# Pursued By Stacks 4: Fiber Products

18 May 2022

Okay, we’re closing in on the definition of a geometric stack over $$\mathbf{Top}$$. Actually, the nLab tells me such stacks are called topological, so I’ll try and say that. But! Before we get there, we need the construction of the 2-fiber product of categories fibered in groupoids over $$\mathbf{Top}$$. The purpose of this post is to discuss this construction, which I think is adorable. I cannot shake the feeling that I’ve talked through the 2-fiber product of categories before, but I have no idea where or why I might have done this except possibly the last time I tried to learn about stacks.

Friends of the blog will recall the categorical definition of a pullback. This notion makes sense for categories (after all, categories assemble into a category), but because of the existence of natural transformations, the universal property looks a little bit different. Namely, suppose we have three categories $$\mathsf{C}$$, $$\mathsf{D}$$ and $$\mathsf{E}$$ and two functors $$F\colon \mathsf{C} \to \mathsf{E}$$ and $$G\colon \mathsf{D} \to \mathsf{E}$$. The 2-fiber product $$\mathsf{C}\times_{\mathsf{E}} \mathsf{D}$$ makes the following diagram commute

$\require{AMScd}\begin{CD} \mathsf{C}\times_{\mathsf{E}}\mathsf{D} @>\pi_1>> \mathsf{C} \\ @V\pi_2VV @VFVV \\ \mathsf{D} @>G>> \mathsf{E} \end{CD}$

up to natural isomorphism $$\eta\colon F\pi_1 \Rightarrow G\pi_2$$ and is universal for this property in the following slightly complicated sense. Given any category $$\mathsf{X}$$, maps $$H\colon \mathsf{X} \to \mathsf{C}$$ and $$K\colon \mathsf{X} \to \mathsf{D}$$ with a natural isomorphism $$\epsilon \colon FH \Rightarrow GK$$, there is a functor $$P\colon\mathsf{X} \to \mathsf{C}\times_{\mathsf{E}}\mathsf{D}$$ and natural isomorphisms $$\alpha\colon H \Rightarrow \pi_1P$$ and $$\beta\colon K\Rightarrow \pi_2P$$ with the property that $$(G\beta)^{-1} \circ \eta P \circ F\alpha = \epsilon$$. This is summarized in the following diagram. Given another functor $$Q\colon \mathsf{X} \to \mathsf{C}\times_{\mathsf{E}}\mathsf{D}$$ and natural isomorphisms $$\gamma\colon H\Rightarrow \pi_1Q$$ and $$\delta\colon K \Rightarrow \pi_2Q$$ as above, there exists a unique natural isomorphism $$\lambda \colon P \Rightarrow Q$$ with the property that $$\pi_1\lambda \colon \pi_1P \Rightarrow \pi_1Q$$ equals $$\gamma\circ\alpha^{-1}$$ and $$\pi_2\lambda = \delta\circ\beta^{-1}$$.

In the case of categories fibered in groupoids over $$\mathbf{Top}$$, the 2-fiber product has the following definition. Objects of $$\mathsf{C}\times_{\mathsf{E}} \mathsf{D}$$ are triples $$(c,d,\alpha)$$, where $$c \in \mathsf{C}$$ and $$d \in \mathsf{D}$$ are objects over the same space $$X$$ and $$\alpha\colon F(c) \to G(d)$$ is an arrow over the identity map of $$X$$. Arrows of $$\mathsf{C}\times_{\mathsf{E}}\mathsf{D}$$ are pairs of arrows $$u\colon c_1 \to c_2$$ and $$v\colon d_1 \to d_2$$ over the same map of spaces $$f\colon X \to Y$$ with the property that the following diagram commutes in $$\mathsf{E}$$

$\begin{CD} F(c_1) @>\alpha_1>> G(d_1) \\ @VVF(u)V @VVG(v)V \\ F(c_2) @>\alpha_2>> G(d_2). \end{CD}$

There is an obvious functor $$\pi\colon \mathsf{C}\times_{\mathsf{E}}\mathsf{D} \to \mathbf{Top}$$ sending an object $$(c,d,\alpha)$$ to the space $$X$$ that $$c$$ and $$d$$ are over and sending an arrow $$(u,v)$$ to the map $$f\colon X \to Y$$ that $$u$$ and $$v$$ are over. We’d like to show that $$\mathsf{C}\times_{\mathsf{E}}\mathsf{D}$$ is fibered in groupoids. Given an object $$\xi = (c,d,\alpha)$$ over $$X$$ and a map $$f\colon Y \to X$$, we need a “pullback” $$f^*\xi$$ over $$Y$$. It would be simplest to define $$f^*\xi = (f^*c,f^*d,f^*\alpha)$$, but we don’t necessarily have $$f^*F(c) = F(f^*c)$$, merely that they are canonically isomorphic via an isomorphism $$\sigma_{F,f}\colon F(f^*c) \to f^*F(c)$$ such that $$F(f_c) = f_{F(c)} \sigma_{F,f}$$. So we will in fact define $$f^\xi = (f^*c, f^*d, \sigma_{G,f}^{-1} f^*\alpha \sigma_{F,f})$$. The map $$f_\xi \colon f^*\xi \to \xi$$ is the map $$(f_c,f_d)$$. We check

$\alpha F(f_c) = \alpha f_{F(c)} \sigma_{F,f} = f_{G(d)} f^*\alpha \sigma_{F,f} = G(f_d) \sigma_{G,f}^{-1} f^*\alpha \sigma_{F,f},$

so this really does give a map $$f_\xi \colon f^*\xi \to \xi$$.

