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if $X$ and $Y$ are Hausdorff spaces, $f:X \to Y$ is a local homeomorphism, $X$ is compact, and $Y$ is connected, is $f$ a covering map?

It seems to be, and I almost have a proof, but I'm stuck at the very end of it:

I've already proved that $f$ is surjective (using the connectedness), and that for each $y \in Y$, $f^{-1}(y)$ is finite. Because $X$ is compact, there exists a finite open cover of $X$ by $ \{ U_i \}$ such that $f(U_i)$ is open and $f |_{U_i}:U_i \to f(U_i) $ is a homeomorphism.
For each $y \in Y$, we choose the subset $ \lbrace U_{i_j} \rbrace $ such that $y \in U_{i_j}$, and then define $V = \bigcap_{j=1}^k f(U_{i_j})$, and $U'_j = U_{i_j} \bigcap f^{-1}(V)$.

... and this is were I got stuck. I really want to write that $f^{-1}(V) = \bigcup_{j=1}^k U'_j$ (more or less proving it's a covering map), but I can't justify that, and I actually think that it's not true. I think I might need an extra step, and to take an even smaller neighborhood of $y$, in order to make sure that extra sets from $ \lbrace U_i \rbrace $ didn't sneak into $f^{-1}(V)$.

Any help would be greatly appreciated as I've already spent several hours working on this problem.

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  • $\begingroup$ If you have already shown that $f$ is surjective, why are surjectivity and local homeomorphism not enough to show that $f$ is a covering map? For $y \in Y,$ we know that there is a neighborhood $U$ around $f^{-1}(y) \in X$ on which the restriction of $f$ is a homeomorphism, so $V = f(U)$ is a neighborhood around $y$ onto which $f^{-1}(V) = U$ is mapped homeomorphically. With surjectivity, isn't this enough? $\endgroup$
    – The Ledge
    Commented Oct 10, 2020 at 3:47
  • $\begingroup$ Why $f:X\rightarrow Y $ is surjective? $\endgroup$
    – Infinity
    Commented Jan 22, 2023 at 4:16

3 Answers 3

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For $y \in Y$, let $\{x_1, \dots, x_n\}= f^{-1}(y)$ (the $x_i$ all being different points). Choose pairwise disjoint neighborhoods $U_1, \dots, U_n$ of $x_1, \dots, x_n$, respectively (using the Hausdorff property).

By shrinking the $U_i$ further, we may assume that each one is mapped homeomorphically onto some neighborhood $V_i$ of $y$.

Now let $C = X \setminus (U_1 \cup \dots \cup U_n)$ and set $$V = (V_1 \cap \dots \cap V_n)\setminus f(C)$$

If I'm not mistaken this $V$ should be an evenly covered nbh of $y$.

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    $\begingroup$ Seems right to me. f is a closed map (compactness+hausdoff), C is closed so f(C) is closed, so V is open, and $f^{-1}(y)\bigcap C = \emptyset$ so $y \in V$, and so it's indeed an open nbh of y. $f^{-1}(V) \subset U_1 \cup \ldots \cup U_n$ by definition, and so $f^{-1}(V) = (U_1 \cap f^{-1}(V)) \cup \ldots \cup (U_1 \cap f^{-1}(V))$ when $U_i \cap f^{-1}(V) \cong V$. $\endgroup$
    – Or Sharir
    Commented Jun 17, 2011 at 23:23
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    $\begingroup$ Right, looks like you finished the proof. :) $\endgroup$
    – Sam
    Commented Jun 17, 2011 at 23:31
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    $\begingroup$ @Sam why $f^{-1}(y)$ is finite? where is the compactness is nedded to argue that? $\endgroup$
    – Myshkin
    Commented Feb 28, 2013 at 6:05
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    $\begingroup$ @CityOfGod: Since $f$ is a local homeomorphism, the set $f^{-1}(y)$ consists of isolated points. Since $X$ is compact, this can only be the case, if $f^{-1}(y)$ is finite. $\endgroup$
    – Sam
    Commented Feb 28, 2013 at 6:13
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    $\begingroup$ @evaristegd: $f^{-1}(y) \subset X$ is a closed subset, hence compact (since $X$ is compact). For each $x \in f^{-1}(y)$ there exists a homeomorphically mapped neighbourhood $U_x$ of $x$. In particular, this implies that $x' \notin U_x$ for $x' \in f^{-1}(y)$, $x'\ne x$ (since $f|_{U_x}$ is injective). But at the same time, finitely many of the $U_x$ cover $f^{-1}(y)$ by compactness, so there can be at most a finite number of points in $f^{-1}(y)$. $\endgroup$
    – Sam
    Commented May 21, 2019 at 16:50
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Here is a complete solution, said slightly differently than, but in the same spirit as, Sam's solution.

