Lemma 7.5.1. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories. Suppose that $\mathcal{C}$ has fibre products and equalizers, and that $u$ commutes with them. Then the categories $(\mathcal{I}_ V)^{opp}$ satisfy the hypotheses of Categories, Lemma 4.19.8.

Proof. There are two conditions to check.

First, suppose we are given three objects $\phi : V \to u(U)$, $\phi ' : V \to u(U')$, and $\phi '' : V \to u(U'')$ and morphisms $a : U' \to U$, $b : U'' \to U$ such that $u(a) \circ \phi ' = \phi$ and $u(b) \circ \phi '' = \phi$. We have to show there exists another object $\phi ''' : V \to u(U''')$ and morphisms $c : U''' \to U'$ and $d : U''' \to U''$ such that $u(c) \circ \phi ''' = \phi '$, $u(d) \circ \phi ''' = \phi ''$ and $a \circ c = b \circ d$. We take $U''' = U' \times _ U U''$ with $c$ and $d$ the projection morphisms. This works as $u$ commutes with fibre products; we omit the verification.

Second, suppose we are given two objects $\phi : V \to u(U)$ and $\phi ' : V \to u(U')$ and morphisms $a, b : (U, \phi ) \to (U', \phi ')$. We have to find a morphism $c : (U'', \phi '') \to (U, \phi )$ which equalizes $a$ and $b$. Let $c : U'' \to U$ be the equalizer of $a$ and $b$ in the category $\mathcal{C}$. As $u$ commutes with equalizers and since $u(a) \circ \phi = u(b) \circ \phi = \phi '$ we obtain a morphism $\phi '' : V \to u(U'')$. $\square$

## Comments (2)

Comment #1119 by Olaf Schnürer on

Typos in first part of proof: $u(a) \circ \phi'= \phi$, $u(b) \circ \phi''=\phi$, $u(c) \circ \phi'''=\phi'$, $u(d) \circ \phi'''=\phi''$.

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