# Pursued By Stacks 5: Topological Stacks

19 May 2022

The purpose of this post is to single out the topological (or geometric or Artin) stacks.

## Atlases

We will say that a stack over $$\mathbf{Top}$$ “is a space” if it is isomorphic as a stack to $$\underline{X}$$ for some topological space $$X$$. Following Lerman, we will in fact just drop the underline, thinking of $$X$$ as the stack $$\underline{X}$$.

An atlas for a stack $$\mathsf{D}$$ is a space $$X$$ and a map $$p\colon X \to \mathsf{D}$$ such that for any map $$f\colon Y \to \mathsf{D}$$ of a space into $$\mathsf{D}$$, the fiber product $$Y \times_{\mathsf{D}} X$$ is a space and the map $$\pi_1 \colon Y \times_{\mathsf{D}} X \to Y$$ is an open surjection which admits local sections.

Apparently it is common to say in the literature that the map $$X \to \mathsf{D}$$ is representable. We say that $$\mathsf{D}$$ is a topological stack if it has an atlas.

## Example: $$\mathsf{B}\mathcal{G}$$

Let $$\mathcal{G}$$ be a topological groupoid, and let $$\mathsf{B}\mathcal{G}$$ be the category of principal $$\mathcal{G}$$-bundles and $$\mathcal{G}$$-equivariant maps. Lerman shows that $$\mathsf{B}\mathcal{G}$$ is fibered in groupoids over $$\mathbf{Top}$$; well, strictly speaking he is talking about Lie groupoids, but the argument works just fine for spaces and topological groupoids. In fact it is a stack: just as, you might think, given a topological group $$G$$ and a principal $$G$$-bundle on an open cover of a space $$X$$ with gluing data, you can glue up to form a principal $$G$$-bundle on $$X$$, you can do the same for groupoids, and moreover isomorphisms form a sheaf.

Given a topological groupoid $$\mathcal{G}$$, we claim that the map $$p\colon \mathcal{G}_0 \to \mathsf{B}\mathcal{G}$$ defined (by the 2-Yoneda lemma) by the principal $$\mathcal{G}$$-bundle $$\omega\colon \mathcal{G}_1 \to \mathcal{G}_0$$ is an atlas.

So given any space $$Y$$ and a map $$E\colon Y \to \mathsf{B}\mathcal{G}$$, we need to show that $$Y\times_{\mathsf{B}\mathcal{G}} \mathcal{G}_0$$ is a space. First, note that by the 2-Yoneda lemma, the functor $$E$$ is determined by a principal $$\mathcal{G}$$-bundle $$E \to Y$$. Given a map $$g\colon Z \to Y$$, the bundle $$E(g)$$ is canonically isomorphic to the pullback $$g^*E$$, so to ease notation, we will assume that it is actually equal to $$g^*E$$. Similarly given $$f\colon Z \to \mathcal{G}_0$$, we will assume the bundle $$p(f)$$ is $$f^*\mathcal{G}_1$$. Thus an object of the fiber product over a space $$Z$$ is a triple $$(g,f,\beta)$$ where $$g\colon Z \to Y$$ and $$f\colon Z \to \mathcal{G}_0$$ are maps of spaces and $$\beta\colon g^*E \to f^*\mathcal{G}_1$$ is a $$\mathcal{G}$$-equivariant map of bundles, hence an isomorphism.

Recall that

$f^*\mathcal{G}_1 = \{(z,h) \in Z \times \mathcal{G}_1 : f(z) = \omega(h) \}.$

The map $$z \mapsto (z,1_{f(z)})$$ defines a global section $$\sigma$$ of this bundle. We claim that the isomorphism $$\beta^{-1}\colon f^*\mathcal{G}_1 \to g^*E$$ is determined uniquely by the image of $$\sigma$$. Indeed, given $$\beta^{-1}(\sigma(z)) = (z,e)$$, we must have $$\beta^{-1}(z,h) = \beta^{-1}(\sigma(z).h) = (z,e).h$$, for all appropriate $$h \in \mathcal{G}_1$$, so fiber by fiber the bundle map $$\beta^{-1}$$ is determined by the point $$\beta^{-1}(\sigma(z))$$. This gives us a map $$\tau\colon Z \to E$$ defined as $$z \mapsto e$$ such that $$\pi \tau = g$$, where $$\pi\colon E \to Y$$ is the defining map of the principal bundle $$E$$. Notice that the anchor map of, for instance, $$g^*E$$ is $$(z,e) \mapsto a(e)$$, where $$a \colon E \to \mathcal{G}_0$$ is the anchor map. We have

