Kerodon

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

Warning 6.2.1.16. The implication $(2) \Rightarrow (1)$ of Proposition 6.2.1.14 generally fails if the functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ does not have a right adjoint. For example, let $X$ be a simply connected Kan complex, let $F: \Delta ^0 \rightarrow X$ be the map corresponding to a vertex $x \in X$, and let $G: X \rightarrow \Delta ^0$ be the projection map. Since $X$ is simply connected, the functors $\mathrm{h} \mathit{F}$ and $\mathrm{h} \mathit{G}$ are equivalences of ordinary categories. In particular, the identity transformation from $\operatorname{id}_{\Delta ^0} = G \circ F$ to itself determines the unit of an adjunction between $\mathrm{h} \mathit{F}$ and $\mathrm{h} \mathit{G}$. However, the functors $F$ and $G$ cannot be adjoint unless the Kan complex $X$ is contractible (see Remark 6.2.1.10)