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

Corollary 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:


There exists an object $\widetilde{Y} \in \operatorname{\mathcal{C}}_{/f}$ satisfying $U( \widetilde{Y} ) = Y$.


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 To prove $(2)$, fix an object $\widetilde{Y} \in \operatorname{\mathcal{C}}_{ / f}$ satisfying $U( \widetilde{Y} ) = Y$. By virtue of Proposition, 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

\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 (4.58) 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 $\square$