
## 4.35 Presheaves of categories

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

Example 4.35.1. Let $\mathcal{C}$ be a category. Suppose that $F : \mathcal{C}^{opp} \to \textit{Cat}$ is a functor to the $2$-category of categories, see Definition 4.28.5. For $f : V \to U$ in $\mathcal{C}$ we will suggestively write $F(f) = f^\ast$ for the functor from $F(U)$ to $F(V)$. From this we can construct a fibred category $\mathcal{S}_ F$ 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{Mor}\nolimits _{\mathcal{S}_ F}((V, y), (U, x)) & = \{ (f, \phi ) \mid f \in \mathop{Mor}\nolimits _\mathcal {C}(V, U), \phi \in \mathop{Mor}\nolimits _{F(V)}(y, f^\ast x)\} \\ & = \coprod \nolimits _{f \in \mathop{Mor}\nolimits _\mathcal {C}(V, U)} \mathop{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 $(U, x) \mapsto U$. Let us check that this is indeed a fibred category. Given $f: V \to U$ in $\mathcal{C}$ and $(U, x)$ a lift of $U$, then we claim $(f, \text{id}_{f^\ast x}): (V, {f^\ast x}) \to (U, x)$ is a strongly cartesian lift of $f$. We have to show a $h$ in the diagram on the left determines $(h, \nu )$ on the right:

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

Just take $\nu = \psi$ which works because $f \circ h = g$ and hence $g^*x = h^*f^*x$. Moreover, this is the only lift making the diagram (on the right) commute.

Definition 4.35.2. Let $\mathcal{C}$ be a category. Suppose that $F : \mathcal{C}^{opp} \to \textit{Cat}$ is a functor to the $2$-category of categories. We will write $p_ F : \mathcal{S}_ F \to \mathcal{C}$ for the fibred category constructed in Example 4.35.1. A split fibred category is a fibred category isomorphic (!) over $\mathcal{C}$ to one of these categories $\mathcal{S}_ F$.

Lemma 4.35.3. Let $\mathcal{C}$ be a category. Let $\mathcal{S}$ be a fibred category over $\mathcal{C}$. Then $\mathcal{S}$ is split if and only if for some choice of pullbacks (see Definition 4.32.6) the pullback functors $(f \circ g)^*$ and $g^* \circ f^*$ are equal.

Proof. This is immediate from the definitions. $\square$

Lemma 4.35.4. Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category. There exists a contravariant functor $F : \mathcal{C} \to \textit{Cat}$ such that $\mathcal{S}$ is equivalent to $\mathcal{S}_ F$ in the $2$-category of fibred categories over $\mathcal{C}$. In other words, every fibred category is equivalent to a split one.

Proof. Let us make a choice of pullbacks (see Definition 4.32.6). By Lemma 4.32.7 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 fibred category over $\mathcal{C}$, and we need to check that $\mathcal{S} \to \mathcal{S}'$ is an equivalence of categories over $\mathcal{C}$ which maps strongly cartesian morphisms to strongly cartesian morphisms. 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 4.32.7 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. $\square$

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