# Kerodon

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Theorem 3.2.7.1. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. Then $f$ is a homotopy equivalence if and only if it satisfies the following pair of conditions:

$(a)$

The map of sets $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is a bijection.

$(b)$

For every vertex $x \in X$ having image $y = f(x)$ in $Y$ and every positive integer $n$, the map of homotopy groups $\pi _{n}(f): \pi _{n}(X,x) \rightarrow \pi _{n}(Y,y)$ is an isomorphism.

Proof of Theorem 3.2.7.1. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. Suppose that $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is a bijection and that the induced map $\pi _{n}(X,x) \rightarrow \pi _{n}(Y, f(x) )$ is an isomorphism for every base point $x \in X$ and every positive integer $n$; we wish to show that $f$ is a homotopy equivalence (the converse follows from Remark 3.2.2.17). Using Proposition 3.1.7.1 (or Example 3.1.7.10), we can factor $f$ as a composition $X \xrightarrow {f'} X' \xrightarrow {f''} Y$, where $f'$ is anodyne and $f''$ is a Kan fibration. Then $X'$ is Kan complex (Remark 3.1.1.11), so that $f'$ is a homotopy equivalence (Proposition 3.1.6.13). It will therefore suffice to show that the Kan fibration $f''$ is a homotopy equivalence, which follows from Proposition 3.2.7.2. $\square$