0
$\begingroup$

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 $V\subseteq f(U)$.

**

EDIT: definitions

**

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 \begin{equation} g\cdot V \cap h\cdot V = \emptyset, \quad \forall g,h\in G, \text{ } g\neq h \end{equation}

If $G$ acts evenly on $Y$, then we have the set of orbits \begin{equation} Y/G=\{ G\cdot y \text{ : } y\in Y\} \end{equation} 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.

$\endgroup$
2
  • $\begingroup$ You should make precise what a G-covering map is. $\endgroup$
    – Paul Frost
    Commented Aug 21, 2023 at 13:22
  • $\begingroup$ i have added the definitions, thanks $\endgroup$
    – dedekind1
    Commented Aug 21, 2023 at 16:41

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $U'$. $\endgroup$ Commented Aug 22, 2023 at 22:59
  • $\begingroup$ @KritikerderElche Ah, good catch. I unconsciously identified (harmlessly) $p$ and $q$ with their restrictions to the single sheet. I'll edit this $\endgroup$
    – FShrike
    Commented Aug 22, 2023 at 23:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .