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Stream: learning: questions

Topic: Homology is "just" the colimit of a constant functor?

Chris Grossack (they/them) (Feb 22 2025 at 07:20):

In a talk I recently attended, the speaker said offhand that given a space (by which I mean an $\infty$-groupoid) $X$, we have

$H_\bullet(X,k) \simeq \varinjlim^h_X k$

I assume by this he means that the homology of $X$, viewed as chain complex with $0$ differential, is the (homotopy) colimit of the constant functor $X \to \mathcal{D}(k)$ sending all of $X$ to the chain complex of just $k$ concentrated in degree $0$.

This sounded like a cool computation to try, and I think I have some idea of how it should work -- my guess is that this colimit will end up being singular chains on $X$, and this is equivalent to homology when $k$ is a field since in that case every chain complex is formal.

1) Does this sound like the right idea of a proof? How might one show that the colimit is singular chains? I think this will have something to do with a tensoring of $\mathcal{D}(k)$ over spaces, but I don't quite see how to make this precise... Maybe we think about the unique colimit preserving functor from spaces to $\mathcal{D}(k)$ induced by sending the point to $k$? (Remembering that spaces are the free cocompletion of a point)

2) In my gesture at an argument, I crucially needed $k$ to be a field to appeal to formality... Is that really necessary? Or does this kind of argument hold for $\mathbb{Z}$, etc.?

3) Is there a standard reference for this? I looked around and wasn't able to find anything, though it's possible I'm missing some obvious place I should have looked

Thanks in advance ^_^

Mike Shulman (Feb 22 2025 at 15:54):

Re: 2, I think the thing to do in general is just stop with the chain complex. Sometimes people blur the line between the homology groups of X and the chain complex (or spectrum) whose homology (or homotopy) groups are those homology groups. The latter is better-behaved as an object, so it makes sense to keep it around as long as possible.

Mike Shulman (Feb 22 2025 at 15:57):

I'm more familiar with the world of spectra, where the formula $H_*(X;A) = \pi_*(X \wedge HA)$ is a well-known dual to the representability of cohomology $H^*(X;A) = \pi_{-*}(\hom(X,HA))$, where $HA$ is the Eilenberg-Mac Lane spectrum. I think one way to prove those facts is to show that these formulas satisfy the Eilenberg-Steenrod axioms and appeal to the uniqueness of (co)homology, and you could presumably also do that with chain complexes.

Mike Shulman (Feb 22 2025 at 15:59):

Another possibility that occurs to me is to use the Dold-Kan correspondence to replace chain complexes by simplicial abelian groups, represent $X$ by a Kan complex, and work at model category level with the simplicial enrichment of simplicial abelian groups.

Kevin Carlson (Feb 23 2025 at 00:22):

But to be clear, the story “should be” something like: $X$ is a colimit of points indexed by itself, so $\Sigma^\infty X$ is the colimit of (the constant diagram of) sphere spectra $\mathbb S$ indexed by $X,$ so $H_*(X,k)$ is the colimit of $k=k\otimes_{\mathbb S} \mathbb S$ indexed the same way, by cocontinuity of stabilization and tensoring with (the spectrum associated to) $k.$

How satisfactory this story is as told depends on how much you’re worried about absorbing the key intuitions versus recovering exact constructions like the singular chains. To that end I think you could probably recover singular chains using an explicit formula for homotopy colimits of chain complexes, say in terms of weighted colimits. This is probably roughly the same thing as Mike was thinking in his second suggestion, though.

Kevin Carlson (Feb 23 2025 at 00:31):

Actually, if you already know that ordinary homology is cocontinuous from the $\infty$-category of spaces to that of chain complexes, which after all is exactly what the Eilenberg-Steenrod axioms say, then we’re done already, having constructed two cocontinuous functors agreeing on the point, just as you suggested.

Chris Grossack (they/them) (Feb 23 2025 at 18:42):

Great, thanks you two! This is plenty to think about, and look closely related to some ideas I was having.