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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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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