I am trying to understand the proof of theorem 3.5 from Hatcher. I am unsure about the last line that $H^n(X;G)\cong H^n(X^{n+1};G)\cong \ker \delta_n \cong \ker d_n/im \delta_{n-1}\cong \ker d_n/im d_{n-1} $.
By definition, i thought $H^n(X^{n+1};G) \cong \ker \delta_n/ im \delta_{n-1}$. It does not seem obvious to me that $im \delta_{n-1}$ is trivial? I am also not sure why that $H^k(X^n;G\cong 0)$ gives that the diagonal sequences are exact.
When Hatcher writes $H^n(X^{n+1};G) \cong \ker \delta_n$, he is referring to the map $\delta_n$ in the long exact sequence for the pair $(X^{n+1}, X^n)$, not the coboundary map in some cochain complex. The relevant portion of the long exact sequence is in the diagram in your post, going diagonally starting from $H^n(X^{n+1}, X^n) = 0$ in the bottom.