Given a pair of arrows $$(u_1,v_1)\colon (c_1,d_1,\alpha_1) \to (c,d,\alpha)$$ and $$(u_2,v_2)\colon (c_2,d_2,\alpha_2) \to (c,d,\alpha)$$ over maps $$f \colon Y \to X$$ and $$g\colon Z \to X$$ such that there exists a map of spaces $$h\colon Y \to Z$$ with $$gh = f$$, we need a unique pair of maps $$h_{\mathsf{C}}\colon c_1 \to c_2$$ and $$h_{\mathsf{D}}\colon d_1 \to d_2$$ such that the relevant diagram commutes. But observe that because $$u_1$$ and $$u_2$$ are over $$f$$ and $$g$$, there does exist a unique map $$h_{\mathsf{C}}\colon c_1 \to c_2$$ and similarly a unique map $$h_{\mathsf{D}}\colon d_1 \to d_2$$ lifting the diagram $$gh = f$$. We need to check that the following diagram commutes

$\begin{CD} F(c_1) @>\alpha_1>> G(d_1) \\ @VVF(h_{\mathsf{C}})V @VVG(h_{\mathsf{D}})V \\ F(c_2) @>\alpha_2>> G(d_2). \end{CD}$

We will show this by the strategy from the last post: fitting both paths around the square into the vertical arrow in the following diagram

$\begin{CD} F(c_1) @>\alpha F(u_1)>> G(d) \\ @VVV @| \\ G(d_2) @>G(v_2)>> G(d); \end{CD}$

uniqueness then assures equality of the paths around the square. We have

$\alpha F(u_1) = G(v_1)\alpha_1 = G(v_2)G(h_{\mathsf{D}})\alpha_1.$

On the other hand we have

$\alpha F(u_1) = \alpha F(u_2)F(h_{\mathsf{C}}) = G(v_2) \alpha_2 F(h_{\mathsf{C}}),$

how satisfying! This shows that the 2-fiber product really is fibered in groupoids.

On the other hand, we have obvious projections $$\mathsf{C}\times_{\mathsf{E}}\mathsf{D}$$ to $$\mathsf{C}$$ and $$\mathsf{D}$$. Call these $$\pi_1$$ and $$\pi_2$$. We do not quite have $$F\pi_1 = G\pi_2$$, merely that there is a natural isomorphism $$\eta\colon F\pi_1 \Rightarrow G\pi_2$$ defined by $$\eta(c,d,\alpha) = \alpha$$! Adorable, right? Showing that $$\eta$$ really is natural is cute, but I’ll leave it to you.

By the way, are you still thinking vector bundles? I know I am; it’s helpful to pretend that there really is a projection from $$\xi \in \mathsf{C}$$ down to the space $$X$$ it is over.

## The universal property

I’d like to conclude this post by showing that the fiber product really has its universal property. Given a category $$\mathsf{X}$$ fibered in groupoids over $$\mathbf{Top}$$ with functors $$H \colon \mathsf{X} \to \mathsf{C}$$ and $$K\colon \mathsf{X} \to \mathsf{D}$$ and a natural isomorphism $$\epsilon\colon FH \Rightarrow GK$$, the obvious functor $$P$$ to $$\mathsf{C}\times_{\mathsf{E}}\mathsf{D}$$ is defined on objects as $$x \mapsto (H(x),K(x),\epsilon(x))$$ and on arrows $$(u\colon x \to y) \mapsto (H(u),K(u))$$. Naturality of $$\epsilon$$ assures that this is really a functor. In this case, the isomorphisms $$\alpha\colon H \Rightarrow \pi_1P$$ and $$\beta\colon K \Rightarrow \pi_2P$$ are the identity: these diagrams commute on the nose. Now suppose we have another functor $$Q\colon \mathsf{X} \to \mathsf{C}\times_{\mathsf{E}}\mathsf{D}$$ and natural isomorphisms $$\gamma \colon H \Rightarrow \pi_1Q$$ and $$\delta\colon K \Rightarrow \pi_2Q$$. Define $$\lambda \colon P \Rightarrow Q$$ by the rule $$\lambda(x) = (\gamma(x),\delta(x))\colon (H(x),K(x),\epsilon(x)) \to Q(x)$$. We write $$Q(x) = (c,d,\alpha)$$. We need the following diagram to commute

$\begin{CD} FH(x) @>\epsilon(x)>> GK(x) \\ @VVF\gamma(x)V @VVG\delta(x)V \\ F(c) @>\alpha>> G(d). \end{CD}$

But this is exactly the condition at $$x$$ for the equality $$(G\delta)^{-1}\eta Q F\gamma = \epsilon$$, so $$\lambda(x)$$ really defines an arrow. Given an arrow $$u\colon x \to y$$ in $$\mathsf{X}$$, we need to check naturality:

$\begin{CD} (H(x),K(x),\epsilon(x)) @>\lambda(x)>> Q(x) \\ @VV(H(u),K(u))V @VVQ(u)V \\ (H(y),K(y),\epsilon(y)) @>\lambda(y)>> Q(y). \end{CD}$

This is true in each factor, for example in $$\mathsf{C}$$ we have $$\pi_1 Q(u) \gamma(x) = \gamma(y) H(u)$$. Therefore both paths around the square give equal arrows. It is clear from the construction that $$\lambda$$ is unique.