I apologize if this is a duplicate. I don't know enough about group cohomology to know if this is just a special case of an earlier post with the same title.
Let $G=\langle\sigma\rangle$ where $\sigma^m=1$. Let $N=1+\sigma+\sigma^2+\cdots+\sigma^{m-1}$. Then it is claimed in Dummit and Foote that $\cdots\mathbb{Z} G \xrightarrow{\;\sigma -1\;} \mathbb{Z} G \xrightarrow{\;N\;} \mathbb{Z} G \xrightarrow{\;\sigma -1\;} \cdots \xrightarrow{\;N\;} \mathbb{Z} G \xrightarrow{\;\sigma -1\;} \mathbb{Z} G \xrightarrow{\;\text{aug}\;} \mathbb{Z} \longrightarrow 0$ is a free resolution of the trivial $G$-module $\mathbb{Z}$. Here $\mathbb{Z} G$ is the group ring and $\text{aug}$ is the augmentation map which sums coefficients. It's clear that $N( \sigma -1) = 0$ so that the composition of consecutive maps is zero. But I can't see why the kernel of a map should be contained in the image of the previous map. any suggestions would be greatly appreciated. Thanks for your time.