Kerodon

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

Proposition 1.3.6.12. Let $\operatorname{\mathcal{C}}$ be a Kan complex. Then every morphism in $\operatorname{\mathcal{C}}$ is an equivalence.

Proof of Proposition 1.3.6.12. Let $f: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{C}}$. Then the tuple $(\bullet , \operatorname{id}_{X}, f)$ determines a map of simplicial sets $\sigma _0: \Lambda ^{2}_{0} \rightarrow \operatorname{\mathcal{C}}$ (Exercise 1.1.2.14), which we depict as

\[ \xymatrix { & Y \ar@ {-->}[dr] & \\ X \ar [ur]^{f} \ar [rr]^{\operatorname{id}_ X} & & X. } \]

If $\operatorname{\mathcal{C}}$ is a Kan complex, then we can extend $\sigma _0$ to a $2$-simplex $\sigma $ of $\operatorname{\mathcal{C}}$. Then $\sigma $ exhibits the morphism $g = d_0(\sigma )$ as a left homotopy inverse to $f$. A similar argument shows that $f$ admits a right homotopy inverse, so that $f$ is an equivalence by virtue of Remark 1.3.6.9. $\square$