$f = f\pi\sigma = \omega f_{\mathcal{G}_1} \sigma = \alpha (\cdot)^{-1} f_{\mathcal{G}_1} \sigma = \alpha f_{\mathcal{G}_1}\sigma,$

since $$\sigma(z) = (z,1_{f(z)})$$. But then

$\alpha f_{\mathcal{G}_1\sigma} = a g_E \beta^{-1} \sigma,$

since the isomorphism $$\beta$$ intertwines the actions and thus preserves anchor maps. Finally,

$a g_E \beta^{-1} \sigma = a \tau.$

Conversely, given any map $$\tau \colon Z \to E$$, we reconstruct $$(g,f,\beta)$$ as follows: the map $$g \colon Z \to Y$$ is $$\pi\tau$$, the map $$f$$ is $$a\tau$$, Finally we need a map $$\beta \colon g^*E \to f^*\mathcal{G}_1$$. By definition we have

$g^*E = \{(z,e) \in Z\times E : \pi\tau(z) = \pi(e) \} \quad\text{and}\quad f^*\mathcal{G}_1 = \{(z,h) \in Z \times \mathcal{G}_1 : a\tau(z) = \omega(h) \}.$

Define $$\beta(z,\tau(z)) = (z,1_{a\tau(z)})$$ and extend equivariantly. It is clear that these constructions are mutually inverse, and we conclude that objects of the 2-fiber product are (isomorphic to) maps of spaces to $$E$$.

Now for arrows! Suppose we have an arrow $$(u,v)\colon (g_1,f_1,\beta_1) \to (g_2,f_2,\beta_2)$$ over a map of spaces $$h\colon Z_1 \to Z_2$$. This is a pair of maps $$u\colon Z_1 \to Z_2$$ and $$v\colon Z_1 \to Z_2$$ with the property that the following diagrams commute

$\require{AMScd}\begin{CD} Z_1 @>g_1>> Y \\ @VVuV @| \\ Z_2 @>g_2>> Y \end{CD}\qquad\begin{CD} Z_1 @>f_1>> \mathcal{G}_0 \\ @VVvV @| \\ Z_2 @>f_2>> \mathcal{G}_0 \end{CD}\qquad \begin{CD} g_1^*E @>\beta_1>> f_1^*\mathcal{G}_1 \\ @VV\tilde uV @VV\tilde vV \\ g_2^*E @>\beta_2>> f_2^*\mathcal{G}_1, \end{CD}$

where $$\tilde u$$ and $$\tilde v$$ are the unique $$\mathcal{G}$$-equivariant maps satisfying $$(g_1)_E = (g_2)_E \tilde u$$ and $$(f_1)_{\mathcal{G}_1} = (f_2)_{\mathcal{G}_1}\tilde v$$. But because each $$\beta_i$$ is over the identity map of $$Z_i$$, we conclude that in fact $$u = v$$. We have

$\tau_1\pi_1 = (g_1)_E \beta_1^{-1} = (g_2)_E \tilde u \beta_1^{-1} = (g_2)_E \beta_2^{-1}\tilde v = \tau_2\pi_2 \tilde v = \tau_2 u \pi_1,$

so because $$\pi_1$$ is an epimorphism, we conclude $$\tau_1 = \tau_2 u$$. Finally, because, for instance, $$(g_1)_E(z,\tau_1(z).h) = \tau_1(z).h$$ and $$(g_2)_E(z,\tau_2(z).h) = \tau_2(z).h$$, we must have $$\tilde u(z,\tau_1(z).h) = (u(z),\tau_2u(z)) = (u(z),\tau_1(z))$$ and similarly $$\tilde v(z,1_{a\tau_1(z)}) = (v(z),1_{a\tau_2(u(z))}) = (v(z),1_{a\tau_1(z)})$$.

Conversely, given $$u \colon Z_1 \to Z_2$$ such that $$\tau_1 = \tau_2u$$, observe that $$g_1 = \pi\tau_1 = \pi\tau_2 u = g_2u$$ and $$f_2 = a\tau_1 = a\tau_2 u = f_2u$$. We have, for any $$z \in Z_1$$ and appropriate $$h \in \mathcal{G}_1$$,

$\begin{gather*} \tilde v \beta_1(z,\tau_1(z).h) = \tilde v(z,h) = (v(z),h) = (u(z),h) \\ = \beta_2(u(z),\tau_2u(z).h) = \beta_2(u(z),\tau_1(z).h) = \beta_2\tilde u(z,\tau_1(z).h). \end{gather*}$

Therefore arrows in the 2-fiber product are (isomorphic to) maps in $$\underline{E}$$, and we conclude that the 2-fiber product is the space $$E$$. The map $$\pi\colon E \to Y$$ is an open surjection which admits local sections because $$E$$ is a principal $$\mathcal{G}$$-bundle. Therefore the map $$p\colon \mathcal{G}_0 \to \mathsf{B}\mathcal{G}$$ is an atlas.