Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 4.4.1.3. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor between categories. Then $F$ is an isofibration if and only if the opposite functor $F^{\operatorname{op}}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \operatorname{\mathcal{D}}^{\operatorname{op}}$ is an isofibration.

Proof. Assume that $F$ is an isofibration; we will show that $F^{\operatorname{op}}$ is also an isofibration (the reverse implication follows by the same argument). Fix an object $C \in \operatorname{\mathcal{C}}$ and an isomorphism $u: F(C) \rightarrow D$ in the category $\operatorname{\mathcal{D}}$. Since $F$ is an isofibration, the inverse isomorphism $u^{-1}: D \rightarrow F(C)$ can be lifted to an isomorphism $v: \overline{D} \rightarrow C$ in the category $\operatorname{\mathcal{C}}$. Then $v^{-1}: C \rightarrow \overline{D}$ satisfies $F( v^{-1} ) = u$. $\square$