Is there a slick proof that precompactness is preserved under continuous images?
By slick I mean analogous to the following proof that compactness is preserved under continuous images. Is it even true in complete generality?
Take an open cover of the image of $f$. Its pullback covers the domain of $f$, hence has a finite subcover of pullbacks. The corresponding opens pulled back are a finite open cover because of the counit $f_\ast f^\ast \Rightarrow 1$.
On
I assume Hausdorff spaces (at least the codomain of the continuous map should be). You want a proof that if $f\colon X\to Y$ is continuous and $Y$ is Hausdorff, then, for every precompact subset $A\subseteq X$, $f(A)$ is precompact (in $Y$).
If $f\colon X\to Y$ is continuous and $A\subseteq X$, then $f(\overline{A})\subseteq \overline{f(A)}$. On the other hand, if $C\subseteq X$ is compact, $f(C)$ is compact in $Y$, hence closed. So, if $A$ is precompact, $f(A)\subseteq f(\overline{A})$ gives that $\overline{f(A)}\subseteq f(\overline{A})$.
Therefore if $A$ is precompact, $\overline{f(A)}=f(\overline{A})$ is compact and therefore $f(A)$ is precompact.
If, on the other hand, you want to prove that if $f\colon X\to Y$ is a continuous map of metric spaces and $X$ has compact completion, then also $f(X)$ has compact completion, then the assertion is false.
On
As the above example shows, one has to assume that the map extends continuously to the closure.
Given this,continuity easily implies that the closure of the image is contained in the image of the closure and hence is the continuous image of a compact set and thus compact.
To spell the argument for $cl(f(X))\subset f(cl(X))$ out: assume $y\in cl(f(X))$, then there is a sequence $x_n\in X$ with $y=\lim f(x_n)$. Some subsequence $x_{n_k}$ converges towards some $x\in cl(X)$. By continuity $f(x)=y$, so $y\in f(cl(X))$.
The interval $(-1, 1)$ has compact closure and $x \mapsto \tan(\pi/2x)$ is continuous, but its image, the whole line, does not.