Fun example: empty colimit does not commute with empty limit

One important property of filtered colimits is that they commute with finite limits in the category of sets.

This statement is used for example to check that a continuous morphism of sites $\mathcal{D}\to \mathcal{C}$ commuting with finite limits induces a morphism of topoi $\mathrm{Shv}(\mathcal{C})\to \mathrm{Shv}(\mathcal{D})$, i.e. the pullback map $f^{-1}$ is exact (note that a continuous morphism of sites does not induce a morphism of topoi in the general case).

The definition of a filtered system has a somewhat strange condition in that it is non-empty. For a while I was not sure why it is essential to require this property. It turns out that without this condition, filtered colimits will not commute with finite limits. Namely, the empty colimit will not commute with the empty limit (and only with it!).

I claim that the limit over an empty diagram in any category is simply a final object in that category.

Indeed, let $F:\varnothing \to X$ be an empty diagram in a category $X$. Then ${\lim }_{\varnothing }F$ is an object in $X$ such that any other object $A\in X$ admits exactly one morphism $A\to {\lim }_{\varnothing }F$. This means that ${\lim }_{\varnothing }F$ is a final object in $X$. Clearly, a final object in $\mathbf{Sets}$ is the one-point set $\{\ast \}$.

Similarly, the colimit on an empty diagram ${\mathrm{colim}}_{\varnothing }F$ is nothing more than an initial object in $X$. Clearly, an initial object in $\mathbf{Sets}$ is the empty set $\varnothing$.

Then we apply these considerations to the empty diagram $F:\varnothing \to \mathbf{Sets}$, the natural map

is the map

that is clearly not an isomorphism!