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It is well known that if a category has all coequalizers and all (small) coproducts then in fact it has all (small) colimits. More important is the proof which shows that every colimit can be built by using coproducts and coequalizers. This implies that if a functor commutes with coproducts and coequalizers, then it must commute with all (small) colimits as well.

Is there homotopical analog of this? If I have a functor which commutes with all small (homotopy) coproducts and all homotopy coequalizers, does it necessarily commute with all homotopy colimits in general?

This question makes sense for general model categories, but I am particularly interested in the usual model structure on spaces.

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  • $\begingroup$ If you replace "homotopy coequalizer" with "homotopy pushout", the answer is in the affirmative: this is proposition 4.4.2.7 of Higher Topos Theory. I think 4.4.3.1 allows you to extend this to coequalizers (see the proof of 4.4.3.2), but I haven't worked through the details. $\endgroup$
    – user13113
    Commented Jul 22, 2019 at 13:50

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There is an analogue, but one should replace coequalizers by geometric realizations (homotopy colimits over Δop). If $F : I \to M$ is a diagram in a model category, one has

$$\operatorname{hocolim}_I F = \operatorname{hocolim}_{k \in \Delta^{\operatorname{op}}} \coprod_{i_0 \to \cdots \to i_k \in I} F(i_0).$$

For instance, see section 2 of https://emilyriehl.github.io/files/hocolimits.pdf for the simplicial model category case.

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