Kerodon

Proof of Proposition 3.2.6.2. Fix an $n$-simplex $\sigma : \Delta ^{n} \rightarrow X$ such that $\sigma |_{ \operatorname{\partial \Delta }^{n} }$ is the constant map carrying $\operatorname{\partial \Delta }^{n}$ to the base point $x \in X$. We wish to show that the homotopy class $[\sigma ]$ belongs to the image of the map $\pi _{n}(X_ s, x) \rightarrow \pi _{n}(X,x)$ if and only if the image $[f(\sigma )]$ is equal to the base point of $\pi _{n}(S,s)$. The “only if” direction is clear, since the composite map $X_{s} \hookrightarrow X \xrightarrow {f} S$ is equal to the constant map taking the value $s$. For the converse, suppose that $[ f(\sigma ) ]$ is the base point of $\pi _{n}(S,s)$. Then there exists a homotopy $h: \Delta ^{1} \times \Delta ^{n} \rightarrow S$ from $f(\sigma )$ to the constant map $\sigma '_0: \Delta ^{n} \rightarrow \{ s\} \subseteq S$, which is constant when restricted to the boundary $\operatorname{\partial \Delta }^ n$. Since $f$ is a Kan fibration, we can lift $h$ to a homotopy $\widetilde{h}: \Delta ^{1} \times \Delta ^{n} \rightarrow X$ from $\sigma $ to another $n$-simplex $\sigma ': \Delta ^{n} \rightarrow X$, where $\widetilde{h}$ is constant along the boundary $\operatorname{\partial \Delta }^{n}$ and $f( \sigma ') = \sigma '_0$ (Remark 3.1.5.3). Then $\sigma '$ represents a homotopy class $[\sigma '] \in \pi _{n}(X_ s, x)$, and the homotopy $\widetilde{h}$ witnesses that $[\sigma ]$ is equal to the image of $[\sigma ']$ in $\pi _{n}(X,x)$. $\square$