Kerodon

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

Corollary 4.6.4.19. Let $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an equivalence of $\infty$-categories and let $F: K \rightarrow \operatorname{\mathcal{C}}$ be a diagram in $\operatorname{\mathcal{C}}$. Then the induced functors

$G': \operatorname{\mathcal{C}}_{/F} \rightarrow \operatorname{\mathcal{D}}_{/(G \circ F)} \quad \quad G'': \operatorname{\mathcal{C}}_{F/} \rightarrow \operatorname{\mathcal{D}}_{(G \circ F)/}$

are equivalences of $\infty$-categories.

Proof. We will show that $G'$ is an equivalence of $\infty$-categories; the analogous statement for $G''$ follows by a similar argument. Note that we have a commutative diagram

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}_{/F} \ar [r]^-{G'} \ar [d] & \operatorname{\mathcal{D}}_{ / (G \circ F) } \ar [d] \\ \operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{ \operatorname{Fun}(K, \operatorname{\mathcal{C}}) } \{ F\} \ar [r]^-{\overline{G}'} & \operatorname{\mathcal{D}}\operatorname{\vec{\times }}_{ \operatorname{Fun}(K, \operatorname{\mathcal{D}}) } \{ G \circ F \} , }$

where the vertical maps are equivalences of $\infty$-categories by virtue of Theorem 4.6.4.17. It will therefore suffice to show that $\overline{G}'$ is an equivalence of $\infty$-categories, which is a special case of Remark 4.6.4.4. $\square$