# Proof that a simplicial group is a Kan complex

I'm trying to understand the following proof that a simplicial group is a Kan complex by Jardine, but I can't understand the bold statements:

Suppose $$S \subset [n]$$ and $$|S| \leq n$$. Write $$\Delta^n \langle S \rangle$$ for the subcomplex of $$\partial \Delta^n$$ generated by the faces $$d_i \iota_n$$ where $$\iota_n$$ denotes the unique non-degenerate $$n$$-simplex in $$\Delta^n$$. Let $$G$$ be a simplicial group and write $$G_{\langle S \rangle} = \mathsf{sSet}(\Delta^n \langle S \rangle, G)$$. There is a homomorphism $$G_n \xrightarrow{d} G_{\langle S \rangle}$$ induced by $$\Delta^n \langle S \rangle \hookrightarrow \Delta^n$$.

Claim: $$d$$ is surjective.

First note that there exists a $$j \in S$$ such that $$j - 1 \notin S$$ or $$j + 1 \notin S$$. So pick a $$j$$ and suppose there is a simplicial set map $$\Delta^n \langle S \rangle \xrightarrow{\theta} G$$ such that $$\theta_i = \theta(d_i \iota_n) = e$$ for $$i \in S, i \neq j$$. Then there exists a $$y \in G_n$$ such that $$d(y) = \theta$$, indeed if $$j + 1 \notin S$$, then set $$y = s_j \theta_j$$, or set $$y = s_{j-1} \theta_j$$ if $$j - 1 \notin S$$.

(I feel like the only way this is possible is if all of the faces of $$\theta_j$$ are $$e$$, and I can imagine a higher dimensional simplex and a choice of $$S$$ such that $$\theta_i = e$$ for all $$i \neq j$$ but that the faces of $$\theta_j$$ are nontrivial, in which case wouldn't $$s_{j-1} \theta_j$$ have nontrivial faces?)

Now suppose that $$\Delta^n \langle S \rangle \xrightarrow{\sigma} G$$ is a simplicial set map, and let $$\sigma^{(j)}$$ denote the composition $$\Delta^n \langle S \setminus \lbrace j \rbrace \rangle \hookrightarrow \Delta^n \langle S \rangle \to \Delta^n$$. Inductively, there is a $$y \in G_n$$ such that $$d(y) = \sigma^{(j)}$$, or such that $$d_i y = \sigma_i$$ for $$i \neq j$$. Let $$y_S$$ be the restriction of $$y$$ to $$\Delta^n \langle S \rangle$$. The product $$(\sigma \cdot y_S^{-1})_i = e$$ for $$i \neq j$$. Thus there is a $$\theta \in G_n$$ such that $$d(\theta) = \sigma \cdot y_S^{-1}$$. Then $$d(\theta \cdot y) = \sigma$$. Thus $$d$$ is surjective and thus every horn has a filler, so it is a Kan complex.

(The base step for $$|S| = 2$$ is clear, but how do we do the inductive step? It would require extending our map by a face, but its not clear to me how. Any advice or clarification would be appreciated.)

For the first part, Jardine says what $$y$$ is. For example, if $$j+1 \not \in S$$, then let $$y = s_j \theta_j$$. In this case, $$d_j (y) = d_j s_j \theta_j = \theta_j$$, by one of the simplicial identities ($$d_j s_j$$ is the identity map). The case when $$j-1 \not \in S$$ is the same.
For the second part, he is giving the inductive step. The statement to be proved is that the function $$d$$ is surjective, and by induction, you can assume that it's surjective in dimension $$n-1$$ — that's what he's using when he says, "Inductively ...". Then he proves that it's surjective in dimension $$n$$, by showing that any $$\sigma \in G_{\langle S \rangle}$$ is in the image of $$d$$.
• I see now, $d y = \theta$ because $d_j y = \theta_j$ clearly and if $k > j$ then $d_k y = d_k s_j \theta = s_j d_{k-1} \theta = s_j \theta_{k-1} = s_j e = e$, similarly if $k < j$ then $d_k y = s_{j-1} d_k \theta = s_{j-1} \theta_k = s_{j-1} e = e$. – Emilio Minichiello Sep 11 '20 at 15:29
• and the inductive step is over $d$ being surjective, not being able to extend to $S$ with greater $|S|$. – Emilio Minichiello Sep 11 '20 at 15:31