# Why must a finite covering map be closed?

A covering map $p:C\to X$ is finite when for each $x\in X$ we have $|p^{-1}(x)|<\infty.$ I have to prove that such a covering map has to be closed. I'm having trouble with it.

When $p$ is a covering map, we can take open neighborhoods $U_x$ of every point $x\in X$ such that $p^{-1}(U_x)$ is a disjoint union of open neighborhoods $U_{x_i}$ of $x_i$, where $\{x_1,\ldots,x_n\}=p^{-1}(x)$, and $p$ maps each $U_{x_i}$ homeomorphically onto $U_x$. I see that when I have a closed subset $A\subseteq C$ such that $p(C)\subseteq U_x$ for a certain $x\in X$, then $p(A)$ is closed. That's because in this case $A$ is a disjoint union of $A\cap U_{x_i}$ and each of this sets is closed. (Well, certainly they're closed in $U_{x_i}.$ I'm not sure why it must be closed in $C$, but I think it must.) Since $p$ is a homeomorphism on $A\cap U_{x_i}$, we have that each $p(A\cap U_{x_i})$ is closed. (Again, certainly in $U_x$, and I think in $X$ too.) Since there are finitely many of them, they're union, that is $p(A)$ is closed too.

I'm sure there are problems with this reasoning, which show how little I understand of topology, but I think the gist of it is right. But I don't see how I can make this local property global. What if $p(A)$ is large? And what if $A$ spans several components of $C$?

This is a homework question.

• isn't there an assumption on $X$ to be locally compact or something like that ? – Glougloubarbaki Mar 21 '13 at 13:09
• @Glougloubarbaki: Non! – Georges Elencwajg Mar 21 '13 at 13:18

Let $(U_i)_{i\in I}$ be an arbitrary open cover of a a topological space $X$.
The crucial remark is that a subset $F\subset X$ is closed in $X$ if and only if for each $i\in I$ the intersection $F\cap U_i$ is closed in $U_i$.

In your situation, if you have a closed subset $A\subset C$ you should apply the above to $F=f(A)$ and to a trivializing cover $(U_i)_{i\in I}$ of your covering $p:C\to X$.
In a nutshell: you may suppose that $p$ is a trivial covering with finite fibers.

Remark
The result is false if the covering has infinite fibers.
A counterexample is given by the covering of the unit circle in the complex plane $$p:\mathbb R\to S^1:t\mapsto \exp (2\pi it)$$ and the closed set $A=\{n+1/n\mid n\in \mathbb N^*\}$, whose image $p(A)\subset S^1$ is not closed [it accumulates at $1\in S^1$, which is not in $p(A)$ ].

Let $A \subset C$ be closed and let $x \notin p(A)$. Because the covering is finite, there $x_1,...,x_n \in C$ such that $p^{-1}(x)=\{x_1,...,x_n\}$. For every $1 \leq i \leq n$, there exists an open neighborhood $U_i$ of $x_i$ such that $U_i \cap A = \emptyset$ and $p$ induces a homeomorphism between $U_i$ and an open neighborhood $p(U_i)$ of $x$.

Now, consider $U = \bigcap\limits_{i=1}^n p(U_i)$: it is an open neighborhood of $x$ not intersecting $p(A)$.