Question: Let $X$ and $Y$ be topological spaces, $X$ compact, $f : X \to Y$ continuous. Then the preimage of each compact subset of $Y$ is compact.

With the stipulation that $X$ and $Y$ are metric spaces, this is a theorem in Pugh's *Real Mathematical Analysis*. The proof uses sequential compactness. Is this theorem true in general (*i.e.* can it be proved with covering compactness alone)?

Answer: 
The claim is true if $Y$ is Hausdorff: if $C\subseteq Y$ is compact, then it is closed; therefore $f^{-1}(C)$ is closed in $X,$ hence compact.

For a counterexample, take $Y=X$ with the indiscrete topology and $f$ the identity map. Then every subset of $Y$ is compact, which can be easily arranged for $X$ not to.
A map $f:X\to Y$ is called *proper* if the preimage of every compact subset is compact. It is called *closed* if the image of every closed subset is closed.

If $X$ is a compact space and $Y$ is a Hausdorff space, then every continuous $f:X\to Y$ is closed and proper.

Here are some examples where $f$ is not proper:

* With $X$ compact: Let $X=[0,1]$ and $f=\text{Id}:(X,\tau)\to(X,\sigma)$ where $\tau$ is the Euclidean topology and $\sigma$ is the cofinite topology.
* With $Y$ Hausdorff: Let $f:\Bbb R\to\{\*\}$ be the constant map.