Let $f:X\to Y$ be a local homeomorphism between Hausdorff spaces such that $f$ is surjective and $f^{-1}(y)$ is compact $\forall y\in Y$. If $|f^{-1}(y_1)|=|f^{-1}(y_2)|,\forall y_1,y_2\in Y$ then show that $f$ is a covering map.
Attempt: I know $f^{-1}(y)$ is finite as it has a finite cover $\cup_{i=1}^nU_i\supseteq f^{-1}(y)$ with each $U_i$ open such that $f_{U_i}$ is a homeomorphism. I considered $V_i\subseteq U_i$ with $V_i\cap V_j=\emptyset$ but the problem is the preimage of $\cap_{i=1}^nf(V_i)$ is not necessarily in $\cup_{i=1}^nU_i$ otherwise I would have found an evenly covered neighborhood of $y$. I need to show that $f$ is closed so I can use the set $(\cap_{i=1}^nf(V_i))\backslash (f(X\backslash \cup_{i=1}^nU_i))$ instead or consider another approach by using the constant preimage assumption which I do not see how it fits there…
Best Answer
It is much simpler than I thought it seems: the sets $V_i$ are the disjoint restrictions of the $U_i$ as defined in my question above. Suppose that $|f^{-1}(y)|=n,\forall y\in Y$. The problem is that taking the set $V=\cup_{i=1}^nf(V_i)$, I want to show that $f^{-1}(V)\subseteq \cup_{i=1}^nV_i$ so I can use the local 1-1 property and conclude that the preimages are exactly the open sets $f^{-1}(V)\cap U_i$. This is straghtforward however: if $y_0\in f^{-1}(V)$ then there exist $v_1\in V_1,...,v_n\in V_n: f(v_i)=y_0$ as we simply have $v_i=f_{V_i}^{-1}(y_0)$. But $|f^{-1}(y_0)|=n \implies f^{-1}(y_0)=\{v_1,...,v_n\}$ and thus $f^{-1}(y_0)\subseteq \cup_{i=1}^nV_i, \forall y_0\in Y$. The result now follows immediately.