Lemma 4.31.11. Let

be a $2$-fibre product of categories. Then there is a canonical isomorphism $\mathcal{C}_3 \cong \mathcal{C}_1 \times _{G_1, \mathcal{S}, G_2} \mathcal{C}_2$.

Lemma 4.31.11. Let

\[ \xymatrix{ \mathcal{C}_3 \ar[r] \ar[d] & \mathcal{S} \ar[d]^\Delta \\ \mathcal{C}_1 \times \mathcal{C}_2 \ar[r]^{G_1 \times G_2} & \mathcal{S} \times \mathcal{S} } \]

be a $2$-fibre product of categories. Then there is a canonical isomorphism $\mathcal{C}_3 \cong \mathcal{C}_1 \times _{G_1, \mathcal{S}, G_2} \mathcal{C}_2$.

**Proof.**
We may assume that $\mathcal{C}_3$ is the category $(\mathcal{C}_1 \times \mathcal{C}_2)\times _{\mathcal{S} \times \mathcal{S}} \mathcal{S}$ constructed in Example 4.31.3. Hence an object is a triple $((X_1, X_2), S, \phi )$ where $\phi = (\phi _1, \phi _2) : (G_1(X_1), G_2(X_2)) \to (S, S)$ is an isomorphism. Thus we can associate to this the triple $(X_1, X_2, \phi _2^{-1} \circ \phi _1)$. Conversely, if $(X_1, X_2, \psi )$ is an object of $\mathcal{C}_1 \times _{G_1, \mathcal{S}, G_2} \mathcal{C}_2$, then we can associate to this the triple $((X_1, X_2), G_2(X_2), (\psi , \text{id}_{G_2(X_2)}))$. We claim these constructions given mutually inverse functors. We omit describing how to deal with morphisms and showing they are mutually inverse.
$\square$

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