The Dold-Thom theorem for infinity categories?

2010-09-03T18:40:59Z

Let $\mathcal{M}$ denote the category of finite sets and monomorphisms, and let $\mathcal T$ denote the category of based spaces. For a based space $X \in \mathcal T$, one has a canonical funtor $S_X : \mathcal M \rightarrow \mathcal T$ defined by $\{n\} \mapsto X^n$. The definition on morphisms is to insert basepoints on the factors which are not in the image of a given monomorphism.

As is well know, the homotopy groups of $\mathrm{colim} S_X = SP^\infty X$ give the homology of $X$ (this is the Dold-Thom theorem), and the homotopy groups of $\mathrm{hocolim} S_X = SP^\infty_h X$ given the stable homotopy of $X$.

Is there a model for $SP^\infty X$, the ordinary infinite symmetric product, as a homotopy colimit as opposed to a categorical colimit?

The motivation for this question comes from thinking about $\infty$-categories. In an $\infty$-category, one does not really have a good notion (at least not one that I am aware of) of strict categorical colimits. So I'm wondering if there is, nonetheless, some easily defined functor on the $\infty$-category of spaces which will let us calculate ordinary homology. In short, is there any $\infty$-categorical analog of the Dold-Thom theorem?

Update: Following up on André's remark it seems using the orbit category is heading in the right direction, at least for the $n$-th approximations. I'll just quickly sketch what I have so far:

Let $\mathcal O(\Sigma_n)$ denote the orbit category. The objects are the homogeneous (discrete) spaces $\Sigma_n/H$ (with left actions) as $H$ runs over all the subgroups of $\Sigma_n$, and the morphisms are the $\Sigma_n$-equivariant maps. There is a canonical functor $$\Sigma_n \rightarrow \mathcal O(\Sigma_n)^{op}$$ where we regard $\Sigma_n$ as a category with one object as usual.

Given a $\Sigma_n$ space $X$, right Kan extension along this inclusion produces a $\mathcal O(\Sigma_n)^{op}$ diagram $\tilde X$ defined by $$\tilde X(\Sigma_n/H) = X^H$$ It turns out that the above inclusion is final so that it induces an isomorphism of colimits. Hence $\mathrm{colim}_{\mathcal O(\Sigma_n)} \tilde X \cong X_{\Sigma_n}$, i.e., the coinvariants. It's also not hard to see that the undercategories are copies of $B\Sigma_n$, hence not contractible, so we don't expect an equivalence of homotopy colimits, which is good.

On the other hand, I can now show that when $X$ is discrete, the canonical map $$\mathrm{hocolim} \tilde X \rightarrow \mathrm{colim} \tilde X$$ is an equivalence. My methods here do not generalize to all spaces, so if someone has a reference for why this is true in general, that would be much appreciated. (I think something like this must appear in May's book on equivariant homotopy theory if it's true, but I did not have it available this weekend.)

The remaining part would be to let $n \rightarrow \infty$, but somehow this seems like it should not be too bad. (Something like: make a functor $\mathcal M \rightarrow \mathcal Cat$ by $n \mapsto \mathcal O(\Sigma_n)$. Take the Grothendieck construction. Some natural diagram on this category might give the right answer.)

Answer by Eric Finster for The Dold-Thom theorem for infinity categories?

2010-09-06T15:20:50Z

It so happens that Emmanuel Dror Farjoun is visiting the EPFL this week. I figured I'd ask him about this problem at lunch today. What a coincidence! He proved exactly this statement using the exact same techniques. In fact, the construction of $SP^n$ as a homotopy colimit is the subject of Chapter 4 in "Cellular Spaces, Null Spaces, and Homotopy Localization," Lecture Notes in Mathematics, 1622.

It turns out, the idea works more generally so that we can always replace strict colimits with homotopy colimits: define an orbit on a category $\mathcal C$ to be a functor $O : \mathcal C \rightarrow \mathcal Set$ such that $\mathrm{colim}_{\mathcal C} O \cong *$. There is a category of such functors which we call the orbit category of $\mathcal C$, denoted $\mathcal O(\mathcal C)$. The Yoneda embedding factors through $\mathcal O(\mathcal C)$, and the right Kan extension along this inclusion always results in a "free" diagram.

I still want to play around with the construction a bit to see if there are any wrinkles with $n \rightarrow \infty$, and if I can use this to give easy calculations of homology in the $\infty$-category $\mathcal S$, but I think it's safe to say at this point that answer to my question is yes.

The Dold-Thom theorem for infinity categories? - MathOverflow

The Dold-Thom theorem for infinity categories?

Answer by Eric Finster for The Dold-Thom theorem for infinity categories?