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 $d$.

  • $\begingroup$ 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 $d_k y = s_{j-1} d_k \theta = s_{j-1} \theta_k = s_{j-1} e = e$. $\endgroup$ – Emilio Minichiello Sep 11 '20 at 15:29
  • $\begingroup$ and the inductive step is over $d$ being surjective, not being able to extend to $S$ with greater $|S|$. $\endgroup$ – Emilio Minichiello Sep 11 '20 at 15:31

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