# Pursued By Stacks 6: Stacks And Groupoids

19 May 2022

In writing this post, I started to see what’s kind of cool about stacks: you can treat them almost as if they were spaces in the sense that, as you’ll see, objects of a stack $$\mathsf{D}$$ over a space $$Y$$ are the same thing as maps of stacks $$Y \to \mathsf{D}$$, just like the situation of $$\underline{X}$$ for a space $$X$$. The difference is that $$\mathsf{D}$$ may have many isomorphisms stacked over $$Y \to \mathsf{D}$$ while $$\underline{X}$$ has only the identity of $$Y$$.

In what might be the final post on stacks for a while, I’d like to show that every topological stack is isomorphic to $$\mathsf{B}\mathcal{G}$$ for some topological groupoid $$\mathcal{G}$$. We could then say that, for instance, a graph of groups is any stack isomorphic to $$\mathsf{B}\mathcal{G}$$, where $$\mathcal{G}$$ is the étale groupoid we constructed in a previous post.

Suppose $$\mathsf{D}$$ is a topological stack with an atlas $$p\colon X \to \mathsf{D}$$. We claim that $$X \times_{\mathsf{D}} X$$ (which is a space since $$X$$ is an atlas) is a space of arrows for a groupoid $$\mathcal{G}$$ with space of objects $$X$$. That finished, we will construct an isomorphism $$\psi\colon \mathsf{D} \to \mathsf{B}\mathcal{G}$$.

Thus we are in search of structure maps for our groupoid. By the 2-Yoneda lemma, if $$X$$ and $$Y$$ are spaces, any map of stacks $$X \to Y$$ is determined up to natural isomorphism by an object of $$\underline{Y}$$ over $$X$$; in other words, a continuous map $$f\colon X \to Y$$. But because the fiber of $$\underline{Y}$$ comprises only the identity arrow, this map $$f$$ is in fact unique. Therefore, it suffices to produce maps of stacks for our structure maps.

We take source and target maps of $$\mathcal{G}$$ to be the two projections from $$X\times_{\mathsf{D}} X$$ to $$X$$. In more words, $$(x,y,f) \mapsto X$$ and $$(x,y,f) \mapsto Y$$. Note that the diagram

$\require{AMScd}\begin{CD} X @>1_X>> X \\ @VV1_XV @VVpV \\ X @>p>> \mathsf{D} \end{CD}$

commutes (on the nose), so there is a map $$X \to X\times_{\mathsf{D}} X$$. If you chase through the definition of maps to the 2-fiber product, you see that on objects the map is $$Y \mapsto (Y,Y,1_Y)$$.

The multiplication map

$m\colon (X\times_{\mathsf{D}}X)\times_X(X\times_{\mathsf{D}}X) \to X \times_{\mathsf{D}} X$

is defined on objects as $$((x,y,f),(y,z,g) = (x,z,gf)$$. This multiplication is associative because composition of arrows in $$\mathsf{D}$$ is associative. The inversion map $$(\cdot)^{-1}$$ is given on objects as

$(x,y,f)^{-1} = (y,x,f^{-1}).$

So far we have not used any of the stack properties, merely that the category is fibered in groupoids. (I cannot help myself: if the category were merely fibered in categories, whatever that might mean, we would end up with a category which is not a groupoid.)

Now we define $$\psi\colon \mathsf{D} \to \mathsf{B}\mathcal{G}$$. This means that to each object of $$\mathsf{D}$$ over a space $$Y$$, we need to produce a principal $$\mathcal{G}$$-bundle over $$Y$$.

Recall that by the 2-Yoneda lemma an object of $$\mathsf{D}$$ over $$Y$$ is the same thing as a map of stacks $$f\colon Y \to \mathsf{D}$$. Since $$X$$ is an atlas, the fiber product $$Y \times_{\mathsf{D}} X$$ is a space and the map $$Y\times_{\mathsf{D}} X \to Y$$ is an open surjection which admits local sections. We let the projection $$Y\times_{\mathsf{D}} X \to X$$ be the anchor map for the right action of $$\mathcal{G}$$ defined on objects as

$\begin{gather*} (Y\times_{\mathsf{D}} X)\times_X(X\times_{\mathsf{D}}X) \to Y \times_{\mathsf{D}} X \\ ((y,x_2,f),(x_1,x_2,g)) \mapsto (y,x_1,g^{-1}f). \end{gather*}$

(This disagrees with Lerman but is the correct definition for a right action.)

The map

$(Y\times_{\mathsf{D}}X)\times_X\mathcal{G}_1 \to (Y\times_{\mathsf{D}}X)\times_Y(Y\times_{\mathsf{D}}X)$

defined by

$((y,x_2,f),(x_1,x_2,g)) \mapsto ((y,x_2,f),(y,x_1,g^{-1}f))$

is clearly (the action on objects of) an isomorphism of stacks, so the action of $$\mathcal{G}$$ on $$(Y\times_{\mathsf{D}}X)$$ is free and transitive. Thus we define $$\psi(f\colon M \to \mathsf{D})$$ to be the principal $$\mathcal{G}$$-bundle $$Y \times_{\mathsf{D}}X \to Y$$.

