference between the 2 notions can be significant.

Example 15** (Openness is not an open property).**

The following examples show that openness of a morphism at a point is not an open property. I assume this is why this notion is not defined in [Sta15, Tag 004E]. Nonetheless, I think that the notion is natural and has several useful properties.

(15.1) Set $X:=(z=0)\cup(y=z-x=0)\subset{\mathbb{C}}^{3}$, $S:=(z=0)$ and $p:X\to S$ the coordinate projection. Then $p$ is universally open at all points of the plane $(z=0)$ but not open at the points of the punctured line $(y=z-x=0)\setminus\{(0,0,0)\}$. It truns out to be a general feature that openness along a fiber says a lot about the maximal dimensional irreducible components of $X$ but very little about the lower dimensional ones; see Theorem 26.

(15.2) Let $Y$ be the pinch point $(x^{2}=y^{2}z)\subset{\mathbb{C}}^{3}$ and $X\cong{\mathbb{C}}^{2}_{uv}$ its normalization. Here $p:X\to Y$ is given by $(u,v)\mapsto(uv,u,v^{2})$. Note that $g$ is universally open at $(0,0)$ (see (17.3) for a more general claim). However it is not open at any other point of the $v$-axis.

Furthermore, although all fibers of $p:X\to Y$ have dimension 0, it does not have pure relative dimension 0. Indeed, the fiber product $X\times_{Y}X$ has 2 irreducible components; one is $X\cong{\mathbb{C}}^{2}$ and the other is isomorphic to ${\mathbb{C}}^{1}$ lying over the $v$-axis.

See Example 40 for other properties of this surface.

16****Basic properties.

The following are some obvious properties.

1. (1)

Let $f:X\to S$ be a morphism and $Z\subset X$ a locally closed subscheme. If $f|_{Z}$ is open at $x\in Z$ then $f$ is open at $x$. 2. (2)

Let $g:Y\to X$ and $f:X\to S$ be morphisms, $y\in Y$.

1. (a)

If $g$ is open at $y$ and $f$ is open at $g(y)$ then $f\circ g$ is open at $y$. 2. (b)

If $f\circ g$ is open at $y$ then $f$ is open at $g(y)$. 3. (3)

Let $g:X\to Y$ be proper and open. Let $X_{i}\subset X$ be a connected component. Then $g(X_{i})$ is closed and open in $Y$, hence a connected component. 4. (4)

If $f:X\to S$ is open at $x\in X$ then $f_{T}:f^{-1}(T)\to T$ is also open at $x$ for every subvariety $f(x)\in T\subset S$. 5. (5)

Let $f:X\to S$ be morphism and $S_{i}\subset S$ the irreducible components. Set $X_{i}:=f^{-1}(S_{i})$. By restriction we get $f_{i}:X_{i}\to S_{i}$. Then $f$ is open (resp. universally open) at $x$ iff $f_{i}$ is open (resp. universally open) at $x$ whenever $x\in X_{i}$. 6. (6)

A flat morphism of finite presentation is universally open [Sta15, Tag 01UA]. This is probably easiest to see using (17.3). 7. (7)

[Sta15, Tag 04R3] Being universally open is étale local on source and target. That is, if we have a commutative diagram

[TABLE]

where $h,h^{\prime}$ are étale, then $g$ is universally open at $x$ iff $g^{\prime}$ is universally open at $x^{\prime}$. 8. (8)

Let $X\to S$ be a proper morphism with Stein factorization $X\to X^{\prime}\to S$. By (2.b), if $X\to S$ is open then so is $X^{\prime}\to S$. The following example shows that $X\to X^{\prime}$ need not be open. Take 2 copies of ${\mathbb{A}}^{1}\times{\mathbb{P}}^{1}$ and and glue them along the points $(0_{i},p)\in{\mathbb{A}}^{1}_{i}\times{\mathbb{P}}^{1}$ for some $p\in{\mathbb{P}}^{1}$ to get $X$. The Stein factorization of the first coordinate projection is

[TABLE]

Note that $\pi$ and $\sigma$ are universally open but $\rho$ is not open.

17****Valuative and base change criteria.

The simplest non-open morphism is the embedding of a closed point into an irreducible curve $\{p\}\lhook\joinrel\to C$. It turns out that this example is quite typical. If there is an irreducible, positive dimensional subvariety $s\in T\subset S$ such that $f^{-1}(T)=f^{-1}(s)$ then, by (16.4), $f^{-1}(T)\to\{s\}\lhook\joinrel\to T$ shows that $f$ is not open at $x$, giving a necessary openness criterion.

We claim that if $f:X\to S$ is of finite type, then the criterion is also sufficient, after a small change.

(17.1) Let $f:X\to S$ be a morphism of finite type, $x\in X$ a point and $s:=f(x)$. Then $f$ is not open at $x$ iff there is an open subset $x\in U\subset X$ and an irreducible subscheme $s\in C\subset S$ such that $s$ has codimension 1 in $C$ and the generic point of $C$ is not contained in $f(U)$. In particular,

[TABLE]

shows that $f$ is not open at $x$.

Proof. Choose $U$ such that $f(U)$ does not contain any open neighborhood of $s$. Since $f(U)$ is constructible, $S\setminus f(U)$ is constructible and its closure contains $s$. There is thus an irreducible component $W\subset S\setminus f(U)$ whose closure contains $s$. Note that $f(U)\cap\bar{W}$ is a nowhere dense constructible subset. Take any irreducible subscheme $\{s\}\in C\subset S$ that is not contained in the closure of $f(U)\cap\bar{W}$. ∎

Since ${\mathcal{O}}_{s,C}$ is dominated by a valuation ring, we can restate (17.1) in the following variant forms.

