Kerodon

Proof of Proposition 3.2.5.2. Let $\tau : \Delta ^{n+1} \rightarrow S$ be an $(n+1)$-simplex for which $\tau |_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $s$. To prove Proposition 3.2.5.2, it will suffice to prove the following:

$(1)$

There exists an $n$-simplex $\sigma : \Delta ^{n} \rightarrow X_ s$ such that $\sigma |_{ \operatorname{\partial \Delta }^{n} }$ is the constant map taking the value $x$ and $\sigma $ is incident to $\tau $.

$(2)$

Let $\sigma ': \Delta ^{n} \rightarrow X_{s}$ and $\tau ': \Delta ^{n+1} \rightarrow S$ have the property that $\sigma '|_{ \operatorname{\partial \Delta }^{n} }$ and $\tau '|_{ \operatorname{\partial \Delta }^{n+1} }$ are the constant maps taking the values $x$ and $s$, respectively, and suppose that $[\tau ] = [\tau ']$ in $\pi _{n+1}(S,s)$. Then $\sigma '$ is incident to $\tau '$ if and only if $[ \sigma ] = [ \sigma ' ]$ in $\pi _{n}(X_ s, x)$.

Assertion $(1)$ follows from the solvability of the lifting problem

where the upper horizontal map is constant taking the value $x$. Let $\sigma '$ and $\tau '$ be as in $(2)$, and let $\widetilde{\tau }'_{0}: \operatorname{\partial \Delta }^{n+1} \rightarrow X_{s}$ be the map given by the tuple of $n$-simplices $(\sigma ', e, \ldots , e)$ (see Proposition 1.1.4.13) where $e: \Delta ^{n} \rightarrow X_{s}$ denotes the constant map taking the value $x$. If $[ \sigma ] = [ \sigma ' ]$ in $\pi _{n}(X_ s, x)$, then we can choose a homotopy from $\sigma $ to $\sigma '$ (in the Kan complex $X_ s$) which is constant along the boundary $\operatorname{\partial \Delta }^{n}$, and therefore a homotopy $\widetilde{h}_0$ from $\widetilde{\tau }|_{ \operatorname{\partial \Delta }^{n+1} }$ to $\widetilde{\tau }'_{0}$ (also in the Kan complex $X_ s$) which is constant along the simplicial subset $\Lambda ^{n+1}_{0} \subset \operatorname{\partial \Delta }^{n+1}$. Let $h: \Delta ^{1} \times \Delta ^{n+1} \rightarrow S$ be a homotopy from $\tau $ to $\tau '$ which is constant on $\operatorname{\partial \Delta }^{n+1}$. Since $f$ is a Kan fibration, the homotopy extension lifting problem

admits a solution $\widetilde{h}: \Delta ^{1} \times \Delta ^{n+1} \rightarrow X$ (Remark 3.1.5.3), which we can regard as a homotopy from $\widetilde{\tau }$ to another $(n+1)$-simplex $\widetilde{\tau }': \Delta ^{n+1} \rightarrow X$. By construction, this $(n+1)$-simplex witnesses that $\sigma '$ is incident to $\tau '$.

For the converse, suppose that $\sigma '$ is incident to $\tau '$, so that there exists an $(n+1)$-simplex $\widetilde{\tau }': \Delta ^{n+1} \rightarrow X$ satisfying $d^{n+1}_0( \widetilde{\tau }') = \sigma '$, $f( \widetilde{\tau }' ) = \tau '$, and $\widetilde{\tau }'|_{ \Lambda ^{n+1}_{0} }$ is the constant map taking the value $x$. Since $f$ is a Kan fibration, the lifting problem

admits a solution, where $\overline{e}: \Delta ^1 \times \Lambda ^{n+1}_0 \rightarrow X$ is the constant map taking the value $x$. Then $\widetilde{h}$ is a homotopy from $\widetilde{\tau }$ to $\widetilde{\tau }'$ (in the Kan complex $X$) which is constant along the horn $\Lambda ^{n+1}_{0} \subseteq \Delta ^{n+1}$, and it restricts to a homotopy from $\sigma = d^{n+1}_0( \widetilde{\tau } )$ to $\sigma ' = d^{n+1}_0( \widetilde{\tau }' )$ (in the Kan complex $X_ s$) which is constant along the boundary $\operatorname{\partial \Delta }^{n}$. It follows that $[\sigma ] = [\sigma ']$ in $\pi _{n}(X_ s,x)$. $\square$

