How I can prove that: If $X$ is compact, then any map $f\colon X \to Y$ is proper?

Question

A continuous map $f\colon X \to Y$ of locally compact spaces is called proper if for any compact $C\subset Y$ the preimage $f^{-1}(C)$ is compact. My question is: How I can prove that: If $X$ is compact, then any map $f\colon X\to Y$ is proper?

Answer 1

Assuming $Y$ is Hausdorff, if $C$ is compact then it is closed. So $f^{-1}(C)$ is closed and any closed subspace of a compact space is compact.

Answer 2

Hint: Assume $Y$ to be hausdorff. Then use that preimages under a continuous map of closed subsets are closed.

As pointed out by StefanH. and Martin in the comments, what you really need is compact subsets of $Y$ to be closed therein, which holds in hausdorff spaces. Locally compact spaces are usually assumed to be hausdorff.

Here’s an example where the criterion fails if $Y$ if compact subsets of $Y$ are not necessarily closed:

Take $Y = \{0,1\}$ with the trivial topology. And let $f : [0..1] → Y$ be the indicator function on $(0..1)$. Since the only open subsets of $Y$ are $∅$ and $Y$ itself, clearly $f$ is continuous, but $f^{-1}(\{1\}) = (0..1)$ is not compact, whereas singletons are definitely compact.