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: 
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.
**Disclaimer:** This answer is intended to compliment the above, and thought it is not fully generally, but is has often been helpful in my experience... 

If $Y$ is $\mathbb{R}$ then its sufficient for $f$ to be [coercive](https://en.wikipedia.org/wiki/Coercive_function). When $X$ is a topological vector space, the coercivity condition reduces to the requirement that
$$
\lim\_{\|x\|\to \infty} f(x)=\infty,
$$
which is typically easy to check.