Let mathcalM denote the category of finite sets and monomorphisms, and let mathcalT denote the category of based spaces. For a based space XinmathcalT, one has a canonical funtor SX:mathcalMrightarrowmathcalT defined by nmapstoXn. 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 mathrmcolimSX=SPinftyX give the homology of X (this is the Dold-Thom theorem), and the homotopy groups of mathrmhocolimSX=SPinftyhX given the stable homotopy of X.
Is there a model for SPinftyX, 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 mathcalO(Sigman) denote the orbit category. The objects are the homogeneous (discrete) spaces Sigman/H (with left actions) as H runs over all the subgroups of Sigman, and the morphisms are the Sigman-equivariant maps. There is a canonical functor SigmanrightarrowmathcalO(Sigman)op
Given a Sigman space X, right Kan extension along this inclusion produces a mathcalO(Sigman)op diagram tildeX defined by tildeX(Sigman/H)=XH
On the other hand, I can now show that when X is discrete, the canonical map mathrmhocolimtildeXrightarrowmathrmcolimtildeX
The remaining part would be to let nrightarrowinfty, but somehow this seems like it should not be too bad. (Something like: make a functor mathcalMrightarrowmathcalCat by nmapstomathcalO(Sigman). Take the Grothendieck construction. Some natural diagram on this category might give the right answer.)
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