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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