# Kerodon

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Proposition 3.2.7.2. Let $f: X \rightarrow Y$ be a Kan fibration between Kan complexes. The following conditions are equivalent:

$(1)$

The morphism $f$ is a trivial Kan fibration.

$(2)$

The morphism $f$ is a homotopy equivalence.

$(3)$

The map of sets $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is a bijection and, for every vertex $x \in X$ and every integer $n > 0$, the map of homotopy groups $\pi _{n}(X,x) \rightarrow \pi _{n}(Y,y)$ is an isomorphism.

$(4)$

For each vertex $y \in Y$, the fiber $X_{y} = \{ y\} \times _{Y} X$ is a contractible Kan complex.

Proof. The implication $(1) \Rightarrow (2)$ follows from Proposition 3.1.6.10 and the implication $(2) \Rightarrow (3)$ from Remark 3.2.2.17. Using Corollary 3.2.6.8 and Variant 3.2.6.9, we can reformulate $(3)$ as follows:

$(3')$

The map $\pi _0(f)$ is surjective and, for every vertex $x \in X$ having image $y = f(x)$, the homotopy groups $\pi _{n}(X_{y}, x)$ vanish for $n > 0$.

The equivalence of $(3') \Leftrightarrow (4)$ follows from Theorem 3.2.4.3. We will complete the proof by showing that $(4)$ implies $(1)$. Assume that condition $(4)$ is satisfied; we wish to show that every lifting problem

$\xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^{m} \ar [r] \ar [d] & X \ar [d]^{f} \\ \Delta ^{m} \ar [r]^-{\sigma } \ar@ {-->}[ur] & Y }$

admits a solution. Since $\sigma$ is nullhomotopic (Exercise 3.2.4.8), we can use the homotopy extension lifting property to to reduce to the special case where $\sigma$ is the constant map $\Delta ^{m} \rightarrow \{ y\} \hookrightarrow Y$ for some vertex $y \in Y$. In this case, the desired result follows from the contractibility of the fiber $X_{y}$ (Theorem 3.2.4.3). $\square$