Inverse image of a compact set is compact

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.

This is not true in general.

Let $X=Y=[0,1]$. Take $X$ with the usual topology. For $Y$, take the topology $$\tau=\left\{\varnothing,Y,(1/2,1]\right\}.$$ Then $id:x\in X\mapsto x\in Y$ is continuous, but $(1/2,1]=id^{-1}(1/2,1]$ is not compact, although $(1/2,1]$ is compact in $Y$.

On the other hand, if $Y$ is Hausdorff, then every compact of $Y$ is closed, so the inverse image of compact sets is closed, hence compact.

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.