Why do filtered colimits commute with finite limits? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T01:59:49Z http://mathoverflow.net/feeds/question/57099 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57099/why-do-filtered-colimits-commute-with-finite-limits Why do filtered colimits commute with finite limits? Steve Lack 2011-03-02T10:34:02Z 2011-03-03T09:42:38Z <p>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.</p> <p><strong>Question 1:</strong> is there a soft proof of this fact?</p> <p>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 </p> <p>$$colim_{j\in J} R(j)\times colim_{k\in J} S(k) \cong colim_{j\in J} colim_{k\in J} R(j)\times S(k)$$ $$\cong colim_{(j,k)\in J\times J} R(j)\times S(k) \cong colim_{j\in J} R(j)\times S(j)$$</p> <p>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. </p> <p>Is there some way to extend this to deal with equalizers and/or pullbacks? (The case of the terminal object is easy.)</p> <p>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.</p> <p>So another way to ask my question might be</p> <p><strong>Question 2:</strong> 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)?</p> <p>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.</p> http://mathoverflow.net/questions/57099/why-do-filtered-colimits-commute-with-finite-limits/57124#57124 Answer by Buschi Sergio for Why do filtered colimits commute with finite limits? Buschi Sergio 2011-03-02T16:28:32Z 2011-03-03T09:42:38Z <p>For a generalization to pullback we have to proof that $colim_i X_i\times_{Y_i} B_i \cong X\times_YB$ (where $X, Y, B$ are the respective colimits). Because $I$ is filtred the triple diagonal $I\to I\times I\times I$ is final and we can make this colimit partially, then we can do the colimit in the $Y_i$ before. </p> <p>Then we have to prove that $colim_i X_i\times_Y B_i \cong X\times_YB$ .</p> <p>Then is enought show that the pullback of any colimit is still a colimit, and then with the some "soft proof" argumentations you done.</p> <p>Is enought to show that: </p> <p>give a $f: X\to Y$ and a cocone $B_i \to Y$ with $I$ a small category (no necessarly filtred), with a colimit $B_i\to B$ and the natural arrow $B\to Y$. Then the pullback with $f$: $B_i\times_Y X \to B\times_Y X$ is a colimit.</p> <p>this is true if the pullbach funtor $(X, f)^\ast: \mathcal{C}\downarrow Y\to \mathcal{C}\downarrow X$ is a left adjoint, and then is cocomplete.</p> <p>This is as said that $\mathcal{C}$ is locally-cartesian-closed.</p> <p>This is true in any topos, and this property is a specific and profound aspect of topoi and their internal logic. </p> <p>We can observe that in my above argomentation $I$ need not be filtred, but for $I$ no filtred the diagonal $I\to I\times I$ could be no final. </p>