(17.2) A finite type morphism $f:X\to S$ is open at $x\in X$ if $f_{T}:X\times_{S}T\to T$ is open along $g_{X}^{-1}(x)$ for every 1-dimensional, irreducible subscheme $s\in T\subset\operatorname{Spec}{\mathcal{O}}_{s,S}$ where $s:=f(x)$. ∎

(17.3) [Sta15, Tag 01TZ]. A finite type morphism $f:X\to S$ is universally open at $x\in X$ if $f_{T}:X\times_{S}T\to T$ is open along $g_{X}^{-1}(x)$ for every $g:T\to S$ where $T$ is the spectrum of a valuation ring and $g$ maps its closed point to $s:=f(x)$. ∎

(17.4) Let $g:X\to S$ be a morphism. Assume that there is a closed subscheme $x\in Z\subset X$ such that $h:=g|_{Z}:Z\to S$ is finite, dominant, $h^{-1}(s)=\{x\}$ and $k(x)/k(s)$ is purely inseparable. Then $g$ is universally open at $x$.

Proof. Let $x\in U\subset X$ be open. Then $Z\setminus U$ is closed, hence so is $h(Z\setminus U)$, which does not contain $s$. So

[TABLE]

shows that $g$ is open at $x$. The assumptions are preserved by base change, so $g$ is universally open at $x$. ∎

(17.5) Let $g:X\to(s,S)$ be a morphism of finite type. Assume that $X,S$ are integral, $g$ is open along $X_{s}$. Then $\dim X=\dim X_{s}+\dim S$.

Proof. We need to prove that the generic fiber also has dimension $X_{s}$. This is clear after base change to $g:T\to S$ as in (17.3), where $g$ maps the closed point to $s$ and the generic point of $T$ to the generic point of $S$. ∎

(17.6) We prove in Theorem 19 that a finite type morphism $f:X\to(s,S)$ is universally open along $X_{s}$ iff $f_{S^{\prime}}:X\times_{S}S^{\prime}\to(s^{\prime},S^{\prime})$ is open along $(X\times_{S}S^{\prime})_{s^{\prime}}$ for every étale $g:(s^{\prime},S^{\prime})\to(s,S)$.

18****Openness and pure dimensional morphisms.

Let $g:X\to S$ be a morphism of finite type.

(18.1) Set $X^{(n)}:=\{x\in X:\dim_{x}X_{g(x)}=n\}$ and $X^{(\leq n)}:=\{x\in X:\dim_{x}X_{g(x)}\leq n\}$.

By the upper semicontinuity of the fiber dimension [Sta15, Tag 02FZ], $X^{(\leq n)}$ is open in $X$ and $X^{(n)}=X^{(\leq n)}\setminus X^{(\leq n-1)}$ is closed in $X^{(\leq n)}$ and locally closed in $X$.

(18.2) Let $g:X\to(s,S)$ be a morphism of finite type. Then $g$ is open (resp. universally open) along $X_{s}$ iff $g^{(n)}:X^{(n)}\to S$ is open (resp. universally open) along $X^{(n)}_{s}$ for every $n$.

Proof. As we noted, $X^{(\leq n)}$ is open in $X$, thus we may as well assume that all fibers have dimension $\leq n$. The formation of $X^{(n)}$ commutes with base change, and over a 1-dimensional base the claims are clear. ∎

Note that the punctual version does not hold. As an example, set $S:=(xy=0)\subset{\mathbb{A}}^{2}$ and $X=(x=0)\cup(y=z=0)\subset{\mathbb{A}}^{3}$. The coordinate projection is open at the origin yet $X^{(1)}\to S$ is not open at the origin. A much worse example is given in Example 25.

(18.3) Let $g:X\to S$ be a morphism of finite type whose fibers have pure dimension $n$. Let $x\in Z\subset X$ be a relative complete intersection. Then $g$ is open (resp. universally open) at $x$ iff $g|_{Z}$ is open (resp. universally open) at $x$.

Proof. Assume that $g|_{Z}$ is open at $x$ and let $x\in U\subset X$ be open. Then $Z\cap U\subset Z$ is open and $g(x)\in g(Z\cap U)\subset g(U)$ shows that $g$ is open at $x$.

Conversely, assume that $g|_{Z}$ is not open at $x$. By (17.1–2) it is enough to check the claim when $S$ is a spectrum of a local ring of dimension 1. Thus, after replacing $X$ with an open neighborhood of $x$ we may asume that $Z_{s}=Z$; this follows from (17.1). Since $\dim_{x}Z=\dim_{x}X-r$ and $\dim_{x}Z_{s}=\dim_{x}X_{s}-r$, we conclude that $\dim_{x}X=\dim_{x}X_{s}$. Thus $X_{s}$ is an irreducible component of $X$, hence $g$ is not open. ∎

(18.4) Let $g:X\to S$ be a morphism of finite type. If $g$ has pure relative dimension $n$ (that is, $X\times_{S}T$ has pure dimension $\dim T+n$ for every local morphism $T\to S$ from an integral scheme to $S$) then $g$ is also universally open. This follows from (17.3). Note, however, that the local version of this is not true; see (15.2).

Next we show that it is enough to use étale base changes in the definition of universal openness.

Theorem 19**.**

Let $g:X\to(s,S)$ be a morphism of finite type. Then $g$ is universally open along $X_{s}$ iff $g^{\prime}$ is open along $X^{\prime}_{s^{\prime}}$ for every étale base change diagram as in (16.7).

Proof. If $g$ is universally open at $x$ then it is open after every base change.

Conversely, pick $x\in X_{s}$ and assume that $g$ is open at $x$ after every étale base change. Set $n=\dim_{x}X_{s}$. By (18.2) $g^{(n)}:X^{(n)}\to S$ is also open at $x$ after every étale base change and by (18.3) the same holds for every relative complete intersection $x\in Z\subset X^{(n)}$ of codimension $n$. Then $g|_{Z}$ is quasi-finite. Take an étale base change $(s^{\prime},S^{\prime})\to(s,S)$ as in Proposition 20 to get

[TABLE]

By construction every $(z^{\prime}_{i},Z^{\prime}_{i})\to(s^{\prime},S^{\prime})$ is finite and $k(z^{\prime}_{i})/k(s^{\prime})$ is purely inseparable. Furthermore, $(z^{\prime}_{i},Z^{\prime}_{i})\to(s^{\prime},S^{\prime})$ is open by assumption, hence dominant. Thus $(z^{\prime}_{i},Z^{\prime}_{i})\to(s^{\prime},S^{\prime})$ is universally open by (17.4) and so is $g|_{Z}:Z\to S$. Thus $g$ is universally open by (18.3) and (18.2). ∎

We used some results on étale localization of quasi-finite morphisms; see [Sta15, Tag 04HF] for proofs and further generalizations.

