Kerodon

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Corollary 4.8.5.25. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of $\infty $-categories and let $n \geq 2$ be an integer. Then $F$ is $n$-full if and only if, for every morphism $u: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ having image $\overline{u}: \overline{X} \rightarrow \overline{Y}$ in $\operatorname{\mathcal{D}}$, the set $\pi _{n-2}(\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\overline{u}}, u )$ consists of a single element. Here $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\overline{u}}$ denotes the fiber $\{ \overline{u} \} \times _{ \operatorname{Hom}_{\operatorname{\mathcal{D}}}( \overline{X}, \overline{Y} )} \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$.

Proof. By virtue of Corollary 4.8.5.22, the functor $F$ is $n$-full if and only if, for every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, the map of Kan complexes

\[ F_{X,Y}: \operatorname{Hom}_{\operatorname{\mathcal{C}}}( X,Y) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Y) ) \]

is $(n-1)$-full. Since $F$ is an inner fibration, $F_{X,Y}$ is a Kan fibration (Proposition 4.6.1.21). The desired result now follows from Remark 4.8.5.21. $\square$