# Filtered colimits commute with finite limits

I am trying to prove the following fact that given $$I$$ filtered index, $$J$$ finite index and diagram $$F:I\times J \rightarrow \it{Sets}$$, $$colim_{i\in I} lim_{j\in J}\;F_{ij}=lim_{j\in J}colim_{i\in I}\; F_{ij}$$; provided all the corresponding limits and co-limits exist.

Now, I have considered $$F_{j}= colim_{i\in I}\;F_{ij}$$ for fixed $$j\in J$$ and $$F_{i}= lim_{j\in J} \; F_{ij}$$ for fixed $$i\in I$$. And I'm trying to find maps $$F_{i}\rightarrow lim_{j\in J}\;F_{j}$$. However I'm completely clueless about how to use the finite index condition.

I can't seem to find a proof of this anywhere and am stuck for some time. So, I would like to see a detailed proof if possible. Thanks in advance!

• There's a proof in "Categories for the Working Mathematician". Note, that you very much need to use properties of $\mathbf{Set}$ to do this. It is not true that filtered colimits commute with finite limits in any category with the requisite (or even all) limits and colimits. – Derek Elkins left SE Nov 28 '18 at 21:04
• There's also a detailed proof in Borceux's Handbook of Categorical Algebra Vol. 1, Theorem 2.13.4, pg. 79 in my copy. – BW. Nov 29 '18 at 10:30

You don't need anything about finiteness or filteredness at this stage. You just compose the projection $$F_i\to F_{ij}$$ with the injection $$F_{ij}\to F_j$$ and check that this passes to the limit over $$j$$.