Understanding the Moore complex and simplicial groups

If $G$ is a simplicial group, then there is a complex of (not necessarily abelian) groups, $NG$, called the Moore complex with $NG_q = \bigcap_{i=1}^q \ker d_i.$ The key result we’d like to understand is the following.

Proposition 1: If $G$ and $H$ are simplicial groups, then $f:G\to H$ is injective (resp. surjective) if and only if $Nf:NG\to NH$ is injective (resp. surjective), where by surjective and injective we mean degreewise surjective and injective.

This result is Lemma 5 of Chapter II, Section 3 of Quillen’s homotopical algebra, and the proof is said to be similar to the proof of (3.17) in Dold and Puppe’s Homologie nicht-additiver Funktoren. Anwendungen. However, since I can’t read German very well, nor find a proof on the internet, since most sources focus on simplicial abelian groups, I thought I’d write up a proof using what I could make out of Dold and Puppe’s argument.

The key fact necessary for the proof is the following. If $G$ is a simplicial abelian group, $g\in G_q$, and $0\le i\le q$ is an index, then we can write $g$ as a product of a $q$-simplex with only the $i$th face being nonzero, and degenerate simplices.

Given this fact, we have the following proof of Proposition 1.

Proof.

First we’ll start with injectivity, which doesn’t require the fact. Certainly if $f$ is injective, then $Nf$ is injective. Conversely, suppose that $Nf$ is injective. We’ll prove that $f$ is injective by inducting on the degree. Certainly $f_0=Nf_0$ is injective. Then if $f(g)=1$, for $g\in G_q$, we have $d_if(g)=1=f(d_ig)$ for all $i$. By the inductive hypothesis, we have $d_ig=1$ for all $i$, so $g\in NG_q$, and thus $g=1$, since $Nf$ is injective.

Now we’ll move on to the surjectivity half. If $f$ is surjective, and $h\in NH_q,$ then we can produce a preimage of $h$ in $NG_q$ in the following manner.

Let $g_q$ be an arbitrary preimage of $h$. Inductively define $g_{i-1}=g_i\cdot (s_{i-1}d_ig_i)^{-1}$. We then have that for $j>i$, $d_j g_i =1,$ since if $j=i,$ we have $d_ig_{i-1} =(d_ig_i)\cdot d_is_{i-1}d_ig_i)^{-1} =1,$ and for $j>i,$ we have

$$d_jg_{i-1} =(d_jg_i)\cdot (d_js_{i-1}d_ig_i)^{-1} =1\cdot (s_{i-1}d_{j-1}d_ig_i)^{-1} =(s_{i-1}d_id_jg_i)^{-1} =1.$$

Moreover, for $i\ge 0,$ $f(g_i)=h,$ since $f(g_q)=h,$ and for $0\le i < q,$

$$f(g_i) =f(g_{i+1}\cdot (s_id_{i+1}g_{i+1})^{-1}) =h\cdot (s_id_{i+1}f(g_{i+1}))^{-1} =h\cdot (s_id_{i+1}h)^{-1} =h\cdot 1,$$

since $h\in NH_q.$

Thus $g_0\in NG_q,$ and $f(g_0)=h,$ so $Nf$ is surjective.

Finally, if we assume $Nf$ is surjective, we can prove that $f$ is surjective by inducting on the degree. As in the injective case, since $Nf_0=f_0,$ we have that $f_0$ is surjective.

Now assume that $f_i$ is surjective for $i<q$. Now we can use the fact stated above. If $h\in H_q,$ we can write $h$ as a product of simplices $h=\prod_i h_i,$ with exactly one of the $h_i$ in $NH_q$, and the rest degenerate.

Then by the assumption that $Nf$ is surjective and the inductive hypothesis, we can find $g_i$ with $f(g_i)=h_i$ for all $i$, and then $f\left(\prod_i g_i\right) = \prod_ih_i=h,$ so $f_q$ is surjective. $\blacksquare$

Now we’d like to prove the fact. It suffices to show the following precise statement, which is well known, and has proofs available elsewhere, like on the $n$Lab for instance, so I’ll omit the proof. It’s also fundamentally the same idea used to construct a preimage of $h$ that lies in $NG_q$ in the proof that surjectivity of $f$ implies surjectivity of $Nf.$

Theorem (J.C. Moore, 1954) The simplicial set underlying any simplicial group is a Kan complex, and moreover, for any horn, horn fillers may be constructed algorithmically as the product of degenerate simplices.

Thus to prove the fact, we can do the following. Let $g$ be a $q$-simplex of a simplicial group $G,$ let $\lambda_i$ be the restriction of $g$ to the $i$-horn, $\Lambda^q_i$, and let $g_i$ be a filler of $\lambda_i$ by a product of degenerate simplices. Then $gg_i^{-1}$ has only its $i$th face nonzero, and $g=(gg_i^{-1})g_i,$ as desired.