## Are coproducts and patchings filtered colimits ?

1) Is the coproduct $\coprod_{i\in I} X_i$ a filtered colimit ?

2) Is the colimit colim($X => Y$) (two arrows from X to Y) a filtered colimit ?

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I think the section "Abstract definition" of this nlab page ncatlab.org/nlab/show/directed+colimit should suffice to answer your question. – Reid Barton Dec 14 2009 at 15:35
But even in Set, binary products do not commute with binary coproducts: (A ∐ B) x (C ∐ D) is not the same as (A x C) ∐ (B x D). – Reid Barton Dec 14 2009 at 16:06
That's not "commuting with," that's "distributing over". – Mike Shulman Dec 14 2009 at 16:28
Only if you change "commuting with" from the accepted terminology to your terminology, and then the relevance to filtered colimits seems to get lost – Yemon Choi Dec 14 2009 at 18:41
I'm voting -1 because although this question is a perfectly sensible one for someone just learning the material, the answer is right there in the wikipedia article on filtered colimits. If you don't understand the wikipedia article, that suggests to me that you're trying to go too fast for your own good: it probably means you haven't got comfortable with the basic formalism of limits and colimits yet. I'd recommend spending some time on that. – Tom Leinster Dec 16 2009 at 21:30

It seems your actual question is something like "Why is $(\coprod_i X_i) \times (\coprod_j Y_j)$ the same as $\coprod_{i,j} X_i \times Y_j$? This relies on the fact that for fixed Z the functor $Z \times -$ commutes with coproducts—in fact all colimits, since it has a right adjoint $\mathrm{Hom}(Z, -)$. Applying this twice yields the result. The analogous statement holds for any two indexing categories I and J: $(\mathrm{colim}_{i \in I} X_i) \times (\mathrm{colim}_{j \in J} Y_j) = \mathrm{colim}_{i,j \in I \times J} (X_i \times Y_j)$.

You are confusing this with an entirely different fact about the category Set, namely that for a filtered category $I$, $(\mathrm{colim}_{i \in I} X_i) \times (\mathrm{colim}_{i \in I} Y_i) = \mathrm{colim}_{i \in I} (X_i \times Y_i)$. Note the right-hand colimit is over $I$, not $I \times I$. This is what is meant by "products commuting with colimits".

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The coproduct is a colimit indexed over a discrete category, namely the one with set of objects equal to the index set, and no arrows (apart from identities) It should be obvious from the definition that this is not a filtered category, so that that colimit is not a filtered one.

Likewise, the category which indexes your second limit looks like $$\bullet \rightrightarrows\bullet$$ and it is again obvious from the definition that this not a filtered category.

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 Thanx for the answer. I agree with you. My idea was to add a point to these categories, that one would sent to the final object (I suppose it exists). But, it seems that it doesn't work. My goal in this question, is to apply a theorem of commutation of finite limits and filtered colimits. For instance, I remembered to have written that the fact that $(\coprod_i X_i) \times (\coprod_j Y_j)$ is the same as $\coprod_{i,j} X_i \times Y_j$ is indeed an instance of such a commutation. – nicojo Dec 15 2009 at 11:12