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Quite a number of contributions in math.stackexchange deal with important aspects of this question. Here are some examples:

The purpose of this question is to give a standard reference in this forum providing a systematic account. Here are the basic definitions.

Let $f : X \to Y$ be a map (= continuous function) between topological spaces $X, Y$.

$f$ is a local homeomorphism if each $x \in X$ has an open neighborhood $U$ which is mapped by $f$ homeomorphically onto an open neighborhood $V$ of $f(x)$. Note that a local homeomorphism is an open map.

An open subset $V \subset Y$ is evenly covered by $f$ if $f^{-1}(U) = \bigcup_{\alpha \in A} U_\alpha$ with pairwise disjoint open $U_\alpha \subset X$ such that the restrictions $f_{\alpha,V} : U_\alpha \xrightarrow{f} V$ are homeomorphisms. Here we allow $A = \emptyset$, i.e. $f^{-1}(V) = \emptyset$. Note that all fibers $f^{-1}(y)$ with $y \in V$ have the same cardinality.

$f$ is a covering projection if it is surjective and each $y \in Y$ has an open neighborhood $V$ which is evenly covered by $f$.

Clearly each covering projection is a local homeomorphism, but the converse fails. Here are some standard counterexamples.

  • The inclusion $i : X \to Y$ of a proper open subspace $X \subset Y$ is a local homeomorphism, but it is not surjective.

  • The map $f : (0,4\pi) \to S^1, f(t) = e^{it}$, is a local homeomorphism. However, $y = 1$ does not have an evenly covered open neighborhood.

Some obvious questions are:

  1. Are there conditions on $X, Y$ assuring that a local homeomorphism is surjective?

  2. A typical feature of a covering projection $f$ is that for each connected component $C \subset Y$ all fibers $f^{-1}(y)$ with $y \in C$ have the same cardinality. What can be said about the fibers of a local homeomorphism?

  3. Which conditions assure that a surjective local homeomorphism is a covering projection?

Paul Frost
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1 Answers1

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Let $f : X \to Y$ be local homeomorphism with $X \ne \emptyset$.

Question 1

Lemma 1. Let $X$ be compact and $Y$ be connected and Hausdorff. Then $f$ is a surjective closed map.

Proof. $f(X)$ is a non-empty open and closed subset of $Y$. Thus $f(X) = Y$ by connectedness. That $f$ is closed is well-known.

Question 2

Lemma 2. All fibers $f^{-1}(y)$ are discrete.

Proof. Let $x \in f^{-1}(y)$. Choose an open neighborhood $U$ of $x$ which is mapped by $f$ homeomorphically onto an open subset $V \subset Y$. Then $U$ cannot contain a point $x'$ of $f^{-1}(y)$ distinct from $x$ because this would contradict injectivity of $f$ on $U$.

Lemma 3. If $X$ is Hausdorff and $f^{-1}(y)$ is finite, then there exists an open neighborhood $V$ of $y$ such that the cardinality of $f^{-1}(y')$ with $y' \in V$ is greater or equal than the cardinality of $f^{-1}(y)$.

Proof. Let $f^{-1}(y)$ consist of $n$ points $x_1,\ldots,x_n$ ($n = 0$ is allowed and is a trivial case). There exist pairwise disjoint open neighborhoods $U'_i$ of $x_i$. See for example Let $x_1,\dots,x_n$ be distinct elements of a Hausdorff space, prove there are pairwise disjoint open sets $U_1,\dots,U_n$ s.t. $x_i \in U_i$. Choose open neighborhoods $U''_i$ of $x_i$ which are mapped by $f$ homeomorphically onto open neighborhoods $V'_i$ of $y$. Then the $U_i = U'_i \cap U''_i$ are mapped by $f$ homeomorphically onto open neighborhoods $V_i$ of $y$. Let $V = \bigcap_{i=1}^n V_i$ and $W_i = f_i^{-1}(V)$, where $f_i : U_i \xrightarrow{f} V_i$. The $W_i$ are pairwise disjoint and $\bigcup_{i=1}^n W_i \subset f^{-1}(V)$. This shows that each fiber $f^{-1}(y')$ with $y' \in V$ has at least $n$ elements.

Warning. In general $\bigcup_{i=1}^n W_i \subsetneqq f^{-1}(V)$. The second counterexample in the question shows that the cardinality of $f^{-1}(y')$ can be greater than that of $f^{-1}(y)$ for all $y' \ne y$.

Corollary 4. If $X$ is Hausdorff and $y \in Y$ has an open neighborhood $V'$ such that all fibers $f^{-1}(y')$ with $y' \in V'$ have the same finite numbers of elements, then there exists an open neighborhood $V$ of $y$ which is evenly covered by $f$.

Proof. Recall the construction of $V$ and $f_i : W_i \to V$ in the preceding proof. We may of course assume that $V_i \subset V'$. Then $\bigcup_{i=1}^n W_i = f^{-1}(V)$ because all fibers over $V$ have the same finite number of elements.

Question 3

Theorem 5. Let $X$ be Hausdorff, $f$ be a closed map and $f^{-1}(y)$ be finite. Then $y$ has an open neighborhood $V$ which is evenly covered by $f$.

Proof. Let $V$ and $f_i : W_i \to V$ as constructed above. Then $U = \bigcup_{i=1}^n W_i$ is an open subset of $X$ containing $f^{-1}(y)$. Hence $C = X \setminus U$ is closed in $X$ so that $f(C)$ is closed in $Y$. We have $y \notin f(C)$ (the assumption $y \in f(C)$ yields $y = f(x)$ with $x \in X \setminus U$ which implies $x \notin f^{-1}(y) \subset U$, hence $y \ne f(x)$, a contradiction).

Therefore $V' = V \setminus f(C)$ is an open neighborhood of $y$ and the $W'_i = f_i^{-1}(V')$ are pairwise disjoint open subsets of $X$ which are mapped by $f$ homeomorphically onto $V'$. We claim that $f^{-1}(V') = \bigcup_{i=1]}^n W'_i$ which shows that $V'$ is evenly covered by $f$. Let $x \in f^{-1}(V')$, i.e. $f(x) \in V'$. Therefore $f(x) \notin f(C)$ so that $x \notin C$. This means $x \in U$ and thus $x \in W_i$ for some $i$. Since $f_i : W_i \to V$ is a homeomorphism and $f_i(x) = f(x) \in V'$, we see that $x \in f_i^{-1}(V') = W'_i$.

Corollary 6. Let $X$ be Hausdorff, $f$ be a closed map and all fibers $f^{-1}(y)$ be finite non-empty sets. Then $f$ is a covering projection.

Remark. Covering projections are in general no closed maps. As an example take $f : \mathbb R \to S^1, f(t) = e^{it}$. Then the set $M = \{ 2\pi n + 1/n \mid n \in \mathbb N \}$ is closed, but $f(M)$ is not. However, covering projections with finite fibers are closed. See Why must a finite covering map be closed?

Corollary 7. Let $X$ be compact Hausdorff and $Y$ be connected and Hausdorff. Then $f$ is a covering projection.

Proof. Lemma 1 shows that $f$ is surjective closed map. Therefore all fibers are non-empty discrete subspaces of $X$. Since $Y$ is Hausdorff, each $\{y\}$ is closed in $Y$ so that $f^{-1}(y)$ is a closed subspace of $X$, hence compact. We conclude that the $f^{-1}(y)$ are finite. Now Corollary 6 applies.

Paul Frost
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