Proper maps between Hausdorff spaces I was asked to prove these facts:
let $f:X \rightarrow Y$ be a continuous map between two Hausdorff spaces.
i) $f$ is proper if and only if it is closed and $f^{-1}(y)$ is compact for any $y \in Y$;
ii) if $f$ is proper and a local homeomorphism, and $Y$ is connected, then $f^{-1}(y)$ is finite of the same cardinality for any $y \in Y$.
I feel like there is something wrong or at least missing in the hypotheses. Maybe I could say something about cardinality in point ii) if $X$ were compact? Is it possible to prove these facts just using the given hypotheses?
 A: i) and the beginning of ii) are true under only these hypotheses. 
You haven't asked about i) so I won't expand on it. For the finiteness part of ii), you can use Daniel Fischer's comment: $f^{-1}(y)$ will be a discrete subspace of $X$ for any $y$; but $f$ is proper so...
However the fact that $|f^{-1}(y)|$ is constant is not true under these hypotheses only. 
Indeed, let $Y=]0;1[$ and $X=\displaystyle\bigsqcup_{n\in\mathbb{N}}]0; \frac{1}{2^n}[$; and let $f:X\to Y$ be the projection. 
Then $f$ is continuous by the universal property of the disjoint sum; $X,Y$ are Hausdorff and $Y$ is connected. $f$ is also a local homeomorphism: if $x\in X$, for instance $x\in ]0,\frac{1}{2^n}[$ then $f_{\mid ]0,\frac{1}{2^n}[}$ is clearly a homeomorphism onto its image. 
Finally, $f$ is proper: let $K\subset Y$ be compact.Then $f^{-1}(K)$ is closed; and moreover there are $a>0, b<1$ such that $K\subset [a,b]$ so there is $n\geq 0$ such that $f^{-1}(K) \subset \displaystyle\bigsqcup_{k=0}^n [a,b]\cap ]0,\frac{1}{2^n}[$ which is compact as a finite union of compact sets; hence $f^{-1}(K)$ is closed in a compact space, so compact. 
However, as you can clearly see, $|f^{-1}(y)|$ is not constant (though finite, this part is indeed true).
A general situation where you get that this cardinal is constant is when you have a covering map and $Y$ is connected. In this situation, $y\mapsto |f^{-1}(y)|$ is locally constant, hence constant since $Y$ is connected (this works even for infinite cardinals !)
