# Kerodon

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Proposition 4.1.2.10. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ be its homotopy category, and let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{N}_{\bullet }( \mathrm{h} \mathit{\operatorname{\mathcal{C}}} )$ be the canonical map. Then the construction $(\operatorname{\mathcal{D}}\subseteq \mathrm{h} \mathit{\operatorname{\mathcal{C}}} ) \mapsto ( F^{-1}( \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})) \subseteq \operatorname{\mathcal{C}})$ induces a bijection

$\{ \textnormal{Subcategories of the ordinary category \mathrm{h} \mathit{\operatorname{\mathcal{C}}}} \} \simeq \{ \textnormal{Subcategories of the \infty -category \operatorname{\mathcal{C}}} \}$

Proof. We first observe that if $\operatorname{\mathcal{D}}$ is a subcategory of the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$, then the nerve $\operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})$ is a subcategory of the $\infty$-category $\operatorname{N}_{\bullet }( \mathrm{h} \mathit{\operatorname{\mathcal{C}}} )$ (Example 4.1.2.4), so that $F^{-1}( \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}}) )$ is a subcategory of the $\infty$-category $\operatorname{\mathcal{C}}$ (Remark 4.1.2.6). Moreover, the subcategory $\operatorname{\mathcal{D}}$ is uniquely determined by its inverse image $F^{-1}( \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})$): this follows from the fact that $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{N}_{\bullet }(\mathrm{h} \mathit{\operatorname{\mathcal{C}}} )$ is an epimorphism of simplicial sets (Remark 1.5.7.10). To complete the proof, it will suffice to show that every subcategory $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$ arises in this way. Note that the inclusion map $\operatorname{\mathcal{C}}' \hookrightarrow \operatorname{\mathcal{C}}$ induces a functor of homotopy categories $G: \mathrm{h} \mathit{\operatorname{\mathcal{C}}'} \hookrightarrow \mathrm{h} \mathit{\operatorname{\mathcal{C}}}$, which is obviously injective at the level of objects. In addition, for every pair of objects $X,Y \in \mathrm{h} \mathit{\operatorname{\mathcal{C}}'}$, the functor $G$ induces a monomorphism $\operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}'}}( X,Y) \rightarrow \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}(X,Y)$: this follows from the observation that a pair of morphisms $f,g: X \rightarrow Y$ are homotopic in the $\infty$-category $\operatorname{\mathcal{C}}'$ if and only if they are homotopic in the $\infty$-category $\operatorname{\mathcal{C}}$ (Remark 4.1.2.8). It follows that the functor $G$ induces an isomorphism from $\mathrm{h} \mathit{\operatorname{\mathcal{C}}'}$ onto a subcategory $\operatorname{\mathcal{D}}\subseteq \mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. We therefore have an inclusion $\operatorname{\mathcal{C}}' \subseteq F^{-1}( \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}}) )$. To complete the proof, it will suffice to show that this inclusion is an equality. In other words, we must show that an $n$-simplex $\sigma : \Delta ^{n} \rightarrow \operatorname{\mathcal{C}}$ is contained in $\operatorname{\mathcal{C}}'$ if and only if the induced map $[n] \rightarrow \mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ factors through the subcategory $\operatorname{\mathcal{D}}\subseteq \mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. Without loss of generality, we may assume that $n > 0$ (the case $n=0$ is trivial). Using Remark 4.1.2.9, we can reduce to the case where $n=1$, so that $\sigma$ can be identified with a morphism $g: X \rightarrow Y$ in the $\infty$-category $\operatorname{\mathcal{C}}$. Our assumption that $F( \sigma )$ belongs to $\operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})$ guarantees that $g$ is homotopic to a morphism $f: X \rightarrow Y$ which belongs to the subcategory $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$ (and, in particular, that the objects $X$ and $Y$ belong to $\operatorname{\mathcal{C}}'$). Invoking Remark 4.1.2.8, we conclude that $g$ also belongs to the subcategory $\operatorname{\mathcal{C}}'$, as desired. $\square$