# Pursued by Stacks 6: Stacks and Groupoids

May 19, 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

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

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

The map

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

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

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

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

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

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.