The Stacks project

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.


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