# Kerodon

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

Corollary 3.5.4.2. Let $X$ be a topological space. Then the counit map $v_{X}: | \operatorname{Sing}_{\bullet }(X) | \rightarrow X$ is a weak homotopy equivalence of topological spaces.

Proof. We must show that $\operatorname{Sing}_{\bullet }(v_ X): \operatorname{Sing}_{\bullet }(| \operatorname{Sing}_{\bullet }(X) | ) \rightarrow \operatorname{Sing}_{\bullet }(X)$ is a homotopy equivalence of Kan complexes. This is clear, since $\operatorname{Sing}_{\bullet }(v_ X)$ is left inverse to the unit map $u_{ \operatorname{Sing}_{\bullet }(X)}: \operatorname{Sing}_{\bullet }(X) \rightarrow \operatorname{Sing}_{\bullet }(| \operatorname{Sing}_{\bullet }(X) | )$, which is a weak homotopy equivalence by virtue of Theorem 3.5.4.1 (and therefore a homotopy equivalence, since both $\operatorname{Sing}_{\bullet }(X)$ and $\operatorname{Sing}_{\bullet }( | \operatorname{Sing}_{\bullet }(X) | )$ are Kan complexes). $\square$