It's not hard to show that this is true in the category Set, and proofs have been written down in many places. But all the ones I know are a bit fiddly.

**Question 1:** is there a soft proof of this fact?

For example, here's a soft proof of the fact that filtered colimits in Set commute with binary products. If $J$ is a filtered category, and $R,S:J\to$ Set are functors, then

$$colim_{j\in J} R(j)\times colim_{k\in J} S(k) \cong colim_{j\in J} colim_{k\in J} R(j)\times S(k)$$ $$\cong colim_{(j,k)\in J\times J} R(j)\times S(k) \cong colim_{j\in J} R(j)\times S(j) $$

where the first isomorphism uses the fact that Set is cartesian closed, so that the functors $X\times-$ and $-\times X$ are cocontinuous; the second isomorphism is the "Fubini theorem"; and the third isomorphism follows from the fact that the diagonal functor $\Delta:J\to J\times J$ is final.

Is there some way to extend this to deal with equalizers and/or pullbacks? (The case of the terminal object is easy.)

For the sort of person who'd rather just prove the fact directly (which after all is not that hard), it's worth pointing out that this proof works not just in Set but for any cartesian closed category with filtered colimits. It works without knowing how to construct colimits in Set.

So another way to ask my question might be

**Question 2:** what is a class of categories in which you can prove that filtered colimits commute with finite limits (without first proving that this is true in Set)?

So yes, I know that the commutativity holds in any locally finitely presentable category, but the only proofs of this I know depend on the fact that it is true in Set.

mustuse something more than what you've used for finite limits. Indeed, all you really used was that $J$ was "sifted" (the diagonal map is cofinal). But sifted colimits do not commute with equalizers in general. – Dylan Wilson Mar 10 '13 at 1:23