Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 9.2.1.16. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a reflective localization functor, and let $W$ be the collection of all morphisms $w$ of $\operatorname{\mathcal{C}}$ such that $F(w)$ is an isomorphism in $\operatorname{\mathcal{D}}$. Then $W$ is localizing.

Proof. Conditions $(1)$ and $(2)$ of Definition 9.2.1.15 follow immediately from the definitions (and do not require any assumptions on $F$). We will verify condition $(3)$. Since $F$ is a reflective localization functor, it admits a fully faithful right adjoint $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$. For every object $Y \in \operatorname{\mathcal{D}}$, the image $G(Y) \in \operatorname{\mathcal{C}}$ is $W$-local (Corollary 9.2.1.14). In particular, if $X$ is an object of $\operatorname{\mathcal{C}}$, then $(G \circ F)(X)$ is a $W$-local object of $\operatorname{\mathcal{C}}$. Let $\eta : \operatorname{id}_{\operatorname{\mathcal{C}}} \rightarrow G \circ F$ be the unit of an adjunction between $F$ and $G$. To complete the proof, it will suffice to show that the unit map $\eta _{X}: X \rightarrow (G \circ F)(X)$ belongs to $W$: that is, that $F(\eta _{X})$ is an isomorphism in $\operatorname{\mathcal{D}}$. Since the functor $G$ is fully faithful, this follows from Remark 6.3.3.5. $\square$