How to show that finite holomorphic maps are proper?

Question

I would like to solve Exercise 1.1.5 of Introduction to Singularities and Deformations, which states

Show that finite maps between complex spaces are proper.

A map is called finite if it is closed, and quasi-finite, i.e. every point has only finitely many preimages, and it is called proper if the preimage of every compact subspace is compact.

So suppose $f: X \to Y$ is finite, and $K \subset Y$ is compact. We would like to show that $f^{-1}(K)$ is compact, and I had some hopes that this goes similar to proving that a finite covering of a compact space is compact. In particular since Lemma 1.54 states

Let $f: X \to Y$ be a finite map of topological spaces where $X$ is hausdorff, let $y \in Y$ and let $f^{-1}(y) = \{x_1, \dotsc, x_s\}$. Further let $U_i' \subset X$ be pairwise disjoint open neigbourhoods of $x_i$. Then, for each open neighbourhood $V'$ of $y$, there exists an open neighbourhood $V \subset V'$ of $y$ such that with $U_i = U_i' \cap f^{-1}(V)$

1. $f^{-1}(V) = U_1 \cup \dotsb \cup U_s$
2. the restrictions $f: U_i \to V$ are finite

I had the hope that those $U_i$ somehow replace the evenly covered neighourhoods in the case of a covering, but the proofs I found all use that a covering is locally homeomorphic.

Answer 1

If $f^{-1}(K)$ is not compact, there is a sequence $(x_n)$ in $f^{-1}(K)$ with no accumulation point in $f^{-1}(K)$ (and hence no accumulation point in $X$ since $f^{-1}(K)$ is closed in $X$). Since $K$ is compact, the sequence $(f(x_n))$ must accumulate at some point $y\in Y$. Now since $(x_n)$ has no accumulation point, there is no value that is repeated infinitely many times among the $x_n$'s. Since $f$ is quasifinite, there is also no value that is repeated infinitely many times among the $f(x_n)$'s. In particular, only finitely many of the $f(x_n)$'s can be equal to $y$; omitting those finitely many values from our sequence, we may assume $f(x_n)\neq y$ for all $n$.

Now consider the set $A=\{x_n:k\in\mathbb{N}\}$. It is closed in $X$ since our sequence $(x_n)$ has no accumulation point. Thus since $f$ is closed, $f(A)$ is also closed. But this is a contradiction, since $(f(x_n))$ accumulates at the point $y$ and $y\not\in f(A)$.

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This argument uses only the assumption that $f:X\to Y$ is a continuous closed map from a metric space to a Hausdorff space whose fibers or finite (in fact, with minor modifications, the assumption that $Y$ is Hausdorff can be dropped and it suffices to assume the fibers are compact rather than finite). More generally, in fact, it can be shown that if $f:X\to Y$ is a continuous closed map of arbitrary topological spaces whose fibers are compact, then $f$ is proper. The proof is a bit trickier though since you can't just use sequences.

Here's how the argument goes. Suppose $K\subseteq Y$ is compact and $f^{-1}(K)$ is not compact. Then there is a family $(C_i)$ of closed subsets of $X$, closed under finite intersections, such that $C_i\cap f^{-1}(K)$ is nonempty for all $i$ but $\bigcap_i C_i\cap f^{-1}(K)$ is empty. Now since $f$ is closed, the sets $f(C_i)$ are all closed, and they also all have nonempty intersection with $K$. Since $K$ is compact, this means there is some point $y\in \bigcap_i f(C_i)\cap K$. But now since $f^{-1}(\{y\})$ is compact and $\bigcap C_i\cap f^{-1}(\{y\})\subseteq\bigcap_i C_i\cap f^{-1}(K)=\emptyset$, there is some $i$ such that $C_i\cap f^{-1}(\{y\})$ is empty. This is a contradiction, since by our choice of $y$, $y\in f(C_i)$.