# Why is the homotopy group of a simplicial abelian group the homology group of the normalized complex?

Given a simplicial abelian group $A$, which is a Kan complex, we may talk about its relative homotopy group $$\pi_n(A,0):= [(\Delta^n,\partial\Delta^n),(NA,0)].$$ In Goerss and Jardine's Simplicial Homotopy Theory, it is claimed without proof that $$\pi(A,0) \cong H_n(NA),$$ which is what I don't understand.

Here is my attempt to understand this: Using the Dold-Kan equivalence of simplicial model categories, let $N: s\mathbf{Ab}\to \operatorname{Ch}_+(\mathbf{Ab})$ denote the normalized complex functor, one has $$[(\Delta^n,\partial \Delta^n),(A,0)] = [(N\Delta^n,N\partial \Delta^n),(NA,0)]$$ where the $[-,-]$ denotes the homomorphism modulo homotopy (rel second component). So I first wrote down $N\Delta^n$ and $N\partial \Delta^n$, let's just take $n = 1$ now for concreteness. We have $$N\Delta^1:\quad \mathbb{Z}\oplus \mathbb{Z} \xleftarrow{\begin{bmatrix}1 & 0\\ 0 & -1\end{bmatrix}} \mathbb{Z}\oplus\mathbb{Z} \leftarrow 0 \leftarrow \cdots$$ and $$N\partial \Delta^1:\quad \mathbb{Z}\oplus \mathbb{Z} \leftarrow 0 \leftarrow \cdots.$$ I am unable to see that why does $[(N\Delta^1,N\partial \Delta^1),(C,0)] = H_1(C)$ for a chain complex $C$. Any help please?

• I think this is true just because of the definition of homotopy groups as homology of Moore complex, which in this case also computes homology of $A$. – Samarkand Jul 28 '17 at 13:11

I suggest that you 'draw' some of the low dimensional simplices that are in the Moore complex. A 1-simplex is there if it starts at 0; a 2-simplex is there if its first and second faces are at zero. The boundary is $d_0$ so is a 1-simplex that forms a loop at 0. The 2-simplex itself acts as a null homotopy of that loop, ... and so on. The picture you build up then is clearly giving both the homotopy of the simplicial set underlying $A$ AND the homology of the Moore complex.