Kerodon

Proof. Let $\widetilde{e}: (X,x) \rightarrow (X,x)$ be a lift of $e$ to a morphism in the $\infty $-category $\operatorname{\mathcal{S}}_{\ast }$. If $\widetilde{e}$ is homotopy idempotent, then it is idempotent (Corollary 8.5.7.6), so that $e$ is also idempotent. For the converse, suppose that $e$ is idempotent: that is, it can be extended to a functor $F: \operatorname{N}_{\bullet }( \operatorname{Idem}) \rightarrow \operatorname{\mathcal{C}}$. Since the $\infty $-category $\operatorname{\mathcal{S}}$ admits small colimits (Corollary 7.4.3.3), the idempotent $F$ splits. Consequently, there is a retraction diagram

in the $\infty $-category of spaces $\operatorname{\mathcal{S}}$, where $e$ is homotopic to the composition $(i \circ r): X \rightarrow X$. Fix vertices $x \in X$ and $y \in Y$. Since $X$ is connected, we can lift $i$ to a morphism $\widetilde{i}: (Y,y) \rightarrow (X,x)$ in the $\infty $-category $\operatorname{\mathcal{S}}_{\ast }$ (see Example 5.5.3.4). Since the forgetful functor $\operatorname{\mathcal{S}}_{\ast } \rightarrow \operatorname{\mathcal{S}}$ is a left fibration, we can lift (8.67) to a retraction diagram

in the $\infty $-category $\operatorname{\mathcal{S}}_{\ast }$. It follows that $e$ can be lifted to a (split) homotopy idempotent in $\operatorname{\mathcal{S}}_{\ast }$, given by any composition of $\widetilde{r}$ with $\widetilde{i}$. $\square$