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Proposition Let $\operatorname{\mathcal{C}}$ be an $\infty $-category, let $B$ be a simplicial set, let $A \subseteq B$ be a simplicial subset which contains every vertex of $B$, and let $f: A \rightarrow \operatorname{\mathcal{C}}$ be a diagram. Then the fiber product $\operatorname{Fun}(B,\operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}(A,\operatorname{\mathcal{C}}) } \{ f \} $ is a Kan complex.

Proof. Corollary guarantees the restriction map $\theta : \operatorname{Fun}(B,\operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}(A, \operatorname{\mathcal{C}})$ is an isofibration, so that the fiber $\operatorname{Fun}(B,\operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}(A,\operatorname{\mathcal{C}}) } \{ f \} $ is an $\infty $-category. To show that it is a Kan complex, it will suffice to show that every morphism $u$ in $\operatorname{Fun}(B,\operatorname{\mathcal{C}}) \times _{ \operatorname{Fun}(A,\operatorname{\mathcal{C}}) } \{ f \} $ is an isomorphism (Proposition By virtue of Corollary, this is equivalent to the assertion that the image of $u$ in the $\infty $-category $\operatorname{Fun}( B, \operatorname{\mathcal{C}})$ is an isomorphism. This follows from Theorem, since for every vertex $b \in B$, the evaluation functor

\[ \operatorname{ev}_ b: \operatorname{Fun}(B, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}( \{ b\} , \operatorname{\mathcal{C}}) \simeq \operatorname{\mathcal{C}} \]

factors through $\operatorname{Fun}(A, \operatorname{\mathcal{C}})$ and therefore carries $u$ to the identity morphism $\operatorname{id}_{ f(b)}$. $\square$