Proposition 20**.**

Let $g:X\to S$ be a quasi-finite morphism and $s\in S$. Then there is an étale morphism $(s^{\prime},S^{\prime})\to(s,S)$ such that

[TABLE]

where, $W$ does not have any points lying over $s^{\prime}$ and, for every $i$, the morphism $g^{\prime}_{i}:(x^{\prime}_{i},X^{\prime}_{i})\to(s^{\prime},S^{\prime})$ is finite, $(g^{\prime}_{i})^{-1}(s^{\prime})=\{x^{\prime}_{i}\}$ and $k(x^{\prime}_{i})/k(s^{\prime})$ is purely inseparable. ∎

Corollary 21**.**

Let $g:(x,X)\to(s,S)$ be a quasi-finite morphism. Then there is a commutative diagram

[TABLE]

where $h,h^{\prime}$ are étale, $g^{\prime}$ finite, ${g^{\prime}}^{-1}(s^{\prime})=\{x^{\prime}\}$ and $k(x^{\prime})/k(s^{\prime})$ is purely inseparable. ∎

We refer to (21.1) as an étale base change diagram of $g$.

Corollary 22**.**

[Sta15, Tag 02LO]** Let $g:(x,X)\to(s,S)$ be a finite morphism. Then there is an étale morphism $(s^{\prime},S^{\prime})\to(s,S)$ such that

[TABLE]

where, for every $i$, the morphism $g^{\prime}_{i}:(x^{\prime}_{i},X^{\prime}_{i})\to(s^{\prime},S^{\prime})$ is finite, $(g^{\prime}_{i})^{-1}(s^{\prime})=\{x^{\prime}_{i}\}$ and $k(x^{\prime}_{i})/k(s^{\prime})$ is purely inseparable. ∎

Geometrically unibranch schemes

Definition 23**.**

A local scheme $(x,X)$ is called geometrically unibranch if for every étale morphism $(x^{\prime},X^{\prime})\to(x,X)$, the local scheme $(x^{\prime},X^{\prime})$ is irreducible. See [Sta15, Tag 06DT] for other definitions and basic properties.

Openness is very well behaved for geometrically unibranch targets; cf. [Sta15, Tag 0F32]

Proposition 24**.**

Let $g:(x,X)\to(s,S)$ be a finite type morphism. Assume that $X$ is irreducible and $S$ is geometrically unibranch at $s$. The following are equivalent.

1. (1)

$\dim X=\dim X_{s}+\dim S$. 2. (2)

$g$* is open along $X_{s}$.* 3. (3)

$g$* is universally open along $X_{s}$.*

Proof. Note that (3) implies (2) by definition, and (2) implies (1) by (17.5). It remains to show that (1) $\Rightarrow$ (3).

Note that $\dim_{x}X\leq\dim_{x}X_{s}+\dim_{s}S$ for every $x\in X_{s}$, thus $X_{s}$ is pure dimensional.

Let $x\in Z\subset X$ be a relative complete intersection of codimension $\dim X_{s}$. By (18.3) it is enough to show that $g|_{Z}$ is universally open at $x$. Thus we may assume that $g$ is quasi-finite. Apply Proposition 20 to get an étale morphism $(s^{\prime},S^{\prime})\to(s,S)$ and a decomposition

[TABLE]

where the $k(x^{\prime}_{i})\supset k(s^{\prime})$ are purely inseparable and the projections $g^{\prime}_{i}:X^{\prime}_{i}\to S^{\prime}$ are finite. The $g^{\prime}_{i}$ are also dominant by assumption, hence universally open at $x^{\prime}_{i}$ by (17.4). Thus $X\times_{S}S^{\prime}$ is universally open along $X^{\prime}_{s^{\prime}}$, and so $X\to S$ is universally open along $X_{s}$. ∎

Example 25**.**

It seems natural to hope that the equivalence (24.2) $\Leftrightarrow$ (24.3) holds pointwise. This is, however, not the case. To see this, we construct below a projective morphism of surfaces $g:(x,X)\to(s,S)$ such that $S$ is normal and $g$ is open at $x$, yet $g$ is not universally open at $x$.

Let $(s,S)$ be a normal surface singularity with a (non-minimal) resolution $\tau:Y\to S$ and exceptional curves $E_{1},\dots,E_{n}$. Assume that

1. (1)

for every $i=1,\dots,n-1$ there is a morphism $Y\to(x_{i},X_{i})$ that contracts only the curve $E_{i}$, 2. (2)

$X_{i}$ is normal and $\pi_{i}:X_{i}\to S$ is projective, 3. (3)

if an algebraic curve $C\subset Y$ is disjoint from $E_{1}\cup\cdots\cup E_{n-1}$ then it is also disjoint from $E_{n}$.

Let $(x,X)$ be obtained from the surfaces $X_{1},\dots,X_{n-1}$ by identifying the points $x_{1},\dots,x_{n-1}$. The morphisms $\pi_{i}$ glue to a projective morphism $\pi:(x,X)\to(s,S)$.

Claim 25.4. $\pi$ is open but not universally open at $x$.

Proof. Let $B\subset Y$ be an algebraic curve that meets $E_{n}$ transversally at a general point $b\in E_{n}$. Then $\operatorname{Spec}{\mathcal{O}}_{b,B}\to S$ shows that $\pi$ is not universally open at $x$.

