# Morphism between covering maps are open

Let $$q:Z\rightarrow X$$, $$p:Y\rightarrow X$$ covering maps, and $$f:Z\rightarrow Y$$ a morphism of covering maps. Then $$f$$ is open.

The definition of covering map that im using is the following:

A map $$p:Y\rightarrow X$$ is a covering map if for all $$x\in X$$ there exists an open neighbourhood $$U\subseteq X$$ and an homeomorphism $$p^{-1}\rightarrow U\times F$$ such that the following diagram conmutes:

$$\require{AMScd}$$

$$\begin{CD} p^{-1}(U) @>>> U\times F\\ @. {_{\rlap{p}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{4em}{0em}]{}}}} @VV\pi_1V\\ @. U \end{CD}$$

My attempt was:

Let $$U\subseteq Z$$ an open subset. Let's to see that $$f(U)\subseteq Y$$ is open. Let $$y\in f(U)$$. Then there exists $$z\in Z$$ such that $$f(z)=y$$. Let $$x=q(z)=p(y)$$. Then there exists open sets $$W_1,W_2\subseteq X$$, with $$z\in W_1\cap W_2$$, and homeomorpisms $$q^{-1}(W_1)\rightarrow W_1\times F_1$$, $$p^{-1}(W_2)\rightarrow W_2\times F_2$$ such that certain diagrams commutes. The candidate for neighbourhood of $$y$$ is, i think, the open subset $$\begin{equation*} V:=p^{-1}\left( q(U)\cap W_1\cap W_2 \right) \subseteq Y \end{equation*}$$ Its follows that $$y\in V$$, but I want to see that $$V\subseteq f(U)$$.

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EDIT: definitions

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Definition: We say that $$G$$ acts evenly over $$Y$$ if for all $$y\in Y$$ there exists an open neighbourhood $$V\subseteq Y$$ such that $$$$g\cdot V \cap h\cdot V = \emptyset, \quad \forall g,h\in G, \text{ } g\neq h$$$$

If $$G$$ acts evenly on $$Y$$, then we have the set of orbits $$$$Y/G=\{ G\cdot y \text{ : } y\in Y\}$$$$ and we can endow iy with the quotient topology given by $$y\mapsto G\cdot y$$.

I already saw that if $$p:Y\rightarrow X$$ is a cover map, with $$G$$ acting evenly over $$Y$$, then the projection $$Y\rightarrow Y/G$$, given by $$y\mapsto G\cdot y$$ its a cover map.

Definition: A $$G$$-cover is a cover $$p:Y\rightarrow X$$ such that comes up from an even action of $$G$$ on $$Y$$, i.e, if there exists an even action $$G\times Y \rightarrow Y$$ such that $$X\cong Y/G$$.

Question: Using the notation of the previous excercise, if the cover maps are $$G$$-covers, then $$f$$ must be an isomorphism (i.e biyective).

Any hint will be welcome. Thanks for read.

• You should make precise what a G-covering map is. Commented Aug 21, 2023 at 13:22
• i have added the definitions, thanks Commented Aug 21, 2023 at 16:41

I think you have the right ideas but it's hard for me to follow the thread of what you're doing.

Fix $$U$$, $$y\in f(U)$$, $$z\in Z$$ with $$f(z)=y$$, $$x=q(z)=p(y)$$ as you say.

Because $$q$$ is a cover and $$U$$ is open, there is a sheet $$z\in U'\subseteq U$$ with $$U'$$ and $$q(U')$$ both open and $$q$$ a homeomorphism $$U'\cong q(U')$$. Because $$p$$ is a cover and $$q(U')$$ a neighbourhood of $$x=p(y)$$, there is a sheet $$y\in V$$ with $$V$$ and $$p(V)$$ both open and $$p$$ a homeomorphism $$V\cong p(V)$$.

Denote by $$W$$ the intersection $$q(U')\cap p(V)$$ and let $$V'$$ denote $$p^{-1}(W)\cap V$$. Let $$y'\in V'$$ be arbitrary; $$x':=p(y')\in p(V')=W$$ has some (an unique, actually) preimage $$z':=q^{-1}(x')\in U'$$; $$f(z')=y'$$, I claim.

Why? Because $$p(f(z'))=q(z')=p(y')$$ and $$p$$ is a homeomorphism twixt $$W$$ and $$V'$$. Since $$y'\in V'$$ was an arbitrary point and such a $$z'\in U$$ always exists, we find $$V'\subseteq f(U)$$ is an open neighbourhood of $$y$$. It follows $$f(U)$$ is open.

For the updated question about $$G$$-covers, I think you mean for there to be left $$G$$-covering actions on $$Z$$ and $$Y$$ such that $$p,q$$ exhibit $$X$$ as $$Y/G,Z/G$$ respectively. I also think you mean for $$f$$ to be $$G$$-equivariant, as an unstated hypothesis. I'm not convinced $$f$$ has to be injective or surjective otherwise.

Suppose $$f(z)=f(z')=:y$$. Put $$x:=p(y)=q(z)=q(z')$$. $$q(z)=q(z')$$ means exactly that there exists $$g,gz=z'$$; if $$z\neq z'$$ then it would follow that $$g\neq1$$. There is a neighbourhood $$V$$ of $$y$$ satisfying $$gV\cap V=\emptyset$$, then; however, $$gV$$ contains $$gy=gf(z)=f(gz)=f(z')$$ which is a member of $$V$$, so this is a contradiction. Therefore $$f$$ is injective. We would like to show $$f$$ is surjective.

Suppose $$y\in Y$$. $$x:=p(y)$$ must lie in the image of $$q$$; pick $$z$$, $$q(z')=x$$. Because $$p(f(z))=x$$ too, $$f(z)$$ must be $$G$$-related to $$y$$; there is some $$g$$ with $$y=g\cdot f(z')=f(g\cdot z')$$. Therefore $$y$$ is in the image of $$f$$, so $$f$$ surjects.

• I do not understand what "and $q^{-1}(W)\subseteq U'$ some open neighbourhood of $z$, $p^{-1}(W)$ some open neighbourhood of $y$" means. Certainly $q^{-1}(W)$ is not contained in the single sheet $U'$. Commented Aug 22, 2023 at 22:59
• @KritikerderElche Ah, good catch. I unconsciously identified (harmlessly) $p$ and $q$ with their restrictions to the single sheet. I'll edit this Commented Aug 22, 2023 at 23:03