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Corollary 7.1.3.11. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an equivalence of $\infty $-categories and let $u: K \rightarrow \operatorname{\mathcal{C}}$ be a morphism of simplicial sets. Then:

$(1)$

The morphism $u$ can be extended to a limit diagram $\overline{u}: K^{\triangleleft } \rightarrow \operatorname{\mathcal{C}}$ if and only if the composite map $(F \circ u): K \rightarrow \operatorname{\mathcal{D}}$ can be extended to a limit diagram $K^{\triangleleft } \rightarrow \operatorname{\mathcal{D}}$.

$(2)$

The morphism $u$ can be extended to a colimit diagram $\overline{u}: K^{\triangleright } \rightarrow \operatorname{\mathcal{C}}$ if and only if the composite map $(F \circ u): K \rightarrow \operatorname{\mathcal{D}}$ can be extended to a colimit diagram $K^{\triangleright } \rightarrow \operatorname{\mathcal{D}}$.

Proof. We will prove $(1)$; the proof of $(2)$ is similar. If $u$ can be extended to a limit diagram $\overline{u}: K^{\triangleleft } \rightarrow \operatorname{\mathcal{C}}$, then Proposition 7.1.3.9 guarantees that $F \circ \overline{u}$ is a limit diagram in $\operatorname{\mathcal{D}}$ extending $F \circ u$. Conversely, suppose that $F \circ u$ can be extended to a limit diagram $\overline{v}: K^{\triangleleft } \rightarrow \operatorname{\mathcal{D}}$. Let $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be an equivalence of $\infty $-categories which is homotopy inverse to $F$, so that $G \circ F$ is isomorphic to the identity functor $\operatorname{id}_{\operatorname{\mathcal{C}}}$. Then $(G \circ \overline{v}): K^{\triangleleft } \rightarrow \operatorname{\mathcal{C}}$ is a limit diagram in $\operatorname{\mathcal{C}}$ (Proposition 7.1.3.9), and the restriction $(G \circ \overline{v})|_{K} = (G \circ F \circ u)$ is isomorphic to $u$ as an object of the $\infty $-category $\operatorname{Fun}(K,\operatorname{\mathcal{C}})$. Applying Corollary 7.1.2.15, we deduce that $u$ can be extended to a limit diagram $\overline{p}: K^{\triangleleft } \rightarrow \operatorname{\mathcal{C}}$. $\square$