Neighborhood of $f^{-1}(K)$ where $K$ is compact and $f$ is proper I met the following question in point set topology (actually in my study of sheaf theory using Joseph Taylor's beautiful textbook 'Several complex variables with connections to algebraic geometry and Lie groups'):
Let $X$ and $Y$ be two locally compact topological spaces, and $f:X\rightarrow Y$ a continuous, proper map (recall: proper means any preimage of a compact set is compact). Let $K\subset Y$ be any compact subset, and $U\subset X$ an arbitrary neighborhood of $f^{-1}(K)$. Prove that we can always find a neighborhood $V\subset Y$ of $K$ such that $f^{-1}(V)\subset U$.
This is a claim on Taylor's textbook, page 156. We can reduce the question to the case $K=\{y\}$ is a single point, for it suffices to find such a neighborhood locally, of course. We may also assume that $U$ is precompact (i.e. its closure in $X$ is compact). Then $f(U)\subset Y$ is a precompact set containing $y$. How to find an open neighborhood of $y$ such that its preimage lies in $U$? I have no idea how to solve this problem.
Any advice or hint would be very helpful, thanks a lot in advance!
 A: Let $V = Y \setminus f(X \setminus U)$.

*

*Let's prove that $K \subset V$ : let $y \in f(X \setminus U)$. Then there exists $x \in X \setminus U$ such that $y = f(x)$. So $x \notin U$, so $x \notin f^{-1}(K)$, and because $f(x)=y$, you deduce that $y \notin K$. So we proved
$$f(X \setminus U) \subset Y \setminus K, \quad \quad \quad \text{i.e.} \quad \boxed{K \subset V}$$


*Let's prove that $f^{-1}(V) \subset U$ : Let $x \in f^{-1}(V)$. Then $f(x) \in V$, so by definition of $V$, one has $f(x) \notin f(X \setminus U)$, so $x \notin X \setminus U$, i.e. $x \in U$. This proves directly
$$\boxed{f^{-1}(V) \subset U}$$


*Let's prove that $V$ is open : Let $y \in V$. Because $Y$ is locally compact, $y$ has a neighbourhood $W$ such that $\overline{W}$ is compact. Since $f$ is proper, then $f^{-1}(\overline{W})$ is compact, and because $U$ is open, then $(X \setminus U) \cap f^{-1}(\overline{W})$ is still compact. But $f$ is continuous, so $f \left[ (X \setminus U) \cap f^{-1}(\overline{W})\right]$ is compact. In particular, it is closed, so
$$Z : = Y \setminus f \left[ (X \setminus U) \cap f^{-1}(\overline{W})\right]$$
is open. It is clear by definition that $y \in Z$ and that $Z \subset V$, which proves finally that $V$ is open.