Assume to the contrary that $\pi$ is not open. Then, by (17.1), there is a curve $0\in C_{S}\subset S$ such that the closure $C_{X}$ of the preimage of $C_{S}\setminus\{s\}$ does not pass through $x$. Note that $C_{X}$ is the union of the birational transforms $(\pi_{i})^{-1}_{*}C_{S}$ and these in turn are the images of $C:=\tau^{-1}_{*}C_{S}\subset Y$. Thus $C$ is disjoint from $E_{1}\cup\cdots\cup E_{n-1}$. By (3) then $C$ is also disjoint from $E_{n}$, which contradicts $0\in C_{X}$. ∎

In order to construct such an $(s,S)$, we start with ${\mathbb{P}}^{2}$ and 3 general lines $L_{1},L_{2},L_{3}$. Let $B\subset{\mathbb{P}}^{2}$ be a general quartic curve. We obtain $Y^{\prime}$ by blowing up the 12 intersection points of $B$ with the lines. The birational transforms of the $L_{i}$ become $E^{\prime}_{i}\subset Y^{\prime}$. Contracting them we get $(s,S)$. Pick next a very general point $p\in E^{\prime}_{3}$ and let $Y=B_{p}Y^{\prime}$ denote the blow-up. The birational transforms of the $E^{\prime}_{i}$ give $E_{1},E_{2},E_{3}$ and $E_{4}$ is the exceptional curve of $Y\to Y^{\prime}$.

Each $E_{i}\subset Y$ is a rational curve with negative self-intersection, so it can be contracted projectively. (This is essentially due to Castelnuovo; the proof in [Har77, V.5.7] is easy to modify.) It remains to check assumption (3). Slightly stronger, we claim that an algebraic curve $C^{\prime}\subset Y^{\prime}$ can not intersect $E^{\prime}:=E^{\prime}_{1}\cup E^{\prime}_{2}\cup E^{\prime}_{3}$ only at $p$. To see this note that $\operatorname{Pic}(Y^{\prime})$ is finitely generated (in fact, isomorphic to ${\mathbb{Z}}^{13}$), hence the image of the restriction map

[TABLE]

is finitely generated. Thus, for a very general point $p\in E^{\prime}$, the intersection of ${\mathbb{Z}}[p]\subset\operatorname{Pic}(E^{\prime})$ with the image is the trivial element $[0]\in\operatorname{Pic}(E^{\prime})$.

As a concrete example, we can take $(s,S)$ to be the projectivisation of the affine surface

[TABLE]

If $X$ is reducible and $g:X\to(s,S)$ is universally open along $X_{s}$ then, as shown by Example 41.2, we can say very little about the lower dimensional irreducible components of $X$. The next result shows that the maximal dimensional irreducible components behave better.

Theorem 26**.**

Let $g:X\to(s,S)$ a finite type morphism that is universally open along $X_{s}$. Set $n:=\dim X_{s}$ and let $X^{\rm max}\subset X^{(n)}$ be the union of all those irreducible components of $X^{(n)}$ that dominate some irreducible component of $S$ and have nonempty intersection with $X_{s}$.

Then $X^{\rm max}\to S$ is universally open along $X_{s}\cap X^{\rm max}$.

Proof. The conclusions can be checked after an étale base change, hence, using Lemma 27, we may assume that every irreducible component of $(s,S)$ is unibranch. Then, by (16.7), we may also assume that $(s,S)$ is irreducible, hence unibranch.

Let $X_{i}\subset X^{\rm max}$ be an irreducible component. By assumption $X_{i}$ dominates $S$ and the generic fiber of $X_{i}\to S$ has dimension $n$ since $X_{i}\subset X^{(n)}$. Thus $\dim X_{i}=n+\dim S$ and so $X_{i}\to S$ is universally open along $X_{s}\cap X_{i}$ by Proposition 24. ∎

Lemma 27**.**

[Sta15, Tag 0CB4]** Let $(s,S)$ be a local scheme. Then there is an étale morphism $(s^{\prime},S^{\prime})\to(s,S)$ such that every irreducible component $(s^{\prime},S^{\prime}_{i})\subset(s^{\prime},S^{\prime})$ is geometrically unibranch. ∎

The following is a partial converse to (14.3).

Lemma 28**.**

Let $S$ be a connected scheme and $g:X\to S$ a dominant morphism of finite type whose fibers have pure dimension $n$. The following are equivalent.

1. (1)

$g:X\to S$* is universally open,* 2. (2)

$g\times g:X\times_{S}X\to S$* is open,* 3. (3)

Every irreducible component of $X$ dominates an irreducible component of $S$ and every irreducible component of $X\times_{S}X$ dominates an irreducible component of $S$.

Proof. If $g:X\to S$ is universaly open then so is $X\times_{S}X\to X$, hence also their composite $X\times_{S}X\to S$. Thus (1) $\Rightarrow$ (2) and (2) $\Rightarrow$ (3) is clear.

It remains to show that if (3) holds then $g$ is universally open along $X_{s}$ for every $s\in S$. Using Lemma 27, after an étale base change we may assume that every irreducible component $(s,S_{i})\subset(s,S)$ is geometrically unibranch.

Set $X_{i}:=g^{-1}(S_{i})$. By (16.7) it is enough to show that each $X_{i}\to S_{i}$ is universaly open along $(X_{i})_{s}$. If this does not hold, then, by Proposition 24, there is an irreducible component $Z_{i}\subset X_{i}$ that does not dominate $S_{i}$. By assumption there is an irreducible component $X_{j}\subset X$ that contains $Z_{i}$ and this $X_{j}$ dominates some irreducible component $S_{j}\subset S$. By assumption $S_{i}\neq S_{j}$.

Let $X^{*}_{i}\subset X$ be an irreducible component that does dominate $S_{i}$. Then $X_{j}\times_{S}X^{*}_{i}$ is an union of irreducible components of $X\times_{S}X$ that lie over $S_{i}\cap S_{j}$, hence they do not dominate any irreducible component of $S$. This is impossible by (3), ∎

Example 29**.**

This example shows that the equidimensionality assumption in the previous Lemma 28 is necessary. Set

[TABLE]

with projection $\pi:X\to S={\mathbb{A}}^{3}_{\mathbf{x}}$. The central fiber has dimension 2, so $\pi$ is not open. The fiber product

[TABLE]

is given by 2 equations, hence its irreducible components have dimension $\geq 7-2=5$. The central fiber of $X\times_{S}X\to S$ has dimension 4, so it can not be an irreducible component. Thus $X\times_{S}X$ is irreducible.

