Kerodon

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Corollary 4.6.7.23. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $Y$ be an object of $\operatorname{\mathcal{C}}$. Then:

$(1)$

The object $Y$ is final if and only if the projection map $F: \operatorname{\mathcal{C}}_{/Y} \rightarrow \operatorname{\mathcal{C}}$ admits a section $G$ satisfying $G(Y) = \operatorname{id}_{Y}$.

$(2)$

The object $Y$ is initial if and only if the projection map $F': \operatorname{\mathcal{C}}_{Y/} \rightarrow \operatorname{\mathcal{C}}$ admits a section $G'$ satisfying $G'(Y) = \operatorname{id}_{Y}$.

Proof. We will prove $(1)$; the proof of $(2)$ is similar. If $Y$ is a final object, then the projection map $F: \operatorname{\mathcal{C}}_{/Y} \rightarrow \operatorname{\mathcal{C}}$ is a trivial Kan fibration (Proposition 4.6.7.10), so the construction $Y \mapsto \operatorname{id}_ Y$ can be extended to a section of $F$. Conversely, suppose that $F$ admits a section $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}_{/Y}$ satisfying $G(Y) = \operatorname{id}_ Y$. Let $X$ be an object of $\operatorname{\mathcal{C}}$: we wish to show that the Kan complex $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ is contractible. The functors $G$ and $F$ induce morphisms of Kan complexes

\[ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) \xrightarrow {G} \operatorname{Hom}_{\operatorname{\mathcal{C}}_{/Y}}( G(X), \operatorname{id}_ Y ) \xrightarrow {F} \operatorname{Hom}_{\operatorname{\mathcal{C}}}( X,Y), \]

whose composition is the identity. In particular, the Kan complex $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ is a retract of $ \operatorname{Hom}_{\operatorname{\mathcal{C}}_{/Y}}( G(X), \operatorname{id}_ Y )$. It will therefore suffice to show that the Kan complex $ \operatorname{Hom}_{\operatorname{\mathcal{C}}_{/Y}}( G(X), \operatorname{id}_ Y )$ is contractible. This is clear, since $\operatorname{id}_{Y}$ is a final object of the slice $\infty $-category $\operatorname{\mathcal{C}}_{/Y}$ (Proposition 4.6.7.22). $\square$