## 4.37 Presheaves of groupoids

In this section we compare the notion of categories fibred in groupoids with the closely related notion of a “presheaf of groupoids”. The basic construction is explained in the following example.

Example 4.37.1. This example is the analogue of Example 4.36.1, for “presheaves of groupoids” instead of “presheaves of categories”. The output will be a category fibred in groupoids instead of a fibred category. Suppose that $F : \mathcal{C}^{opp} \to \textit{Groupoids}$ is a functor to the category of groupoids, see Definition 4.29.5. For $f : V \to U$ in $\mathcal{C}$ we will suggestively write $F(f) = f^\ast$ for the functor from $F(U)$ to $F(V)$. We construct a category $\mathcal{S}_ F$ fibred in groupoids over $\mathcal{C}$ as follows. Define

$\mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ F) = \{ (U, x) \mid U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), x\in \mathop{\mathrm{Ob}}\nolimits (F(U))\} .$

For $(U, x), (V, y) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ F)$ we define

\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_ F}((V, y), (U, x)) & = \{ (f, \phi ) \mid f \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(V, U), \phi \in \mathop{\mathrm{Mor}}\nolimits _{F(V)}(y, f^\ast x)\} \\ & = \coprod \nolimits _{f \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(V, U)} \mathop{\mathrm{Mor}}\nolimits _{F(V)}(y, f^\ast x) \end{align*}

In order to define composition we use that $g^\ast \circ f^\ast = (f \circ g)^\ast$ for a pair of composable morphisms of $\mathcal{C}$ (by definition of a functor into a $2$-category). Namely, we define the composition of $\psi : z \to g^\ast y$ and $\phi : y \to f^\ast x$ to be $g^\ast (\phi ) \circ \psi$. The functor $p_ F : \mathcal{S}_ F \to \mathcal{C}$ is given by the rule $(U, x) \mapsto U$. The condition that $F(U)$ is a groupoid for every $U$ guarantees that $\mathcal{S}_ F$ is fibred in groupoids over $\mathcal{C}$, as we have already seen in Example 4.36.1 that $\mathcal{S}_ F$ is a fibred category, see Lemma 4.35.2. But we can also prove conditions (1), (2) of Definition 4.35.1 directly as follows: (1) Lifts of morphisms exist since given $f: V \to U$ in $\mathcal{C}$ and $(U, x)$ an object of $\mathcal{S}_ F$ over $U$, then $(f, \text{id}_{f^\ast x}): (V, {f^\ast x}) \to (U, x)$ is a lift of $f$. (2) Suppose given solid diagrams as follows

$\xymatrix{ V \ar[r]^ f & U & (V, y) \ar[r]^{(f, \phi )} & (U, x) \\ W \ar@{-->}[u]^ h \ar[ru]_ g & & (W, z) \ar@{-->}[u]^{(h, \nu )} \ar[ru]_{(g, \psi )} & \\ }$

Then for the dotted arrows we have $\nu = (h^\ast \phi )^{-1} \circ \psi$ so given $h$ there exists a $\nu$ which is unique by uniqueness of inverses.

Definition 4.37.2. Let $\mathcal{C}$ be a category. Suppose that $F : \mathcal{C}^{opp} \to \textit{Groupoids}$ is a functor to the $2$-category of groupoids. We will write $p_ F : \mathcal{S}_ F \to \mathcal{C}$ for the category fibred in groupoids constructed in Example 4.37.1. A split category fibred in groupoids is a category fibred in groupoids isomorphic (!) over $\mathcal{C}$ to one of these categories $\mathcal{S}_ F$.

Lemma 4.37.3. Let $p : \mathcal{S} \to \mathcal{C}$ be a category fibred in groupoids. There exists a contravariant functor $F : \mathcal{C} \to \textit{Groupoids}$ such that $\mathcal{S}$ is equivalent to $\mathcal{S}_ F$ over $\mathcal{C}$. In other words, every category fibred in groupoids is equivalent to a split one.

Proof. Make a choice of pullbacks (see Definition 4.33.6). By Lemmas 4.33.7 and 4.35.2 we get pullback functors $f^*$ for every morphism $f$ of $\mathcal{C}$.

We construct a new category $\mathcal{S}'$ as follows. The objects of $\mathcal{S}'$ are pairs $(x, f)$ consisting of a morphism $f : V \to U$ of $\mathcal{C}$ and an object $x$ of $\mathcal{S}$ over $U$, i.e., $x\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$. The functor $p' : \mathcal{S}' \to \mathcal{C}$ will map the pair $(x, f)$ to the source of the morphism $f$, in other words $p'(x, f : V\to U) = V$. A morphism $\varphi : (x_1, f_1: V_1 \to U_1) \to (x_2, f_2 : V_2 \to U_2)$ is given by a pair $(\varphi , g)$ consisting of a morphism $g : V_1 \to V_2$ and a morphism $\varphi : f_1^\ast x_1 \to f_2^\ast x_2$ with $p(\varphi ) = g$. It is no problem to define the composition law: $(\varphi , g) \circ (\psi , h) = (\varphi \circ \psi , g\circ h)$ for any pair of composable morphisms. There is a natural functor $\mathcal{S} \to \mathcal{S}'$ which simply maps $x$ over $U$ to the pair $(x, \text{id}_ U)$.

At this point we need to check that $p'$ makes $\mathcal{S}'$ into a category fibred in groupoids over $\mathcal{C}$, and we need to check that $\mathcal{S} \to \mathcal{S}'$ is an equivalence of categories over $\mathcal{C}$. We omit the verifications.

Finally, we can define pullback functors on $\mathcal{S}'$ by setting $g^\ast (x, f) = (x, f \circ g)$ on objects if $g : V' \to V$ and $f : V \to U$. On morphisms $(\varphi , \text{id}_ V) : (x_1, f_1) \to (x_2, f_2)$ between morphisms in $\mathcal{S}'_ V$ we set $g^\ast (\varphi , \text{id}_ V) = (g^\ast \varphi , \text{id}_{V'})$ where we use the unique identifications $g^\ast f_ i^\ast x_ i = (f_ i \circ g)^\ast x_ i$ from Lemma 4.35.2 to think of $g^\ast \varphi$ as a morphism from $(f_1 \circ g)^\ast x_1$ to $(f_2 \circ g)^\ast x_2$. Clearly, these pullback functors $g^\ast$ have the property that $g_1^\ast \circ g_2^\ast = (g_2\circ g_1)^\ast$, in other words $\mathcal{S}'$ is split as desired. $\square$

We will see an alternative proof of this lemma in Section 4.42.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).