Note also that if $X\to S$ is not pure dimensional then the fiber product $X\times_{S}\cdots\times_{S}X$ of more than $\dim S$ copies of $X$ always has a non-dominant irreducible component.

3. Path lifting in the Euclidean topology

Let $g:X\to Y$ be a morphism of ${\mathbb{C}}$-schemes of finite type. In this section we compare scheme-theoretic properties of $g$ with properties of $g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$ in the Euclidean topology.

It is easy to see that universal openness in the Zariski topology (Definition 14) is equivalent to openness in the Euclidean topology; see Lemma 32. Next we study 3 versions of the path lifting property. We see in Theorems 31 and 35 that 2 of them have very satisfactory scheme-theretic descriptions.

Definition 30** (Arc and path lifting).**

Let $h:M\to N$ be a continuous map of topological spaces and $\gamma:[0,1]\to N$ a path. A lift of $\gamma$ with starting point m\in h^{-1}\bigl{(}\gamma(0)\bigr{)} is a continuous map $\gamma^{\prime}:[0,1]\to M$ such that $\gamma^{\prime}(0)=m$ and $\gamma=h\circ\gamma^{\prime}$.

1. (1)

We say that $h$ has the path lifting property if a lift $\gamma^{\prime}$ exists for every $\gamma:[0,1]\to N$ and m\in h^{-1}\bigl{(}\gamma(0)\bigr{)}. We do not require $\gamma^{\prime}$ to be unique. If this holds then every path $\gamma:[0,1)\to N$ also lifts. 2. (2)

We say that $h$ has the arc lifting (or local path lifting) property if the lift $\gamma^{\prime}$ exists over some subinterval $[0,\epsilon]$, where $\epsilon>0$ may depend on m\in h^{-1}\bigl{(}\gamma(0)\bigr{)}. 3. (3)

We say that $h$ has the 2-point path lifting property if given $\gamma:[0,1]\to N$ and $m_{i}\in h^{-1}(\gamma(i))$ for $i=0,1$, there is a lifting $\gamma^{\prime}$ such that $\gamma^{\prime}(0)=m_{0}$ and $\gamma^{\prime}(1)=m_{1}$. The basic example is a fiber bundle $h:M\to N$ with path-connected fiber $F$.

Note also that if $N$ is path-connected, $M\neq\emptyset$ and $h:M\to N$ has the path lifting property then $h$ is surjective.

Lifting constant paths shows that if $h$ has the 2-point path lifting property then all fibers of $h$ are path-connected.

The concept of path lifting occurs most frequently in the topological literature, but the next result shows that, from the scheme-theoretic point of view, arc lifting is the most natural property.

Theorem 31**.**

Let $g:X\to Y$ be a morphism of ${\mathbb{C}}$-schemes of finite type. The following are equivalent.

1. (1)

$g:X\to Y$* is universally open (Definition 14).* 2. (2)

$g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$* is open in the Euclidean topology.* 3. (3)

$g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$* has the arc lifting property (30.2).*

Proof. The equivalence of (31.1) and (31.2) is established in Lemma 32.

Assume next that $g$ is not universally open at $x$. By Theorem 19, after a suitable étale base change, it is not open at $x$. An étale base change is a local homeomorphism in the Euclidean topology, thus it does not alter the liftability of arcs.

Thus, by (17.2) there is a curve $g(x)\in C\subset Y$ such that $f(U)\cap C=\{g(x)\}$. Choosing a path in this curve $C$ shows that (31.3) $\Rightarrow$ (31.2). Thus it remains to prove that (31.2) $\Rightarrow$ (31.3). This done in 3 steps.

If $g$ is finite, arc lifting is proved in Lemmas 33–34. If $X_{g(x)}$ has pure dimension $n$ at $x$, then we take a relative complete intersection $x\in Z\subset X$ of codimension $n$. Then $g|_{Z}:Z\to Y$ is quasi-finite at $x$ and open by (18.3). An étale base change as in Proposition 20 reduces this to the already discussed finite case. Thus an arc $\gamma:[0,\epsilon]\to Y$ has a lift $\gamma^{\prime}:[0,\epsilon]\to Z$. This is also a lift to $X$.

Finally, in general set $n=\dim_{x}X_{g(x)}$. By (18.2) the restriction of $g$ to $X^{(n)}$ is universally open. The fibers of $X^{(n)}\to Y$ have pure dimension $n$, thus $X^{(n)}\to Y$ has the arc lifting property and so does $X$. ∎

Lemma 32**.**

$g:X\to Y$* be a morphism between ${\mathbb{C}}$-schemes of finite type. Then $g$ is universally open at $x$ in the Zariski topology iff $g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$ is open at $x$ in the Euclidean topology.*

Proof. Assume that $g({\mathbb{C}})$ is open at $x$ in the Euclidean topology and let $x\in U\subset X$ be a Zariski open neighborhood. Then $U$ is also Euclidean open, hence $g(U)$ contains a Euclidean open neighborhood of $g(x)$. Since $g(U)$ is also constructible, it also contains a Zariski open neighborhood of $x$. Thus $g$ is open at $x$ in the Zariski topology. An étale base change is a local homeomorphism in the Euclidean topology, hence $g$ is also universally open at $x$ in the Zariski topology by Theorem 19.

Conversely, assume that $g$ is not universally open at $x$. By Theorem 19, after a suitable étale base change, it is not open at $x$. Thus, by (17.1) this means that there is a curve $g(x)\in C\subset Y$ such that $f(U)\cap C=\{g(x)\}$. This shows that $g({\mathbb{C}})$ is not open at $x$ in the Euclidean topology either. ∎

Lemma 33**.**

Let $g:X\to Y$ be a finite morphisms of ${\mathbb{C}}$-schemes of finite type. Then one can choose triangulations on them such that $g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$ is simplicial.

Proof. Set $n:=\dim Y$, $X_{n}:=X$ and $Y_{n}:=Y$. If $g_{r}:X_{r}\to Y_{r}$ is already defined then let $U_{r}\subset Y_{r}$ be the largest smooth, open subset of pure dimension $r$ such that $g_{r}$ is étale. Set $Y_{r-1}:=Y_{r}\setminus U_{r}$ and $X_{r-1}:=\operatorname{red}g^{-1}(Y_{r-1})$.

