Kerodon

Proof of Proposition 3.2.6.6. Fix an $(n+1)$-simplex $\tau : \Delta ^{n+1} \rightarrow S$ for which $\tau |_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $s$. By construction, the connecting homomorphism $\partial : \pi _{n+1}(S,s) \rightarrow \pi _{n}(X_ s, x)$ carries $[ \tau ]$ to the base point of $\pi _{n}(X_ s,x)$ if and only if the constant map $e: \Delta ^{n} \rightarrow \{ x\} \hookrightarrow X_{s}$ is incident to $\tau $, in the sense of Definition 3.2.5.1. This is equivalent to the requirement that $\tau $ can be lifted to a map $\widetilde{\tau }: \Delta ^{n+1} \rightarrow X$ for which $\widetilde{\tau }|_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $x$, which clearly implies that that $[\tau ]$ belongs to the image of the map $\pi _{n+1}(f): \pi _{n+1}(X,x) \rightarrow \pi _{n+1}(S,s)$. To prove the reverse implication, suppose that $[\tau ]$ belongs to the image of $\pi _{n+1}(f)$, so that we can write $[ \tau ] = [ f( \widetilde{\tau }' ) ]$ for some map $\widetilde{\tau }': \Delta ^{n+1} \rightarrow X$ for which $\widetilde{\tau }'|_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $x$. It follows that there is a homotopy $h: \Delta ^{1} \times \Delta ^{n+1} \rightarrow S$ from $f( \widetilde{\tau }' )$ to $\tau $ which is constant along the boundary $\operatorname{\partial \Delta }^{n+1}$. Since $f$ is a Kan fibration, we can lift $h$ to a map $\widetilde{h}: \Delta ^{1} \times \Delta ^{n+1} \rightarrow X$ such that $h|_{ \{ 0\} \times \Delta ^{n+1} } = \widetilde{\tau }'$ and $h|_{ \Delta ^{1} \times \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $x$ (Remark 3.1.5.3). The restriction $\widetilde{\tau } = h|_{ \{ 1\} \times \Delta ^{n+1} }$ then satisfies $f( \widetilde{\tau } ) = \tau $ and $\widetilde{\tau }|_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $x$. $\square$