4.36 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 4.36.1. 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 4.29.5. 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). 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{\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. 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 4.36.2. 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 4.36.1. A split fibred category is a fibred category isomorphic (!) over \mathcal{C} to one of these categories \mathcal{S}_ F.
Lemma 4.36.3. 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 4.33.6) the pullback functors (f \circ g)^* and g^* \circ f^* are equal.
Proof.
This is immediate from the definitions.
\square
Lemma 4.36.4. 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 4.33.6). By Lemma 4.33.7 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 4.33.7 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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