Triangulate $Y$ by starting with $Y_{0}$, extending it to a triangulation of $Y_{1}$ (possibly adding new vertices), then extending it to a triangulation of $Y_{2}$ (possibly after a refinement) and so on. At the end we get a triangulation of $Y$ such that the interior $\Delta^{\circ}$ of every simplex lies in some $U_{r}$. Thus the preimages of the simplices give a triangulation of $X$ and $g$ is simplicial. ∎

Lemma 34**.**

Let $M,N$ be connected simplicial complexes and $g:M\to N$ a proper, open, simplicial map with finite fibers. Then $g$ has the path lifting property.

Proof. We are given a path $\gamma:[0,1]\to N$ and a lifting $\gamma^{\prime}(0)$. Choose a maximal lifting $\gamma^{\prime}_{c}:[0,c)\to N$. (At the begining we allow $c=0$.) First we define $\gamma^{\prime}(c)$. Set $n=\gamma(c)$ and let $m_{i}\in M$ be the preimages of $n$. Choose an open neighborhood $n\in U\subset N$ such that $g^{-1}(U)=\cup_{i}V_{i}$ is a disjoint union of connected neighborhoods of the $m_{i}$. Then $\gamma^{\prime}(c-\epsilon,c)$ is contained in one of them, say $V_{j}$, and setting $\gamma^{\prime}(c)=m_{j}$ is the unique continuous extension of $\gamma^{\prime}:[0,c)\to M$ to $\gamma^{\prime}:[0,c]\to M$.

In order to extend beyond $c$, we may as well assume that $n:=\gamma(c)$ is a vertex. Then $m:=\gamma^{\prime}(c)$ is also a vertex. We can choose the neighborhood $n\in U\subset N$ to be a cone over the link $L_{n}$. Then $m\in V\subset M$ is also a cone over its link $L_{m}$ and $g|_{L_{m}}:L_{m}\to L_{n}$ is finite, open and simplicial. If $\gamma$ maps $[c,c+\epsilon]$ to $V$ then we define $\gamma^{\prime}$ on $[c,c+\epsilon]$ as follows. If $\gamma(x)=n$ then set $\gamma^{\prime}(x)=m$. The complement of this set, denoted by $B\subset[c,c+\epsilon]$, is a countable union of open intervals and $\gamma$ maps $B$ to

[TABLE]

Correspondingly we can write $\gamma|_{B}=(\alpha_{B},\gamma_{B})$ where $\gamma_{B}$ maps $B$ to $L_{n}$. Since $g|_{L_{m}}:L_{m}\to L_{n}$ is simplicial, by induction on the dimension, $\gamma_{B}$ lifts to $\gamma^{\prime}_{B}:B\to L_{m}$. (As we noted in (30.1), open paths also lift.) Then set $\gamma^{\prime}|_{B}:=(\alpha_{B},\gamma^{\prime}_{B})$.∎

Next we consider the scheme-theretic description of the 2-point path lifting property.

Theorem 35**.**

Let $g:X\to Y$ be a morphism of ${\mathbb{C}}$-schemes of finite type. The following are equivalent.

1. (1)

$g:X\to Y$* is universally open, surjective and has connected fibers.* 2. (2)

$g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$* is open in the Euclidean topology, surjective and has connected fibers.* 3. (3)

$g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$* is surjective, has connected fibers and satisfies the arc lifting property (30.2).* 4. (4)

$g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$* has the 2-point path lifting property (30.3).*

Proof. The equivalence of (1), (2) and (3) follows from Theorem 31, once we note that a ${\mathbb{C}}$-scheme of finite type is connected in the Zariski topology iff it is connected in the Euclidean topology. (This is usually proved using Chow’s theorem, but one can use Bertini’s hyperplane section theorem to reduce it to the 1-dimensional case, which was known to Riemann.)

We noted at the end of Definition 30 that if (4) holds then $g$ is surjective and has connected fibers. Thus (4) implies (1–3) and it remains to show that (1–3) imply (4).

We have $\gamma:[0,1]\to Y({\mathbb{C}})$ and liftings $m_{0}$ of $\gamma(0)$ and $m_{1}$ of $\gamma(1)$. Assume that we already have a lifting $\gamma^{\prime}$ defined on $[0,c]$. By (3) we have a lifting $\gamma^{\prime\prime}_{c}:[c,c+\epsilon]\to X({\mathbb{C}})$ such that $\gamma^{\prime\prime}_{c}(c)=\gamma^{\prime}(c)$. The concatenation of $\gamma^{\prime}$ on $[0,c]$ with $\gamma^{\prime\prime}_{c}$ defines $\gamma^{\prime}$ on $[0,c+\epsilon]$.

Assume next that we have $\gamma^{\prime}$ defined on $[0,c)$. We do not claim that it extends to $[0,c]$, but we show that the restriction of $\gamma^{\prime}$ to $[0,c-\epsilon]$ extends to $\gamma^{\prime\prime}:[0,c+\epsilon]$ for some $\epsilon>0$. Moreover, if $c=1$ then we can also arrange that $\gamma^{\prime\prime}(1)=m_{1}$. For other values of $c$ pick any lifting $m_{c}$ of $\gamma(c)$. Applying (3) gives a lifting

1. (5)

$\gamma^{\prime\prime}_{c}:[c-\epsilon,c+\epsilon]\to X$ such that $\gamma^{\prime\prime}_{c}(c)=m_{c}$.

