# Kerodon

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Warning 8.5.6.6. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $R: \operatorname{Fun}( \operatorname{N}_{\bullet }( \operatorname{Idem}), \operatorname{\mathcal{C}}) \rightarrow \operatorname{End}_{\operatorname{\mathcal{C}}}^{\mathrm{idm}}$ be the restriction functor. Proposition 8.5.6.4 asserts that there exists a functor $S: \operatorname{End}_{\operatorname{\mathcal{C}}}^{\mathrm{idm}} \rightarrow \operatorname{Fun}( \operatorname{N}_{\bullet }( \operatorname{Idem}), \operatorname{\mathcal{C}})$ for which the composition

$\operatorname{Fun}( \operatorname{N}_{\bullet }( \operatorname{Idem}), \operatorname{\mathcal{C}}) \xrightarrow {R} \operatorname{End}_{\operatorname{\mathcal{C}}}^{\mathrm{idm}} \xrightarrow {S} \operatorname{Fun}( \operatorname{N}_{\bullet }( \operatorname{Idem}),\operatorname{\mathcal{C}}).$

is isomorphic to the identity functor. Let $e: X \rightarrow X$ be an idempotent endomorphism in $\operatorname{\mathcal{C}}$, so that $e$ can be extended to a morphism $F: \operatorname{N}_{\bullet }( \operatorname{Idem}) \rightarrow \operatorname{\mathcal{C}}$. Then $S(e) = (S \circ R)(F)$ is isomorphic to $F$, so there is an isomorphism of $(R \circ S)(e)$ with $e$ in the category $\operatorname{End}_{\operatorname{\mathcal{C}}}^{\mathrm{idm}}$. Beware that this isomorphism usually cannot be chosen to depend functorially on $e$. In general, the functor $R$ is not an equivalence of $\infty$-categories, so the composition

$\operatorname{End}_{\operatorname{\mathcal{C}}}^{\mathrm{idm}} \xrightarrow {S} \operatorname{Fun}( \operatorname{N}_{\bullet }( \operatorname{Idem}),\operatorname{\mathcal{C}}) \xrightarrow {R} \operatorname{End}_{\operatorname{\mathcal{C}}}^{\mathrm{idm}}$

is not isomorphic to the identity functor on $\operatorname{End}_{\operatorname{\mathcal{C}}}^{\mathrm{idm}}$.