I have this sentence from the article Resolutions for extensions of groups by C.T.C. Wall:
Let $Z(K)$ denote the group ring of the group $K$ over the ring $Z$ of the integers. Let $\otimes_K$ denote tensor product over the ring $Z(K)$. Let $Z$ also denote the chain complex in which the $0$-th chain group is $Z$ and all the other are zero. Let $B$ be a chain complex of free $K$-modules $$ 0\leftarrow B_0 \leftarrow B_1 \leftarrow B_r \leftarrow \dots $$ and $\epsilon \colon B\rightarrow Z$ a $K$-map of chain complexes ($K$ acts trivially over $Z$) inducing an isomorphism of homology. Then we call $B$ a free resolution for $K$ and $\epsilon$ its augmentation.
My question is this:
- The chain complex $Z$ will be $Z\leftarrow 0 \leftarrow 0\leftarrow \dots$ with $Z=\mathbb{Z}$?
- If the chain complex $Z$ is like in 1, then how is $\epsilon$ and why it induces an isomorphism in homology?
It's actually $0\leftarrow Z \leftarrow 0\leftarrow 0 \leftarrow \dots$ You want $\epsilon$ to induce a map $B_0\to Z$ (otherwise it couldn't induce an ismorphism in homology, because you would have $0\to Z$ which isn't an isomorphism)
$\epsilon$ is given, there's no statement that there's always such an $\epsilon$ (for starters, for such an $\epsilon$ to exist, the complex $B$ must be exact in positive degrees) : we're assuming that such an $\epsilon$ exists, and then we call $B$ a free resolution