Starting from $X\to Y$, we get a stratification $\{Y_{i}\}$ of $Y$ as in (36.1). Applying Lemma 34 to $\amalg_{i}Y_{i}\to Y$ we get a triangulation of $Y$ with $\gamma(c)$ as a vertex. Let $L$ be the smallest subcomplex that contains \gamma\bigl{(}(c-\delta,c)\bigr{)} for some $\delta>0$ and let $\Delta\subset L$ be a maximal dimensional simplex with interior $\Delta^{\circ}$. Then $\gamma^{-1}(\Delta^{\circ})$ is open in $(c-\delta,c)$ and its closure contains $c$. In particular, we see that

1. (a)

there is a homeomorphism $\beta:g^{-1}(\Delta^{\circ})\sim\Delta^{\circ}\times M$ that commutes with projection to $\Delta^{\circ}$ for some $M$ and 2. (b)

there are $0<\eta_{1}<\eta_{2}<\epsilon$ such that $\gamma$ maps $[c-\eta_{2},c-\eta_{1}]$ to $\Delta^{\circ}$.

As we noted in (30.3), the restriction of $\gamma$ to $[c-\eta_{2},c-\eta_{1}]$ has a lifting $\bar{\gamma}$ such that

[TABLE]

Thus the concatenation of $\gamma^{\prime}$ on $[0,c-\eta_{2}]$, $\bar{\gamma}$ on $[c-\eta_{2},c-\eta_{1}]$ and $\gamma^{\prime\prime}_{c}$ on $[c-\eta_{1},c+\epsilon]$ defines a lifting of $\gamma$ over $[0,c+\epsilon]$. ∎

36****Stratification of maps.

Let $g:X\to Y$ be a morphism of varieties over ${\mathbb{C}}$. As in [GM88, Sec.I.1.7], $Y$ has a stratification by closed subvarieties

[TABLE]

where $Y_{i}\setminus Y_{i-1}$ has pure dimension $i$ and each

[TABLE]

is a topologically locally trivial fiber bundle. (There may be different fibers over different connected components.) Thus path lifting could fail only at points where a path moves from one stratum to another.

Let $g:X\to Y$ be a quasi-finite, universally open morphism. Using Lemma 34 we see that $g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$ has the path lifting property iff $g$ is proper. However, as shown by Example 41, the path lifting property does not seem to have an equivalent scheme-theoretic version for morphisms with positive dimensional fibers. Nonetheless, the following sufficient condition is quite natural and useful.

Theorem 37**.**

Let $g:X\to Y$ be a proper, universally open, pure dimensional morphism of ${\mathbb{C}}$-schemes of finite type. Then $g({\mathbb{C}}):X({\mathbb{C}})\to Y({\mathbb{C}})$ has the path lifting property

Proof. As (16.8) shows, we can not use Stein factorization to reduce Theorem 37 directly to Theorem 35, but a suitable modification of the proof will work.

We follow the proof of (35.3) $\Rightarrow$ (35.4), but we need to make a different choice of $\gamma^{\prime\prime}_{c}$ in (35.5). Note that \gamma^{\prime}\bigl{(}[c-\eta_{1},c-\eta_{2}]\bigr{)} is contained in a connected component $\Delta^{\circ}\times M_{1}\subset\Delta^{\circ}\times M$, and everything works as before if we can ensure that \gamma^{\prime\prime}_{c}\bigl{(}[c-\eta_{1},c-\eta_{2}]\bigr{)} is also contained in $\Delta^{\circ}\times M_{1}$.

To do this, let $X\to Y_{1}\to Y$ be the Stein factorization. The choice of the connected component $\Delta^{\circ}\times M_{1}\subset\Delta^{\circ}\times M$ defines a section $\sigma:\Delta^{\circ}\to Y_{1}$. By Lemma 38, after an étale base change we may assume to have a relative complete intersection $Z\subset X$ such that the induced map $g_{1}:Z\to Y_{1}$ is finite and surjective. In particular, there is an $m_{z}\in Z$ such that g_{1}(m_{z})=\sigma\bigl{(}\gamma(c-\eta_{2})\bigr{)}. Since $Z\to Y$ is finite and open, it satisfies the path lifting property by Lemma 34. Thus the restriction of $\gamma$ to $[c-\eta_{2},c]$ has a lifting $\gamma^{\prime\prime}_{c}$ such that $\gamma^{\prime\prime}_{c}(c-\eta_{2})=m_{z}$. By construction $\gamma^{\prime\prime}_{c}(c-\eta_{2})$ and $\gamma^{\prime}(c-\eta_{2})$ are in the same commected component of the fiber and the rest of the proof now works as before. ∎

Lemma 38**.**

Let $g:X\to Y$ be a proper, universally open morphism of finite type and of relative dimension $n$. Let $X\to Y_{1}\to Y$ denote its Stein factorization. Let $y\in Y$ be a point and $P\subset X_{y}$ a finite subset that has nonempty intersection with every irreducible component of $X_{y}$. Let $P\subset Z\subset X$ be a relative complete intersection of codimension $n$. Then there is an étale neighborhood $(y^{\prime},Y^{\prime})\to(y,Y)$ such that the induced morphism $Z^{\prime}\to Y^{\prime}_{1}$ is finite and surjective.

Proof. By construction $Z\to Y$ is quasi-finite, hence there is an étale neighborhood $(y^{\prime},Y^{\prime})\to(y,Y)$ such that $Z^{\prime}\to Y^{\prime}$ is finite. Thus $Z^{\prime}\to Y^{\prime}_{1}$ is also finite. We need to prove that it is surjective.

Let $\tilde{Y}^{\prime}_{1}$ be the disjoint union of the irreducible components of $Y^{\prime}_{1}$. By base change we get $\tilde{X}^{\prime}_{1}\to\tilde{Y}^{\prime}_{1}$, \tilde{P}^{\prime}_{1}\subset\bigl{(}\tilde{X}^{\prime}_{1})_{y^{\prime}} and $\tilde{P}^{\prime}_{1}\subset\tilde{Z}^{\prime}_{1}\subset\tilde{X}^{\prime}_{1}$. An irreducible component of \bigl{(}\tilde{X}^{\prime}_{1})_{y^{\prime}} has pure dimension $n$, thus its image in $X_{y}$ is an irreducible component. Thus $\tilde{P}^{\prime}_{1}$ has nonempty intersection with every irreducible component of \bigl{(}\tilde{X}^{\prime}_{1})_{y^{\prime}}. Thus $\tilde{Z}^{\prime}_{1}$ dominates $\tilde{Y}^{\prime}_{1}$ and hence $Z^{\prime}$ dominates $Y^{\prime}_{1}$. Since $Z^{\prime}\to Y^{\prime}_{1}$ is finite, this implies that it is surjective. ∎

Next we establish the Euclidean version of Corollary 9.

