Kerodon

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

Remark 6.2.2.20. In the situation of Corollary 6.2.2.19, suppose that $\eta : \operatorname{id}_{\operatorname{\mathcal{C}}} \rightarrow G \circ F$ is the unit of an adjunction between $F$ and $G$. Then an object $C \in \operatorname{\mathcal{C}}$ belongs to the essential image of $G$ if and only if the unit map $\eta _{C}: C \rightarrow (G \circ F)(C)$ is an isomorphism. The “if” direction is obvious. To prove the converse, we may assume without loss of generality that $C = G(D)$, for some object $D \in \operatorname{\mathcal{D}}$. In this case, the morphism $\eta _{C} = \eta _{G(D)}$ fits into a commutative diagram

\[ \xymatrix { & (G \circ F \circ G)(D) \ar [dr]^{ G( \epsilon _{D} ) } & \\ G(D) \ar [ur]^{ \eta _{G(D)} } \ar [rr]^{\operatorname{id}_{G(D)}} & & G(D), } \]

where $\epsilon : F \circ G \rightarrow \operatorname{id}_{\operatorname{\mathcal{D}}}$ is compatible with $\epsilon $ up to homotopy. Since $\epsilon $ is an isomorphism, it follows that $\eta _{C} = \eta _{G(D)}$ is also an isomorphism.