7
$\begingroup$

Say C is a left proper model structure. I have a diagram where all maps are cofibrations. Is its colimit a homotopy colimit?

I know this is true for pushouts. Is it true for sequential colimits? Filtered colimits? Sifted colimits? I am most interested in the sequential case.

This question is obviously related to, and is a slight generalization of, this previous question.

$\endgroup$
1

1 Answer 1

8
$\begingroup$

In general, this is certainly not true. Take for example a space $X$ with an action by a group $G$. As a group acts by isomorphisms, it acts in particular by cofibrations. But the map $X/G \to X_{hG}\simeq EG \times_G X$ from the orbits to the homotopy orbits is usually not an equivalence if the action is not free.

A sequential colimit of cofibrations is always a homotopy colimit if the first object is cofibrant. This follows for example by the general theory of Reedy model structures: In the Reedy model structure a sequential diagram as above is cofibrant. If your model category is left proper, you can indeed drop the assumption that the first object is cofibrant.

In general, if you have a direct diagram category $\mathcal{D}$ such that for every $X \in \mathcal{D}$ the category of $Y\neq X$ mapping to $X$ has a terminal object or is empty, then the colimit of every $\mathcal{D}$-shaped diagram of cofibrant objects and cofibrations is a homotopy colimit. This follows again by the Reedy model structure on $\mathcal{D}$-shaped diagrams (as the Latching objects are very easy to determine).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.