For simplicial abelian group quotient map to the quotient chain complex is homotopy equivalence. Let $A$ be a simplicial abelian group. Define chain complex: $C_n(A)=A([n])$ for $n \geq 0$  and $C_n(A)=0$ for $n<0$, boundary: $\partial_n:C_n(A)\rightarrow C_{n-1}(A),$ $\partial_n=\sum_{i=1}^n (-1)^i d_i.$ Let $D_n(A)\leq C_n(A)$ be subgroup generated by degenerate simplices.

Prove that quotient map $C(A) \rightarrow C(A)/D(A)$ is homotopy equivalence.

I know that this map is quasi-isomorphism, but it doesn't imply that it is homotopy equivalence. I read somewhere that $D(A)$ is contractible and it is a theorem that quotient map by contractible chain is homotopy equivalence, but I can't find it in other materials and any book - I don't know a proofs and I can't prove it alone. 
If anyone has an idea how to prove it and could give me a hint or give me a reference to the materials, I would be very grateful. 
 A: You can find a proof of this in Goerss and Jardine's Simplicial Homotopy Theory (Theorems III.2.1 and III.2.4).  Here's an outline of the argument.  First, let $N_n(A)\subseteq C_n(A)$ consist of all $a\in C_n(A)$ such that $d_i(a)=0$ for all $i<n$.  Then $N(A)$ is a subcomplex of $C(A)$ (the "normalized chains"), and the composition of the inclusion $i:N(A)\to C(A)$ with the the projection $p:C(A)\to C(A)/D(A)$ can be shown to be an isomorphism (Theorem III.2.1).
So, to show that $p$ is a homotopy equivalence, it suffices to show that $i$ is a homotopy equivalence (Theorem III.2.4).  This is proved by breaking $i$ into simpler pieces: for each $j$, we let $N^j_n(A)$ consist of all $a\in C_n(A)$ such that $d_i(a)=0$ for all $i<\min(n,j)$.  It then suffices to show the inclusion $N^{j+1}(A)\to N^{j}(A)$ is a homotopy equivalence for each $j$, since $N^0(A)=C(A)$ and $N(A)$ is the limit of all these inclusions (and the inclusions eventually stabilize in each degree).  Finally you can show $N^{j+1}(A)\to N^j(A)$ is a chain equivalence by just writing down a simple explicit formulas for the chain inverse and chain homotopy (the chain inverse is the map $x\mapsto x-s_jd_j(x)$ and the homotopy is $x\mapsto (-1)^{j+1}s_j(x)$, where $s_j$ and $d_j$ are taken to be $0$ if $j$ is too large for them to be defined).
