Condition that a local homeomorphism be a covering map.

Question

Let be $f:Y\to X$ a local homeomorphism, with $Y$ a compact space and $X$ a Hausdorff connected space. How can I show that, for each $x\in X$, $f^{-1}(x)\subset Y$ is finite?

So, is clear that $f$ is surjective and I know to do this with the hypothesis that $Y$ is a metric space, but not when $Y$ is just a topological space.

The whole question is:

Show that: If $f:Y\to X$ a local homeomorphism, with $Y$ a compact space and $X$ a Hausdorff connected space then $f$ is a covering map.

Answer 1

Suppose $F_y:= p^{-1}[\{x\}] \subseteq Y$ were infinite. Then $Y$ is compact, so there is a cluster point $y \in Y$, and as $F_y$ is closed (follows from Hausdorff plus continuity), $y \in F_y$. If $U$ is a neighbourhood of $y$ then there is some $y' \neq y$ in $U$ such that $y' \in F_y$ as well. So both $y$, $y'$ map to $x$ under $f$ so $f$ is not even 1-1 on $U$. This contradicts being a local homeomorphism.

So $F_y$ is closed and finite.

Answer 2

Hint: preimage of a point by a local homeomorphism is discrete.