Proof. To avoid confusion in the case $n=1$, let us use multiplicative notation for the group structures on both $\pi _{n+1}(S,s)$ and $\pi _{n}(X_ s,x)$. It is easy to see that the constant map $\Delta ^{n} \rightarrow \{ x\} \subseteq X_{s}$ is incident to the constant map $\Delta ^{n+1} \rightarrow \{ s\} \subseteq S$, so the map $\partial $ carries the identity element of $\pi _{n+1}(S,s)$ to the identity element of $\pi _{n}(X_{s}, x)$. To complete the proof, it will suffice to show that if $(\eta _0, \eta _1, \ldots , \eta _{n+1})$ is an $(n+2)$-tuple of elements of $\pi _{n+1}(S,s)$ for which the product $\eta _0^{-1} \eta _1 \eta _{2}^{-1} \cdots \eta _{n+1}^{(-1)^{n}}$ vanishes in $\pi _{n+1}(S,s)$, then the product $\partial ( \eta _0)^{-1} \partial ( \eta _1) \partial ( \eta _2)^{-1} \cdots \partial ( \eta _{n+1})^{ (-1)^{n}}$ vanishes in $\pi _{n}(X_ s, x)$. To prove this, choose simplices $\tau _{i}: \Delta ^{n+1} \rightarrow S$ for which each restriction $\tau _{i}|_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $s$ and $[ \tau _ i ] = \eta _ i$. Using our assumption that $f$ is a Kan fibration, we can lift each $\tau _ i$ to a simplex $\widetilde{\tau }_{i}: \Delta ^{n+1} \rightarrow X$ carrying the horn $\Lambda ^{n+1}_{0}$ to the vertex $x \in X$, so that $\partial ( \eta _ i ) = [ d^{n+1}_0( \widetilde{\tau }_{i} ) ]$. Since $\pi _{n+1}(S,s)$ is abelian, the vanishing of the product $\eta _0^{-1} \eta _1 \eta _{2}^{-1} \cdots \eta _{n+1}^{(-1)^{n}}$ guarantees that we can choose an $(n+2)$-simplex $\rho : \Delta ^{n+2} \rightarrow S$ such that $d^{n+2}_{0}(\rho )$ is the constant map taking the value $s$ and $d^{n+2}_{i}( \rho ) = \tau _{i-1}$ for $1 \leq i \leq n+2$. Let $\widetilde{\rho }_{0}: \Lambda ^{n+2}_{0} \rightarrow X$ be the map given by the tuple of $(n+1)$-simplices $(\bullet , \widetilde{\tau }_{0}, \widetilde{\tau }_{1}, \ldots , \widetilde{\tau }_{n+1} )$ (see Proposition 1.2.4.7). Since $f$ is a Kan fibration, the lifting problem

admits a solution. Then $\sigma = d^{n+2}_0( \widetilde{\rho } )$ is an $(n+1)$-simplex of $X_ s$ satisfying $d^{n+1}_ i( \sigma ) = d^{n+1}_0( \widetilde{\tau }_{i} )$ for $0 \leq i \leq n+1$, and therefore witnesses that the product

vanishes in the homotopy group $\pi _{n}(X_ s, x)$. $\square$

Proof. We first show that the function $a$ is well-defined: that is, that the map $\partial _{x}: \pi _{1}(S,s) \rightarrow \pi _0( X_ s )$ depends only on the image of $x$ in $\pi _0( X_ s )$. Fix an element $\eta \in \pi _{1}(S,s)$, which we can write as the homotopy class of an edge $v: s \rightarrow s$ in the Kan complex $S$. Let $x$ and $x'$ be vertices belonging to the same connected component of $X_{s}$, so that there exists an edge $u: x' \rightarrow x$ of $X$ satisfying $f(u) = \operatorname{id}_{s}$. We wish to show that $\partial _{x}( \eta ) = \partial _{x'}(\eta )$ in $\pi _0(X_ s)$. Since $f$ is a Kan fibration, we can lift $v$ to an edge $\widetilde{v}: x \rightarrow y$ in $X$. Using the fact that $f$ is a Kan fibration, we can solve the lifting problem

to obtain a $2$-simplex $\sigma $ of $X$ depicted in the diagram

The edges $\widetilde{v}$ and $\widetilde{v}'$ then witness the identities $\partial _{x}(\eta ) = [y] = \partial _{x'}(\eta )$ in $\pi _{0}(X_ s)$.

We now complete the proof by showing that the function $a: \pi _{1}(S,s) \times \pi _0(X_ s) \rightarrow \pi _0( X_ s)$ determines a left action of $\pi _{1}(S,s)$ on $\pi _0(X_ s)$. Note that the identity element of $\pi _{1}(S,s)$ is given by the homotopy class of the degenerate edge $\operatorname{id}_{s}: s \rightarrow s$ of $S$. For each $x \in X_{s}$, we can lift $\operatorname{id}_{s}$ to the edge $\operatorname{id}_{x}: x \rightarrow x$ of $X$, which witnesses the identity $a( [\operatorname{id}_ s], [x] ) = \partial _{x}( [\operatorname{id}_ s]) = [x]$ in $\pi _0(X_ s)$. To complete the argument, it will suffice to show that for every pair of edges $g,g': s \rightarrow s$ of $S$ and every vertex $x \in X_ s$, we have an equality $a( [g'] [g], [x] ) = a( [g'], a( [g], [x] ))$ in $\pi _0(X_ s)$. Since $f$ is a Kan fibration, we can lift $g$ to an edge $\widetilde{g}: x \rightarrow y$ in $X$, and $g'$ to an edge $\widetilde{g}': y \rightarrow z$ in $X$. Since $X$ is a Kan complex, the map $( \widetilde{g}', \bullet , \widetilde{g} ): \Lambda ^2_1 \rightarrow X$ can be completed to a $2$-simplex $\sigma $ of $X$, as depicted in the diagram

The edges $\widetilde{g}$, $\widetilde{g}'$, and $\widetilde{g}''$ then witness the identities $a( [g], [x] ) = [y]$, $a( [g'], [y]) = [z]$, and $a( [g'] [g], [x]) = [z]$ (respectively), so that we have an equality

as desired. $\square$