Lemma 8.6.5.14 (04E6)

Lemma 8.6.5.14. Let $\kappa $ be an uncountable cardinal, let $\operatorname{\mathcal{E}}$ be an $\infty $-category which is locally $\kappa $-small, and let $\mathscr {F}: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{S}}^{< \kappa }$ be a functor. Suppose we are given a functor $U: \operatorname{\mathcal{E}}\rightarrow \Delta ^1$ and a $U$-cocartesian morphism $e: X \rightarrow Y$ of $\operatorname{\mathcal{E}}$ satisfying $U(X) = 0$ and $U(Y) = 1$. Write $\operatorname{\mathcal{E}}_0 = \{ 0\} \times _{ \Delta ^1} \operatorname{\mathcal{E}}$ and $\operatorname{\mathcal{E}}_1 = \{ 1\} \times _{\Delta ^1} \operatorname{\mathcal{E}}$, and let $\eta \in \mathscr {F}(X)$ be a vertex which exhibits the functor $\mathscr {F}_0 = \mathscr {F}|_{ \operatorname{\mathcal{E}}_0 }$ as corepresented by the object $X$. The following conditions are equivalent:

$(1)$: The functor $\mathscr {F}$ is left Kan extended from $\operatorname{\mathcal{E}}_0$.
$(2)$: The vertex $\eta $ exhibits the functor $\mathscr {F}$ as corepresented by $X$.
$(3)$: The vertex $\mathscr {F}(e)(\eta ) \in \mathscr {F}(Y)$ exhibits the functor $\mathscr {F}_1 = \mathscr {F}|_{ \operatorname{\mathcal{E}}_1 }$ as corepresented by the object $Y \in \operatorname{\mathcal{E}}_1$.

Proof. The equivalence of $(1)$ and $(2)$ follows from Lemma 8.6.5.13. We will show that $(2)$ and $(3)$ are equivalent. Fix an object $Z \in \operatorname{\mathcal{E}}_1$. Then the diagram of Kan complexes

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{E}}_1}(Y,Z) \ar [rr]^{ \circ [e] } \ar [dr] & & \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Z) \ar [dl] \\ & \mathscr {F}(Z) & } \]

commutes up to homotopy, where the right vertical map is determined by $\eta \mathscr {F}(X)$ and the left vertical map is determined by $\mathscr {F}(e)(\eta ) \in \mathscr {F}(Y)$. Our assumption that $e$ is $U$-cocartesian guarantees that the horizontal map is a homotopy equivalence (Corollary 5.1.2.3). It follows that the left vertical map is a homotopy equivalence if and only if the right vertical map is a homotopy equivalence. The desired result now follows by allowing the object $Z$ to vary. $\square$