Lemma 4.32.3. Let $\mathcal{C}$ be a category. The $(2, 1)$-category of categories over $\mathcal{C}$ has 2-fibre products. Suppose that $F : \mathcal{X} \to \mathcal{S}$ and $G : \mathcal{Y} \to \mathcal{S}$ are morphisms of categories over $\mathcal{C}$. An explicit 2-fibre product $\mathcal{X} \times _\mathcal {S}\mathcal{Y}$ is given by the following description

1. an object of $\mathcal{X} \times _\mathcal {S} \mathcal{Y}$ is a quadruple $(U, x, y, f)$, where $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, $x\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)$, $y\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$, and $f : F(x) \to G(y)$ is an isomorphism in $\mathcal{S}_ U$,

2. a morphism $(U, x, y, f) \to (U', x', y', f')$ is given by a pair $(a, b)$, where $a : x \to x'$ is a morphism in $\mathcal{X}$, and $b : y \to y'$ is a morphism in $\mathcal{Y}$ such that

1. $a$ and $b$ induce the same morphism $U \to U'$, and

2. the diagram

$\xymatrix{ F(x) \ar[r]^ f \ar[d]^{F(a)} & G(y) \ar[d]^{G(b)} \\ F(x') \ar[r]^{f'} & G(y') }$

is commutative.

The functors $p : \mathcal{X} \times _\mathcal {S}\mathcal{Y} \to \mathcal{X}$ and $q : \mathcal{X} \times _\mathcal {S}\mathcal{Y} \to \mathcal{Y}$ are the forgetful functors in this case. The transformation $\psi : F \circ p \to G \circ q$ is given on the object $\xi = (U, x, y, f)$ by $\psi _\xi = f : F(p(\xi )) = F(x) \to G(y) = G(q(\xi ))$.

Proof. Let us check the universal property: let $p_\mathcal {W} : \mathcal{W}\to \mathcal{C}$ be a category over $\mathcal{C}$, let $X : \mathcal{W} \to \mathcal{X}$ and $Y : \mathcal{W} \to \mathcal{Y}$ be functors over $\mathcal{C}$, and let $t : F \circ X \to G \circ Y$ be an isomorphism of functors over $\mathcal{C}$. The desired functor $\gamma : \mathcal{W} \to \mathcal{X} \times _\mathcal {S} \mathcal{Y}$ is given by $W \mapsto (p_\mathcal {W}(W), X(W), Y(W), t_ W)$. Details omitted; compare with Lemma 4.31.4. $\square$

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