Kerodon

Proof. Let $x$ denote the unique vertex of $X$. Suppose, for a contradiction, that $e$ is idempotent. Then we can lift $e$ to a homotopy idempotent morphism $\widetilde{e}: (X,x) \rightarrow (X,x)$ in the $\infty $-category $\operatorname{\mathcal{S}}_{\ast }$ (Corollary 8.5.7.7). Passing to fundamental groups, we obtain an idempotent homomorphism $\beta $ from the Thompson group $\operatorname{Aut}(_{ \mathrm{Dy} }( [0,1] ) = \pi _{1}(X,x)$ to itself. Since the forgetful functor $\operatorname{\mathcal{S}}_{\ast } \rightarrow \operatorname{\mathcal{S}}$ carries $\widetilde{e}$ to $e$, $\beta $ is conjugate to the homomorphism $\alpha $ of Construction 8.5.7.13. Since $\alpha $ is a monomorphism, it follows that $\beta $ is also a monomorphism. The equation $\beta ^2 = \beta $ then implies that $\beta $ is the identity map. This is a contradiction, since $\beta $ is conjugate to the homomorphism $\alpha $ (which is not the identity morphism). $\square$