if X and Y are Hausdorff spaces, f:X→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∈Y, f−1(y) is finite. Because X is compact, there exists a finite open cover of X by {Ui} such that f(Ui) is open and f|Ui:Ui→f(Ui) is a homeomorphism.

For each y∈Y, we choose the subset {Uij} such that y∈Uij, and then define V=⋂kj=1f(Uij), and U′j=Uij⋂f−1(V).… and this is were I got stuck. I really want to write that f−1(V)=⋃kj=1U′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 {Ui} didn’t sneak into f−1(V).

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

**Answer**

For y∈Y, let {x1,…,xn}=f−1(y) (the xi all being different points). Choose pairwise disjoint neighborhoods U1,…,Un of x1,…,xn, respectively (using the Hausdorff property).

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

Now let C=X∖(U1∪⋯∪Un) and set V=(V1∩⋯∩Vn)∖f(C)

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

**Attribution***Source : Link , Question Author : Or Sharir , Answer Author : Sam*