$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Corollary 4.6.7.12. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category, let $f: K \rightarrow \operatorname{\mathcal{C}}$ be a diagram, let $U: \operatorname{\mathcal{C}}_{/f} \rightarrow \operatorname{\mathcal{C}}$ be the projection map, and let $Y$ be an initial object of $\operatorname{\mathcal{C}}$. Then:
- $(1)$
There exists an object $\widetilde{Y} \in \operatorname{\mathcal{C}}_{/f}$ satisfying $U( \widetilde{Y} ) = Y$.
- $(2)$
If $\widetilde{Y}$ is any object of $\operatorname{\mathcal{C}}_{/f}$ satisfying $U( \widetilde{Y} ) = Y$, then $\widetilde{Y}$ is an initial object of $\operatorname{\mathcal{C}}_{/f}$.
Proof.
Assertion $(1)$ is equivalent to the statement that $f$ can be lifted to a map $\widetilde{f}: K \rightarrow \operatorname{\mathcal{C}}_{Y/}$. This is clear, since the projection map $\operatorname{\mathcal{C}}_{Y/} \rightarrow \operatorname{\mathcal{C}}$ is a trivial Kan fibration (Proposition 4.6.7.10). To prove $(2)$, fix an object $\widetilde{Y} \in \operatorname{\mathcal{C}}_{ / f}$ satisfying $U( \widetilde{Y} ) = Y$. By virtue of Proposition 4.6.7.10, it will suffice to show that the projection map $(\operatorname{\mathcal{C}}_{/f})_{\widetilde{Y}/ } \rightarrow \operatorname{\mathcal{C}}_{/f}$ is a trivial Kan fibration. Equivalently, we wish to show that every lifting problem
4.60
\begin{equation} \begin{gathered}\label{equation:coslices-create-final} \xymatrix@R =50pt@C=50pt{ A \ar [r] \ar [d] & (\operatorname{\mathcal{C}}_{/f})_{\widetilde{Y}/ } \ar [d] \\ B \ar [r] \ar@ {-->}[ur] & \operatorname{\mathcal{C}}_{/f} } \end{gathered} \end{equation}
admits a solution, provided that the left vertical map is a monomorphism. Unwinding the definitions, we can rewrite (4.60) as a lifting problem
\[ \xymatrix@R =50pt@C=50pt{ A \star K \ar [r] \ar [d] & \operatorname{\mathcal{C}}_{Y/} \ar [d] \\ B \star K \ar [r] \ar@ {-->}[ur] & \operatorname{\mathcal{C}}. } \]
Our assumption that the object $Y \in \operatorname{\mathcal{C}}$ is initial guarantees that this lifting problem has a solution (Proposition 4.6.7.10).
$\square$