Lemma 14.31.6. Let $X$ be a simplicial group. Then $X$ is a Kan complex.

Proof. The following proof is basically just a translation into English of the proof in the reference mentioned above. Using the terminology as explained in the introduction to this section, suppose $f : \Lambda _ k[n] \to X$ is a morphism from a horn. Set $x_ i = f(\sigma _ i) \in X_{n - 1}$ for $i = 0, \ldots , \hat k, \ldots , n$. This means that for $i < j$ we have $d_ i x_ j = d_{j - 1} x_ i$ whenever $i, j \not= k$. We have to find an $x \in X_ n$ such that $x_ i = d_ ix$ for $i = 0, \ldots , \hat k, \ldots , n$.

We first prove there exists a $u \in X_ n$ such that $d_ i u = x_ i$ for $i < k$. This is trivial for $k = 0$. If $k > 0$, one defines by induction an element $u^ r \in X_ n$ such that $d_ i u^ r = x_ i$ for $0 \leq i \leq r$. Start with $u^0 = s_0x_0$. If $r < k - 1$, we set

$y^ r = s_{r + 1}((d_{r + 1}u^ r)^{-1}x_{r + 1}),\quad u^{r + 1} = u^ r y^ r.$

An easy calculation shows that $d_ iy^ r = 1$ (unit element of the group $X_{n - 1}$) for $i \leq r$ and $d_{r + 1}y^ r = (d_{r + 1}u^ r)^{-1}x_{r + 1}$. It follows that $d_ iu^{r + 1} = x_ i$ for $i \leq r + 1$. Finally, take $u = u^{k - 1}$ to get $u$ as promised.

Next we prove, by induction on the integer $r$, $0 \leq r \leq n - k$, there exists a $x^ r \in X_ n$ such that

$d_ i x^ r = x_ i\quad \text{for }i < k\text{ and }i > n - r.$

Start with $x^0 = u$ for $r = 0$. Having defined $x^ r$ for $r \leq n - k - 1$ we set

$z^ r = s_{n - r - 1}((d_{n - r}x^ r)^{-1}x_{n - r}),\quad x^{r + 1} = x^ rz^ r$

A simple calculation, using the given relations, shows that $d_ iz^ r = 1$ for $i < k$ and $i > n - r$ and that $d_{n - r}(z^ r) = (d_{n - r}x^ r)^{-1}x_{n - r}$. It follows that $d_ ix^{r + 1} = x_ i$ for $i < k$ and $i > n - r - 1$. Finally, we take $x = x^{n - k}$ which finishes the proof. $\square$

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