Kerodon

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

Remark 7.7.6.4. In the situation of Lemma 7.7.6.3, the assumption that $\Gamma $ is an isofibration is essentially harmless. More precisely, we can always factor $\Gamma $ as a composition

\[ \operatorname{\mathcal{E}}' \xrightarrow {T} \operatorname{\mathcal{E}}'_0 \xrightarrow { \Gamma _0} \operatorname{\mathcal{E}}, \]

where $T$ is an equivalence of left fibrations over $\operatorname{\mathcal{C}}$ and $\Gamma _0$ is an isofibration (and therefore a left fibration, by virtue of Lemma 7.7.6.3). To prove this, we can use Corollary 5.6.7.3 to reduce to the case where $\operatorname{\mathcal{C}}$ is an $\infty $-category. In this case, $\operatorname{\mathcal{E}}$ and $\operatorname{\mathcal{E}}'$ are also $\infty $-categories. We can therefore use Corollary 4.5.2.23 to factor $\Gamma $ as a composition $\operatorname{\mathcal{E}}' \xrightarrow {T} \operatorname{\mathcal{E}}'_0 \xrightarrow { \Gamma _0} \operatorname{\mathcal{E}}$, where $\Gamma _0$ is an isofibration and $T$ is an equivalence of $\infty $-categories. In this case, both $U'$ and $U \circ \Gamma _0$ are isofibrations, so $T$ is an equivalence of inner fibrations over $\operatorname{\mathcal{C}}$ (Proposition 5.1.7.5); Proposition 5.1.7.14 then guarantees that $(U \circ \Gamma _0): \operatorname{\mathcal{E}}_0 \rightarrow \operatorname{\mathcal{C}}$ is a left fibration.