Kerodon

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Proposition 1.3.5.7. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $u: \operatorname{\mathcal{C}}\rightarrow \operatorname{N}_{\bullet }( \mathrm{h} \mathit{\operatorname{\mathcal{C}}} )$ be as in Construction 1.3.5.6. Then $u$ exhibits $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ as a homotopy category of the simplicial set $\operatorname{\mathcal{C}}$, in the sense of Definition 1.2.5.1. In other words, for every category $\operatorname{\mathcal{D}}$, the composite map

\[ \operatorname{Hom}_{ \operatorname{Cat}}( \mathrm{h} \mathit{\operatorname{\mathcal{C}}}, \operatorname{\mathcal{D}}) \rightarrow \operatorname{Hom}_{ \operatorname{Set_{\Delta }}}( \operatorname{N}_{\bullet }( \mathrm{h} \mathit{\operatorname{\mathcal{C}}} ), \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}}) ) \xrightarrow { \circ u} \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{\mathcal{C}}, \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}}) ) \]

is a bijection.

Proof. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})$ be a map of simplicial sets. Then $F$ induces a functor of homotopy categories $G: \mathrm{h} \mathit{\operatorname{\mathcal{C}}} \rightarrow \mathrm{h} \mathit{ \operatorname{N}}_{\bullet }(\operatorname{\mathcal{D}}) \simeq \operatorname{\mathcal{D}}$ (where the second identification comes from Example 1.3.5.4). By construction, the map of simplicial sets

\[ \operatorname{\mathcal{C}}\xrightarrow {u} \operatorname{N}_{\bullet }( \mathrm{h} \mathit{\operatorname{\mathcal{C}}} ) \xrightarrow { \operatorname{N}_{\bullet }(G) } \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}}) \]

agrees with $F$ on the vertices and edges of $\operatorname{\mathcal{C}}$, and therefore coincides with $F$ (since a simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})$ is determined by its $1$-dimensional facets; see Remark 1.2.1.3). We leave it to the reader to verify that $G$ is the unique functor with this property. $\square$