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I'm working on solving as many of the exercises in Lenstra's Galois Theory for Schemes as possible, but there is one problem I'm partially stuck on. The statement of the problem is:

$\textbf{3.12}:$ Let $X$ and $Y$ be topological spaces, and $f:Y\rightarrow X$ a finite covering. Prove that $f$ is open and closed.

Showing that $f$ is open is no problem (in fact, it doesn't even depend on the finite condition), and showing that $f$ is closed seems like it should be simple as well, but there is one point which I can only justify after making the additional assumption that $f$ is injective (and since this is a finite covering, injectivity makes the whole problem trivial and uninteresting). The proof I came up with is given below:

$\textit{Proof.}$ Let $A\subseteq Y$ be closed, and consider $x\not\in f(A)$. Then we need to show that $x$ is not a limit point of $f(A)$. We can write the fiber over $x$ as

$$f^{-1}(x)=\{y_1,\cdots,y_n\}.$$

Since $x\not\in f(A)$, we know that $y_i\not\in A$ for $1\leq i\leq n$. But $A$ is closed, so none of the $y_i$ are limit points of $A$; therefore, we can find open neighborhoods $U_i$ of $y_i$ for each $i$ such that $A\cap U_i=\emptyset$. Since $f$ is a open from the first portion of the exercise, the set

$$U=\bigcap_{i=1}^nf(U_i)$$

is an open neighborhood of $x$, and since $U_i\cap A=\emptyset$ we have that

$$U\cap f(A)=f\left(\bigcap_{i=1}^nU_i\cap A\right)=f(\emptyset)=\emptyset.$$

Therefore, $x$ is not a limit point of $f(A)$. This means that $f(A)$ contains all of its limit points; hence it is closed.

The important thing to note is: I required $f$ to be injective in order to factor it out of the intersection. In other words, if $f$ is not injective, then we can only say

$$U\cap f(A)\supset f\left(\bigcap_{i=1}^nU_i\cap A\right)=f(\emptyset)=\emptyset,$$

which doesn't help at all. So the question is: is there any way to adapt this proof to get around this problem?

I'm sure that finiteness comes to the rescue somehow, but I'm not sure how... the finiteness condition cannot be removed (consider the classic covering map $\rho:\mathbb{R}\longrightarrow S^1$ given by $\rho(t)=e^{2\pi it}$: the set $\{\tfrac{n^2+1}{n} \ | \ n\in\mathbb{Z}\setminus 0\}$ is discrete and so it is closed in $\mathbb{R}$, but its image $\{e^{\pm 2\pi i/n} \ | \ n\in\mathbb{Z}\setminus 0\}$ doesn't contain its limit point 1, so it is not closed)… and I needed finiteness to get that $U$ is open… but I cannot think of any way to get rid of the injectivity assumption…

Any help would be greatly appreciated!

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1 Answer 1

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Let $W$ be an evenly covered neighbourhood of $x$, and $V = W\cap U$. Let $V_i = f^{-1}(V)\cap U_i$. Then $f^{-1}(V) = \bigcup\limits_{i=1}^n V_i$, since $W \supset V$ is evenly covered, hence also $V$. Thus $A\cap f^{-1}(V) = \varnothing$, and therefore $f(A)\cap V = \varnothing$.

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I think that should do it! Thanks! I figured that only a little more was needed, but just wasn't seeing it.. –  user101616 Dec 19 '13 at 13:02

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