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.