Lemma 4.31.4. In the $(2, 1)$-category of categories $2$-fibre products exist and are given by the construction of Example 4.31.3.

**Proof.**
Let us check the universal property: let $\mathcal{W}$ be a category, let $a : \mathcal{W} \to \mathcal{A}$ and $b : \mathcal{W} \to \mathcal{B}$ be functors, and let $t : F \circ a \to G \circ b$ be an isomorphism of functors.

Consider the functor $\gamma : \mathcal{W} \to \mathcal{A} \times _\mathcal {C}\mathcal{B}$ given by $W \mapsto (a(W), b(W), t_ W)$. (Check this is a functor omitted.) Moreover, consider $\alpha : a \to p \circ \gamma $ and $\beta : b \to q \circ \gamma $ obtained from the identities $p \circ \gamma = a$ and $q \circ \gamma = b$. Then it is clear that $(\gamma , \alpha , \beta )$ is a morphism from $(W, a, b, t)$ to $(\mathcal{A} \times _\mathcal {C} \mathcal{B}, p, q, \psi )$.

Let $(k, \alpha ', \beta ') : (W, a, b, t) \to (\mathcal{A} \times _\mathcal {C} \mathcal{B}, p, q, \psi )$ be a second such morphism. For an object $W$ of $\mathcal{W}$ let us write $k(W) = (a_ k(W), b_ k(W), t_{k, W})$. Hence $p(k(W)) = a_ k(W)$ and so on. The map $\alpha '$ corresponds to functorial maps $\alpha ' : a(W) \to a_ k(W)$. Since we are working in the $(2, 1)$-category of categories, in fact each of the maps $a(W) \to a_ k(W)$ is an isomorphism. We can use these (and their counterparts $b(W) \to b_ k(W)$) to get isomorphisms

It is straightforward to show that $\delta $ defines a $2$-isomorphism between $\gamma $ and $k$ in the $2$-category of $2$-commutative diagrams as desired. $\square$

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