Corollary 39**.**

Let $g:(x,X)\to(y,Y)$ be a proper, universally open morphism of pointed, connected ${\mathbb{C}}$-schemes. Then \operatorname{im}[\pi_{1}\bigl{(}X({\mathbb{C}}),x\bigr{)}\to\pi_{1}\bigl{(}Y({\mathbb{C}}),y\bigr{)}] has finite index in \pi_{1}\bigl{(}Y({\mathbb{C}}),y\bigr{)}.

Proof. Let $n$ denote the maximal fiber dimension. Then $g^{(n)}:X^{(n)}\to Y$ is also proper and universally open by (18.2). We may choose $x\in X^{(n)}$ and then $g^{(n)}_{*}$ factors as

[TABLE]

Thus it is enough to show that \operatorname{im}[\pi_{1}\bigl{(}X^{(n)}({\mathbb{C}}),x\bigr{)}\to\pi_{1}\bigl{(}Y({\mathbb{C}}),y\bigr{)}] has finite index in \pi_{1}\bigl{(}Y({\mathbb{C}}),y\bigr{)}. The advantage is that $g({\mathbb{C}}):X^{(n)}({\mathbb{C}})\to Y({\mathbb{C}})$ has the path lifting property by Theorem 37. The rest follows from (39.1).

Claim 39.1. Let $h:M\to N$ be a continuous map of path-connected topological spaces that has the path lifting property. Assume that $h^{-1}(n)$ has finitely many path-connected components for some $n\in N$. Then the image of $\pi_{1}(M,m)\to\pi_{1}(N,n)$ has finite index in $\pi_{1}(N,n)$.

Proof. Every loop $\gamma$ starting and ending in $n$ lifts to a path that starts at $m$ and ends in $h^{-1}(n)$. If 2 loops $\gamma_{1},\gamma_{2}$ end at the same path-connected component then $\gamma_{1}^{-1}\gamma_{2}$ lifts to a loop on $M$. This shows that the index of the image of $\pi_{1}(M,m)\to\pi_{1}(N,n)$ is bounded by the number of path-connected components of the fiber. ∎

Examples

The first example shows that Theorem 35 does not have a pointwise version.

Example 40**.**

Let $Y:=(x^{2}=y^{2}z)$ be the pinch point as in Example 15.1 with normalization $p:\bar{Y}\to Y$. We saw that $p$ is universally open at the origin. However, it does not have the arc lifting property at the origin.

To see this consider the real curves \gamma^{\pm}:t\mapsto\bigl{(}t,\pm t^{2}\sin(t^{-1})\bigr{)}\subset{\mathbb{R}}^{2}. Note that $p\circ\gamma^{+}$ and $p\circ\gamma^{-}$ intersect at the points $t=(m\pi)^{-1}$. Thus the arc

[TABLE]

has no lifting.

Example 41** (Path lifting and properness).**

It is natural to hope that arc lifting plus properness should imply path lifting, but this is not the case.

(41.1) Let $X$ be obtained from $X_{1}:=B_{(0,0)}{\mathbb{C}}^{2}$ and of $X_{2}:={\mathbb{P}}^{1}\times{\mathbb{C}}^{2}$ by identifying the exceptional divisor of the blow-up with ${\mathbb{P}}^{1}\times\{(0,0)\}$. The projection $X\to{\mathbb{C}}^{2}$ is universally open, and so is $X_{2}\to{\mathbb{C}}^{2}$, but $X_{1}\to{\mathbb{C}}^{2}$ is not even open. The path

[TABLE]

does not lift to $B_{(0,0)}{\mathbb{C}}^{2}$. Thus $X\to{\mathbb{C}}^{2}$ is universally open and proper, it has the arc lifting property but not the path lifting property.

In the next 2 variants of the above construction, gluing of 2 irreducible components has the opposite effect on path lifting.

(41.2) Set $Y_{1}:=(ux=vy)\subset{\mathbb{P}}^{2}_{xyz}\times{\mathbb{C}}^{2}_{uv}$. Note that $\Sigma:=(x=y=0)\subset Y_{1}$ is a section of $\pi_{2}:Y_{1}\to{\mathbb{C}}^{2}_{uv}$. Thus every path starting in ${\mathbb{C}}^{2}_{uv}\setminus\{(0,0)\}$ can be lifted to $Y_{1}$ but $Y_{1}\to{\mathbb{C}}^{2}_{uv}$ does not have the path lifting property.

Let $Y$ be obtained from $Y_{1}$ and of $Y_{2}:={\mathbb{P}}^{2}\times{\mathbb{C}}^{2}_{uv}$ by identifying the fibers over the origin. Then $p:Y\to{\mathbb{C}}^{2}_{uv}$ has the path lifting property.

(41.3) Let $Z_{1}$ be obtained from ${\mathbb{P}}^{1}\times{\mathbb{C}}^{2}_{uv}$ by blowing up ${\mathbb{P}}^{1}\times\{(0,0)\}$. The projection to ${\mathbb{C}}^{2}_{uv}$ factors as $Z_{1}\to B_{(0,0)}{\mathbb{C}}^{2}\to{\mathbb{C}}^{2}$, thus not all paths starting in ${\mathbb{C}}^{2}_{uv}\setminus\{(0,0)\}$ can be lifted to $Z_{1}$. Let $Z$ be obtained from $Z_{1}$ and of $Z_{2}:={\mathbb{P}}^{1}\times{\mathbb{P}}^{1}\times{\mathbb{C}}^{2}_{uv}$ by identifying the fibers over the origin. Thus $Z\to{\mathbb{C}}^{2}$ is universally open and proper, it has the arc lifting property but not the path lifting property.

Acknowledgments**.**

Partial financial support was provided by the NSF under grant numbers DMS-1362960 and DMS-1440140 while the author was in residence at MSRI during the Spring 2019 semester.