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Given a based simplicial group, you can find its reduced homology with coefficients in a field, homotopy, and geometric realization. These are functors. If I have a free product of based simplicial groups, does these associated functors split into coproducts? More generally, does these functors preserve colimits?

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The answer to all of your questions is "yes", in general, if you're talking about filtered colimits -but the free product is NOT a colimit of that kind. More specifically,

  1. As for the homology functor, it preserves filtered colimits.
  2. Homotopy groups preserve filtered colimits too. This you can find in J.P. May's "A Concise Course in Algebraic Topology".
  3. The realization functor preserves all kind of colimits. This follows from the fact that it is left adjoint to the total singular complex functor -see, for instance, Simplicial objects in Algebraic Topology, page 61, also by J.P. May-, and functors which are left adjoints preserve all colimits (S. Mac Lane, "Categories for the working mathematician", first edition, page 115).
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Does the forgetful functor from the category of simplicial groups to the category of simplicial sets preserve colimits? What I want to ask is that is seems your response also holds true whether for the category of simplicial groups or simplicial sets. – Colin Tan Oct 22 '11 at 14:39
For general colimits, the answer is "no". For filtered colimits, the answer is "yes". Since limits and colimits in the category of simplicial "something" are computed pointwise, the answer depends just on the preservation of colimits by the forgetful functor from groups to sets. General colimits are not preserved by this forgetful functor (the coproduct of two groups is NOT just the union of the underlying sets), but filtered colimits are preserved: see for instance the book "Abstract and concrete categories", . – a.r. Oct 22 '11 at 16:11
Thank you Augsti. It seems that "filtered" is the key word here. – Colin Tan Oct 23 '11 at 5:44
@a.r. where can I find the result that homology preserves all kinds of colimit? – Exterior Feb 18 '15 at 20:17
@Exterior Nowhere, because it doesn't. Consider the pushout that constructs $S^1$ as the interval with its endpoints glued. $H_1$ does obviously not preserve that pushout. – archipelago Feb 1 at 16:24

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