Kerodon

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Remark 3.4.0.6 (Homotopy Fibers). Let $f: X \rightarrow Y$ be a morphism of Kan complexes. Then $f$ is a homotopy equivalence if and only if, for each vertex $y \in Y$, the homotopy fiber $X \times ^{\mathrm{h}}_{Y} \{ y\} $ is a contractible Kan complex. To see this, we observe that $f$ factors as a composition

\[ X \xrightarrow {\delta } X \times _{Y}^{\mathrm{h}} Y \xrightarrow {\pi } Y, \]

where $\delta $ is a homotopy equivalence and $\pi $ is a Kan fibration (Example 3.1.7.10). It follows that $f$ is a homotopy equivalence if and only if $\pi $ is a homotopy equivalence, which is equivalent to the requirement that each fiber $\pi ^{-1} \{ y\} = X \times _{Y}^{\mathrm{h}} \{ y\} $ is contractible (Proposition 3.3.7.6).