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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)