Is $f$ necessarily a covering?

Question

Let $f : X \rightarrow Y$ be a continuous map of spaces where $X$ is compact Hausdorff, $Y$ is Hausdorff and both spaces are path-connected and locally path-connected. Suppose that for every $x \in X$ there exists an open neighborhood $U$ such that $f(U)$ is open in $Y$ and $f|U : U \rightarrow f(U)$ is a homeomorphism. Is $f$ necessarily a covering?

@HennoBrandsma You're right. I missed the fact that $X$ is compact. Comment deleted ;). — Abel, May 06 '13 at 19:15

score 3 · Answer 1 · edited Apr 13 '17 at 12:20

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Some remarks that might help:

First, $f$ is clearly an open map, because it is a local homeomorphism. So $f[X]$ is open, compact (and so closed, as $Y$ is Hausdorff) in a (path-)connected space, so $f[X] = Y$ and $f$ is surjective.

Also, for every $x \in X$, let's call the $U$ from the assumptions $U_x$, so that $f$ restricted to $U_x$ is a homeomorphism between open sets. If $y$ is in $Y$, then $y = f(x)$ for some $x$, and for every such $x$, $U_x$ contains no other points of $F_y:= f^{-1}[\{y\}]$, the fibre of $y$, because $f$ is injective on $U_x$. So for every $x \in F_y$, $U_x \cap F_y = \{x\}$, so $F_y$ is a discrete subspace of $X$, and as it is a closed discrete subspace in a compact $X$, it is finite. So all fibres are finite.

The final part is provided in this answer.

Noteworthy: no local connectedness is needed in either space. Connectedness only to get surjectivity.

edited Apr 13 '17 at 12:20

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answered May 06 '13 at 19:26

Henno Brandsma

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So it is indeed a covering map ... – Stefan Hamcke May 06 '13 at 19:32
@StefanH. I believe so, but my remarks do not yet make a proof. More is needed... – Henno Brandsma May 06 '13 at 19:34
I know, but it is just a short step from here. – Stefan Hamcke May 06 '13 at 19:37
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Indeed, let $f^{-1}(y)={x_1,\ldots,x_n}$ and let $U_{x_i}$ be an open set in $X$ containing $x_i$ as stated in the problem. Let $W=\cap_{i=1}^n f(U_{x_i})$. Then $W$ is an evenly covered neighborhood of $Y$. – Cheerful Parsnip May 06 '13 at 20:00
I don't see anywhere where the locally path connected property was used though. – Cheerful Parsnip May 06 '13 at 20:01
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@GrumpyParsnip I'm, not sure whether $W$ is evenly covered. Its inverse image could be more than what's inside the $U_{x_i}$, I think. – Henno Brandsma May 06 '13 at 20:23
You are right. That's a good point. – Cheerful Parsnip May 06 '13 at 20:36

Is $f$ necessarily a covering?

1 Answers1