Kerodon

Proof. We first show that $(1)$ implies $(2)$. By virtue of Corollary 4.5.2.23, we may assume without loss of generality that $F$ is an isofibration of $\infty $-categories. In this case, it follows from Exercise 7.6.3.13 that the diagonal inclusion $\delta : \operatorname{\mathcal{C}}\hookrightarrow \operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}$ (formed in the ordinary category of simplicial sets) can be identified with the relative diagonal of $F$ in the $\infty $-category $\operatorname{\mathcal{QC}}$. Combining this observation with Remark 9.3.4.18, we deduce that $F$ is a monomorphism (in the $\infty $-category $\operatorname{\mathcal{QC}}$) if and only if $\delta $ is an equivalence of $\infty $-categories. In particular, if $F$ is a monomorphism, then $\delta $ is fully faithful: that is, for every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, the induced map

is a homotopy equivalence. Our assumption that $F$ is an isofibration guarantees that the map $F_{X,Y}: \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Y) )$ is Kan fibration (Proposition 4.6.1.21). Applying Corollary 3.5.1.31, we deduce that $F_{X,Y}$ restricts to a homotopy equivalence of $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ with a summand of $\operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Y) )$. To complete the proof, it will suffice to show that this summand contains every isomorphism from $F(X)$ to $F(Y)$. In fact, we will prove something more precise: the induced map of cores $F^{\simeq }: \operatorname{\mathcal{C}}^{\simeq } \rightarrow \operatorname{\mathcal{D}}^{\simeq }$ is a trivial Kan fibration from $\operatorname{\mathcal{C}}^{\simeq }$ to a summand of $\operatorname{\mathcal{D}}^{\simeq }$. This follows again from Corollary 3.5.1.31, since $F^{\simeq }$ is a Kan fibration (Proposition 4.4.3.7).

We now show that $(2)$ implies $(3)$. As above, we may assume that $F$ is an isofibration. Let $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ and $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}$ denote the homotopy categories of $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$, respectively. We define a subcategory $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}_{0} \subseteq \mathrm{h} \mathit{\operatorname{\mathcal{D}}}$ as follows:

• An object $\overline{X}$ of $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}$ belongs to the subcategory $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}_{0}$ if and only if it is the image of an object $X$ of $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$.

• A morphism $\overline{u}: \overline{X} \rightarrow \overline{Y}$ of $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}$ belongs to the subcategory $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}_0$ if and only if it is the image of a morphism $u$ of $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$.

We first claim that the subcategory $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}_0$ is well-defined: that is, if $\overline{u}: \overline{X} \rightarrow \overline{Y}$ and $\overline{v}: \overline{Y} \rightarrow \overline{Z}$ are composable morphisms of $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}$ which can be lifted to morphisms $u: X \rightarrow Y$ and $v: Y' \rightarrow Z$ of $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$, then the composite morphism $\overline{v} \circ \overline{u}$ has the same property. Assumption $(2)$ guarantees that the identity morphism $\operatorname{id}_{ \overline{Y} }$ belongs to the image of the map

That is, there exists a morphism $e: Y \rightarrow Y'$ in $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ satisfying $F(e) = \operatorname{id}_{ \overline{Y} }$. Replacing $v$ by the composition $v \circ e$, we can arrange that $Y = Y'$: that is, that $u$ and $v$ are composable morphisms in the category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. It then follows that $\overline{v} \circ \overline{u} = F( v \circ u)$ is also morphism of $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}_0$, as desired.

By virtue of Proposition 4.1.2.10, the subcategory $\mathrm{h} \mathit{\operatorname{\mathcal{D}}}_0 \subseteq \mathrm{h} \mathit{\operatorname{\mathcal{D}}}$ is the homotopy category of a (unique) subcategory $\operatorname{\mathcal{D}}_0 \subseteq \operatorname{\mathcal{D}}$. Using condition $(2)$, we see that the subcategory $\operatorname{\mathcal{D}}_0$ is replete. By construction, the functor $F$ factors as a composition $\operatorname{\mathcal{C}}\xrightarrow {F_0} \operatorname{\mathcal{D}}_0 \hookrightarrow \operatorname{\mathcal{D}}$. For every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, we can identify $\operatorname{Hom}_{\operatorname{\mathcal{D}}_0}( F(X), F(Y) )$ with the summand of $\operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Y) )$ given by the essential image of $F_{X,Y}$. Invoking assumption $(2)$, we see that the functor $F_0$ is fully faithful. By construction, $F_0$ is also surjective on objects, and is therefore an equivalence of $\infty $-categories (Theorem 4.6.2.21). This completes the proof of the implication $(2) \Rightarrow (3)$.

We now show that $(3)$ implies $(1)$. Assume that $F$ induces an equivalence from $\operatorname{\mathcal{C}}$ to a replete subcategory $\operatorname{\mathcal{D}}_0 \subseteq \operatorname{\mathcal{C}}$; we wish to show that $F$ is a monomorphism. By virtue of Remark 9.3.4.15 (and Example 9.3.4.9), it will suffice to show that the inclusion map $\iota : \operatorname{\mathcal{D}}_0 \hookrightarrow \operatorname{\mathcal{D}}$ is a monomorphism in $\operatorname{\mathcal{QC}}$. Fix an $\infty $-category $\operatorname{\mathcal{B}}$, so that composition with the homotopy class $[\iota ]$ induces a map of Kan complexes $\theta : \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{B}}, \operatorname{\mathcal{D}}_0 ) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{B}}, \operatorname{\mathcal{D}})$. We wish to show that $\theta $ induces a homotopy equivalence from $\operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{B}}, \operatorname{\mathcal{D}}_0 )$ to a summand of $\operatorname{Hom}_{\operatorname{\mathcal{QC}}}( \operatorname{\mathcal{B}}, \operatorname{\mathcal{D}})$. By virtue of Remark 5.5.4.5, it will suffice to prove the analogous assertion for the inclusion map $\operatorname{Fun}( \operatorname{\mathcal{B}}, \operatorname{\mathcal{D}}_0)^{\simeq } \hookrightarrow \operatorname{Fun}(\operatorname{\mathcal{B}}, \operatorname{\mathcal{D}})^{\simeq }$, which follows immediately from Corollary 4.4.3.13. $\square$