Kerodon

Proof. Let $n$ be a positive integer and $\vec{\sigma }: \Lambda ^{n}_{i} \rightarrow G_{\bullet }$ be a map of simplicial sets for some $0 \leq i \leq n$, which we will identify with a tuple $( \sigma _0, \sigma _1, \ldots , \sigma _{i-1}, \bullet , \sigma _{i+1}, \ldots , \sigma _ n)$ of elements of the group $G_{n-1}$ (Proposition 1.2.4.7). We wish to prove that there exists an element $\tau \in G_{n}$ satisfying $d^{n}_{j} \tau = \sigma _ j$ for $j \neq i$. Let $e$ denote the identity element of $G_{n-1}$. We first treat the special case where $\sigma _{i+1} = \cdots = \sigma _{n} = e$. If, in addition, we have $\sigma _{0} = \sigma _1 = \cdots = \sigma _{i-1} = e$, then we can take $\tau $ to be the identity element of $G_{n}$. Otherwise, there exists some smallest integer $j < i$ such that $\sigma _{j} \neq e$. We proceed by descending induction on $j$. Set $\tau '' = s^{n-1}_ j \sigma _ j \in G_{n}$, and consider the map $\vec{\sigma }': \Lambda ^{n}_{i} \rightarrow G_{\bullet }$ given by the tuple $( \sigma '_0, \sigma '_1, \ldots , \sigma '_{i-1}, \bullet , \sigma '_{i+1}, \ldots , \sigma '_ n)$ with $\sigma '_ k = \sigma _ k (d^{n}_ k \tau '')^{-1}$. We then have $\sigma '_0 = \sigma '_1 = \cdots = \sigma '_ j = e$ and $\sigma '_{i+1} = \cdots = \sigma '_{n} = e$. Invoking our inductive hypothesis we conclude that there exists an element $\tau ' \in G_{n}$ satisfying $d^{n}_ k \tau ' = \sigma '_{k}$ for $k \neq i$. We can then complete the proof by taking $\tau $ to be the product $\tau ' \tau ''$.

If not all of the equalities $\sigma _{i+1} = \cdots = \sigma _{n} = e$ hold, then there exists some largest integer $j > i$ such that $\sigma _ j \neq e$. We now proceed by ascending induction on $j$. Set $\tau '' = s^{n-1}_{j-1} \sigma _ j$ and let $\vec{\sigma }': \Lambda ^{n}_{i} \rightarrow G_{\bullet }$ be the map given by the tuple $( \sigma '_0, \sigma '_1, \ldots , \sigma '_{i-1}, \bullet , \sigma '_{i+1}, \ldots , \sigma '_ n)$ with $\sigma '_ k = \sigma _ k (d^{n}_ k \tau '')^{-1}$, as above. We then have $\sigma '_{j} = \sigma '_{j+1} = \cdots = \sigma '_{n} = e$, so the inductive hypothesis guarantees the existence of an element $\tau ' \in G_{n}$ satisfying $d^{n}_ k \tau ' = \sigma '_{k}$ for $k \neq i$. As before, we complete the proof by setting $\tau = \tau ' \tau ''$. $\square$