Now, if an object of $$\mathsf{D}$$ over a space $$Y$$ is a map $$f\colon Y \to \mathsf{D}$$ and an object of $$\mathsf{D}$$ over a space $$Z$$ is a map $$g \colon Z \to \mathsf{D}$$, then an arrow of $$\mathsf{D}$$ over the map $$h \colon Y \to Z$$ should in fact be the map $$h \colon Y \to Z$$ making the diagram

$\begin{CD} Y @>f>> \mathsf{D} \\ @VVhV @| \\ Z @>g>> \mathsf{D} \end{CD}$

commute—at least, up to a specified natural isomorphism $$\eta\colon f \Rightarrow gh$$. I think this is a missing naturality statement in the 2-Yoneda lemma, so expect an appendix. If we believe that for now, let’s continue.

Now notice that we get a commutative diagram

$\begin{CD} Y \times_{\mathsf{D}} X @>>> X \\ @VVh\pi V @VVpV \\ Z @>g>> \mathsf{D} \end{CD}$

up to the composite natural transformation $$\epsilon\eta$$ (where $$\epsilon$$ is the defining natural transformation for the fiber product $$Y\times_{\mathsf{D}} X$$). Therefore by the universal property of the 2-fiber product, we get a map $$\tilde h\colon Y\times_{\mathsf{D}}X \to Z\times_{\mathsf{D}}X$$ such that the following diagram commutes

$\begin{CD} Y\times_{\mathsf{D}}X @>\tilde h>> Z\times_{\mathsf{D}}X \\ @VV\pi V @VV\pi V \\ Y @>h>> Z \end{CD}$

up to a natural isomorphism, which must be the identity, because every object in the diagram is a space. Specifically, chasing through the map as described when we considered 2-fiber products, on objects we have $$\tilde h(y,x,f) = (h(y),x,f)$$. It is clear from this definition that $$\tilde h$$ is $$\mathcal{G}$$-equivariant, and that $$\psi$$ is functorial and fully faithful. We still haven’t used that $$\mathsf{D}$$ is a stack!

To complete the proof, we need to show that $$\psi$$ is essentially surjective. First, we show it for $$\mathcal{G}$$-bundles with a global section. By an argument I think I failed to give—if so, sorry—such a bundle is (canonically isomorphic to) the pullback bundle $$f^*\mathcal{G}_1$$ by a map $$f\colon Y \to X$$. Now, given such a map $$f\colon Y \to X$$, notice that the silly diagram

$\begin{CD} Y @>pf>> \mathsf{D} \\ @VVfV @| \\ X @>p>> \mathsf{D} \end{CD}$

commutes on the nose, so certainly up to natural transformation. The bundle $$\psi(pf)$$ is $$Y\times_{\mathsf{D}}X$$ over $$Y$$, which has an equivariant map $$\psi(f)$$ to $$\mathcal{G}_1$$ and hence an isomorphism $$Y\times_{\mathsf{D}}X \cong f^*\mathcal{G}_1$$. Therefore $$Y\times_{\mathsf{D}}X$$ has a global section $$\sigma$$ with the property, if you chase it through, that $$\pi_2\sigma \colon Y\times_{\mathsf{D}}X \to X$$ is $$k$$.

Similarly, if we have a commuting diagram of spaces

$\begin{CD} Y @>f>> X \\ @VVhV @| \\ Z @>g>> X \end{CD}$

then composing with $$p\colon X \to \mathsf{D}$$ yields a commuting map of stacks, i.e. a map between the bundles $$\psi(f)$$ and $$\psi(g)$$. This map is (isomorphic to) $$\tilde h\colon f^*\mathcal{G}_1 \to g^*\mathcal{G}_1$$. Therefore the full subcategory of $$\mathsf{B}\mathcal{G}$$ spanned by the bundles which have global sections is in the image of $$\psi$$.

Now suppose $$E$$ is a principal $$\mathcal{G}$$-bundle over a space $$Y$$. Then $$Y$$ has an open cover $$\mathcal{U} = \{U_i\}_{i\in I}$$ with the property that each $$\iota_i^*E$$ has a global section. The pullbacks of intersections define a descent datum for this open cover, which lives in the image of $$\psi$$. Thus since $$\mathsf{D}$$ is a stack, the descent datum is effective, giving us an element $$\xi$$ of $$\mathsf{D}$$. By functoriality of $$\mathsf{D}$$ and since isomorphisms form a sheaf, we conclude that $$\psi(\xi)$$ is isomorphic to $$E$$, showing that $$\psi$$ is essentially surjective.

Let us remark that atlases are super not unique, but that if we have two atlases $$p\colon X \to \mathsf{D}$$ and $$q\colon Y \to \mathsf{D}$$ we can construct a bibundle $$P = X\times_{\mathsf{D}} Y$$ which is principal for both resulting groupoids. Therefore the bibundle is invertible. Put another way, the groupoids are equivalent.