  1. Show that $f$ is surjective. We use the fact that $Y$ is connected and Hausdorff. Local homeomorphisms are open, so $U=f(X)$ is an open subset of $Y$. Since $X$ is compact, $f(X)$ is compact, and $Y$ Hausdorff implies that compact subsets are closed. So, $V=Y\setminus f(X)$ is also open. If $f$ were not surjective, then $V\neq \emptyset$, and $U,V$ would be separating sets for $Y$, contradicting connectedness of $Y$. We conclude that $f$ is surjective.

  2. For each $y\in Y$, $f^{-1}(y)$ is finite. Again using $Y$ Hausdorff, $\{y\}$ is closed, so $f^{-1}(y)$ is a closed subset of the compact space $X$, hence compact. For each $x\in f^{-1}(y)$, let $U_x$ be a neighborhood of $x$ where $f$ restricts to a homeomorphism. Such neighborhoods exist by the assumption that $f$ is a local homeomorphism. Then $\{U_x : x\in f^{-1}(y)\}$ is an open cover of $f^{-1}(y)$, hence has a finite subcover which we label $\{U_i\}_{i=1}^n$. The map $f$ is injective on each $U_i$, thus only contains one pre-image of $y$. Hence $y$ has finitely many pre-images in $X$.

  3. Get an evenly covered neighborhood of $y$. Keeping the cover $\{U_i\}$ from the previous step, $V = \bigcap_{i=1}^n{f(U_i)}$ is an open neighborhood of $y$. Then $\{f^{-1}(V)\cap U_i\}$ is a disjoint collection of open neighborhoods, each homeomorphic to $V$ under $f$ since the restriction of a homeomorphism to a subspace is a homeomorphism. Thus, $V$ is an evenly covered neighborhood of $y$.

Therefore, $f$ is a covering map.

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    $\begingroup$ Comment after five years ;-) There is no reason based on the construction why the $U_i$ should be pairwise disjoint. However, you can achieve this by shrinking them (using the fact that $X$ is regular). But even then I do not see why $f^{-1}(V) \subset \bigcup U_i$. $\endgroup$
    – Paul Frost
    Commented Sep 17, 2019 at 22:05
  • $\begingroup$ Paul is correct. The proof given in step 3 is incorrect, the same line can be used to show every surjective local homeomorphism is a covering map, which is wrong. See Sam's answer for the correct construction of $V$. $\endgroup$
    – GK1202
    Commented Feb 13, 2022 at 12:41
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cp. Fulton, Algebraic Topology, Proposition 19.3, p.266. He uses the compactness of X. But a problem in the John Lee's book Introduction to Topological Manifolds is this (Problem 11-9): Show that a proper local homeomorphism between connected, locally path-connected, compactly generated Hausdorff spaces is a covering map.

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    $\begingroup$ I think the second part of your answer would better fit as a separate question. If you ask it, just link here and say that it is related but not exactly the same. $\endgroup$
    – t.b.
    Commented Mar 31, 2012 at 14:03
  • $\begingroup$ Fulton's proof only applies when $Y$ is first countable and $X$ is sequentially compact. This is less general than the question. $\endgroup$
    – Paul Frost
    Commented Sep 17, 2019 at 22:13
  • $\begingroup$ The proof is really not that different. Let $f: X \rightarrow Y$ be the proper local homeomorphism. $f$ is open and continuous by Proposition 2.31, and closed by Theorem 4.95. Since $Y$ is connected, $f$ is surjective. $A \equiv f^{-1}(y)$ is compact because $f$ is proper. Cover each point $x$ in $A$ with $U_x$ and there's a finite cover. But each $U_x$ can contain no point in $A$ except $x$ itself, because of local homeomorphism. Thus $f^{-1}(y)$ is finite. $\endgroup$ Commented May 16 at 20:50

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