# Chain complexes question between free $K$-modules and almost zero chain.

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:

1. The chain complex $$Z$$ will be $$Z\leftarrow 0 \leftarrow 0\leftarrow \dots$$ with $$Z=\mathbb{Z}$$?
2. If the chain complex $$Z$$ is like in 1, then how is $$\epsilon$$ and why it induces an isomorphism in homology?

1. 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)
2. $$\epsilon$$ 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