# Homotopy coherent generalization of classifying space theory

Classically, given a compact Lie group $$G$$, there is a topological space $$BG$$ which classifies principal $$G$$-bundles. This means that there is an equality of sets {principal $$G$$-bundles up to isomorphism} = {maps $$X \to BG$$ up to homotopy}.

Question: is there a "infinity categorical" refinement of this statement? (And if so, what are good references?)

For example, is there a statement along the lines of " every principal $$G$$-bundle over a space $$X$$ is "coherently isomorphic" (whatever this means...) to the pullback of the universal bundle over $$BG$$ under some map $$X \to BG$$ which is unique "up to coherent homotopy"?


• In your first sentence, you probably want $\mathcal S_{/BG}$ :) Aug 6 at 5:15
• Hi! This is very interesting, but I'm a bit confused about whether you are describing maps $BG \to X$ rather than $X \to BG$ (see maybe also @Maxime Ramzi's comment). I am interested in the later (which hopefully involves some homotopy-coherent notion of $G$-bundles over $X$) rather than the former. Aug 6 at 7:13
• @LaurentCote It is saying, informally, that the collection of $X$ along with a map $X\to BG$ is same as functors $BG\to\mathcal S$. The precise reference is given in this answer. Maybe you could also consult Kerodon. If you are not familiar with $\infty$-categories, you could think of the following analogue: given a topological space $X$, the collection of topological spaces $Y$ along with a local homeomorphism $Y\to X$ is same as the collection of sheaves on $X$.
– Z. M
Aug 6 at 11:46
• @MaximeRamzi edited, thanks!
– skd
Aug 6 at 13:33

I think it is worth expliciting skd's answer.

There is a chain of equivalences $$\mathcal S_{/BG} \simeq Fun(BG, \mathcal S) \simeq Mod_G(\mathcal S)$$ each of which is, at an informal level, easy to define.

The first one takes your space $$p: X\to BG$$ to the functor $$BG\to \mathcal S$$ that sends the basepoint $$*\to BG$$ to the fiber $$*\times_{BG} X = p^{-1}(*)$$ and the various paths in $$BG$$ are sent to the equivalences they induce in the fibers, and the higher homotopies are sent to homotopies that witness that these equivalences are all coherent - basically what one would prove at a very truncated level in an AlgTop class about the fibers of a fibration. The inverse of this functor is also easy to describe, and quite informative : it takes a functor $$F : BG\to\mathcal S$$, observes that it has a canonical map to the trivial functor $$*$$ and takes colimits : $$colim_{BG} F \to colim_{BG}* \simeq BG$$. So the way you can think of $$X\to BG$$ is as $$colim_{BG}Y\to BG$$, where $$Y$$ is the fiber of $$X\to BG$$ over the point.

The standard picture is : $$Y\to Y_{hG}\to BG$$. Leaving the world of $$\infty$$-categories for a second, if you think of $$Y$$ as a topological space with a strict, free $$G$$-action (and the usual hypotheses for bundles), then this is exactly the classical picture, because then $$Y_{hG}$$ is then $$Y/G$$, and $$Y\to Y/G$$ becomes $$Y/G \times_{BG} EG$$ - but $$EG$$ is just a point ! So back in the homotopy world, this pullback becomes a (homotopy) pullback $$Y_{hG}\times_{BG} *$$, i.e. exactly the fiber, so we land on our feet.

The second equivalence I wanted to mention because it clarifies that functors $$BG\to \mathcal S$$ can really be thought of as $$G$$-bundles, even for non-discrete $$G$$. If $$F$$ is a functor $$BG\to \mathcal S$$, then it induces a map $$G\simeq \Omega BG \simeq map_{BG}(*,*) \to map_\mathcal S(F(*), F(*))$$, which is a monoid map, and hence corresponds to a $$G$$-action on $$F(*)$$. Conversely, this action is all you need to define such a functor.

So all in all, the composite equivalence sends $$X\to BG$$ to the fiber $$Y$$ with its $$G$$-action induced from that, and conversely given a $$G$$-action on $$Y$$, you get the homotopy orbits $$Y_{hG}$$ with their canonical map to $$BG$$. So somehow the whole thing goes through.

I guess you could ask for what happens if you want to focus on $$X$$, so view $$X$$ as fixed. Then, you can view $$X$$ as a trivial $$G$$-module, and then by adjunction a map of $$G$$-modules $$Y\to X$$ is the same as a map of spaces $$Y_{hG}\to X$$. The former has a canonical map to $$BG$$ so if this map is an equivalence, $$X$$ gets a map to $$BG$$.

Making that precise (e.g. by using un/straightening, this time over $$X$$; or comparing the fiber of $$\mathcal S_{/BG}\to \mathcal S$$ over $$X$$ to that of $$Mod_G(\mathcal S)\to \mathcal S$$; or observing some colimit-preservation thing), you obtain an equivalence between "$$Bun_G(X)$$", by which I mean the full suhspace/sub-$$\infty$$-groupoid of $$(Mod_G(\mathcal S)_{/X})^\simeq$$ spanned by those $$Y\to X$$ that induce an equivalence $$Y_{hG}\to X$$ and the mapping space $$map(X, BG)$$.

If you want more detail, I can sketch a proof of that last equivalence !

• this is really helpful, thanks! A quick question: what do you mean by $(Mod_G(S)_{/X})^{\simeq}$? (i'm not familiar with the $(-)^{\simeq}$ notation). Aug 6 at 12:47
• $C^\simeq$ is what's called the core groupoid of $C$, it's the sub-$\infty$-category obtained by only remembering the equivalences from $C$ (in particular it's an $\infty$-groupoid) Aug 6 at 12:51