Proposition 6.3.1.13 (01MS)

Proposition 6.3.1.13. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a morphism of simplicial sets and let $W$ be a collection of edges of $\operatorname{\mathcal{C}}$. The following conditions are equivalent:

$(1)$: The morphism $F$ exhibits $\operatorname{\mathcal{D}}$ as a localization of $\operatorname{\mathcal{C}}$ with respect to $W$ (Definition 6.3.1.9).
$(2)$: For every $\infty $-category $\operatorname{\mathcal{E}}$, the functor $\operatorname{Fun}( \operatorname{\mathcal{D}},\operatorname{\mathcal{E}}) \xrightarrow { \circ F} \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})$ factors through the full subcategory $\operatorname{Fun}( \operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}})$ and induces an equivalence of $\infty $-categories $\operatorname{Fun}( \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) \rightarrow \operatorname{Fun}( \operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}})$.
$(3)$: For every $\infty $-category $\operatorname{\mathcal{E}}$, the functor $\operatorname{Fun}( \operatorname{\mathcal{D}},\operatorname{\mathcal{E}}) \xrightarrow { \circ F} \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})$ factors through the full subcategory $\operatorname{Fun}( \operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}})$ and induces a homotopy equivalence of Kan complexes $\operatorname{Fun}( \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}})^{\simeq } \rightarrow \operatorname{Fun}( \operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}})^{\simeq }$.
$(4)$: For every $\infty $-category $\operatorname{\mathcal{E}}$, the functor $\operatorname{Fun}( \operatorname{\mathcal{D}},\operatorname{\mathcal{E}}) \xrightarrow { \circ F} \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})$ factors through the full subcategory $\operatorname{Fun}( \operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}})$ and induces a bijection of sets $\pi _0( \operatorname{Fun}( \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}})^{\simeq } ) \rightarrow \pi _0( \operatorname{Fun}( \operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}})^{\simeq } )$.

Proof. The equivalence $(1) \Leftrightarrow (2)$ follows from Corollary 4.6.2.22 (and the repleteness of the full subcategory $\operatorname{Fun}( \operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}}) \subseteq \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})$). The implication $(2) \Rightarrow (3)$ follows from Remark 4.5.1.19 and the implication $(3) \Rightarrow (4)$ from Remark 3.1.6.5. We will complete the proof by showing that $(4) \Rightarrow (2)$. Assume that $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ satisfies condition $(4)$, and let $\operatorname{\mathcal{E}}$ be an $\infty $-category; we wish to show that the precomposition functor $\operatorname{Fun}(\operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) \xrightarrow { \circ F} \operatorname{Fun}(\operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}})$ is an equivalence of $\infty $-categories. For this, it will suffice to show that for every simplicial set $\operatorname{\mathcal{B}}$, the induced map

\[ \theta : \pi _0( \operatorname{Fun}(\operatorname{\mathcal{B}}, \operatorname{Fun}(\operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}))^{\simeq }) \rightarrow \pi _0( \operatorname{Fun}(\operatorname{\mathcal{B}}, \operatorname{Fun}(\operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}}) )^{\simeq } ) \]

is a bijection. Using Remark 6.3.1.7, we can identify $\theta $ with the map

\[ \pi _0( \operatorname{Fun}(\operatorname{\mathcal{D}}, \operatorname{Fun}(\operatorname{\mathcal{B}}, \operatorname{\mathcal{E}}) )^{\simeq } ) \rightarrow \pi _0( \operatorname{Fun}( \operatorname{\mathcal{C}}[W^{-1}], \operatorname{Fun}(\operatorname{\mathcal{B}}, \operatorname{\mathcal{E}}) )^{\simeq } ), \]

which is bijective by virtue of assumption $(4)$. $\square$