Proposition 10.3.4.15. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category, let $q: K \rightarrow \operatorname{\mathcal{C}}$ be a diagram, and let $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ be a morphism in the $\infty $-category $\operatorname{\mathcal{C}}_{/q}$ having image $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$. If $f$ is a universal quotient morphism in $\operatorname{\mathcal{C}}$, then $\widetilde{f}$ is a universal quotient morphism in $\operatorname{\mathcal{C}}_{/q}$.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. Set $\widetilde{\operatorname{\mathcal{C}}} = \operatorname{\mathcal{C}}_{/q}$, so that we have a commutative diagram of forgetful functors
\[ \xymatrix@R =50pt@C=50pt{ \widetilde{\operatorname{\mathcal{C}}}_{ / \widetilde{Y} } \ar [d]^{ \widetilde{U} } \ar [r]^-{V'} & \operatorname{\mathcal{C}}_{/Y} \ar [d]^{U} \\ \widetilde{\operatorname{\mathcal{C}}} \ar [r]^-{V} & \operatorname{\mathcal{C}}. } \]
Let $\operatorname{\mathcal{C}}^{0}_{ / Y } \subseteq \operatorname{\mathcal{C}}_{/Y}$ denote the sieve generated by $f$. Since $f$ is a universal quotient morphism, the functor $U$ is left Kan extended from $\operatorname{\mathcal{C}}^{0}_{/Y}$. Note that $V$ is a right fibration (Proposition 4.3.6.1), so that $V'$ is a trivial Kan fibration (Corollary 4.3.7.13). In particular, $V'$ is a right fibration, so that the functor $U \circ V' = V \circ \widetilde{U}$ is left Kan extended from the subcategory $\widetilde{\operatorname{\mathcal{C}}}^{0}_{ / \widetilde{Y} } = V'^{-1} \operatorname{\mathcal{C}}^{0}_{/Y}$ (Corollary 7.3.8.5). Since the functor $V$ is conservative and creates colimits (Proposition 7.1.4.20), it follows that $\widetilde{U}$ is also left Kan extended from $\widetilde{\operatorname{\mathcal{C}}}^{0}_{ / \widetilde{Y} }$. We conclude by observing that $\widetilde{\operatorname{\mathcal{C}}}^{0}_{ / \widetilde{Y} }$ is the sieve generated by $\widetilde{f}$, so that $\widetilde{f}$ is a universal quotient morphism in $\widetilde{\operatorname{\mathcal{C}}}$. $\square$