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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!

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    $\begingroup$ 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. $\endgroup$ – Derek Elkins left SE Nov 28 '18 at 21:04
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    $\begingroup$ There's also a detailed proof in Borceux's Handbook of Categorical Algebra Vol. 1, Theorem 2.13.4, pg. 79 in my copy. $\endgroup$ – BW. Nov 29 '18 at 10:30
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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 $j$.

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