# Pursued by Stacks 5: Topological Stacks

May 19, 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

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

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.