Let $0 \rightarrow X^{*} \rightarrow Y^{*} \rightarrow Z^{*} \rightarrow 0$ be a short exact sequence of chain complexes of $R$ modules.
Suppose $Z^{*} =\dots \rightarrow Z^{i-1} \rightarrow Z^{i} \rightarrow Z^{i+1} \rightarrow \dots $ is a long exact sequence.
How do I show that $H^{i}(X^{*}) \cong H^{i}(Y^{*})$, where $H^{i}(X^{*})$ is the $i^{th}$ homology group, where $H^{i}(X^{*}) = \ker(d_{i}) / Im(d_{i-1})$
Comments:
I know that when $Z^{*}$ is long exact $H^{i}(Z^{*}) = 0$
By the Snake Lemma, there is a long exact sequence $\dots \rightarrow H^{i}(X^{*}) \rightarrow H^{i}(Y^{*}) \rightarrow H^{i}(Z^{*}) \rightarrow H^{i+1}(X^{*}) \rightarrow H^{i+1}(Y^{*}) \rightarrow H^{i+1}(Z^{*}) \rightarrow \dots$
I can deduce $H^{i}(Y^{*}) / Im(H^{i}(X^{*})) = 0$, but is it the case that the map $H^{i}(X^{*}) \rightarrow H^{i}(Y^{*})$ is injective, and if so why?
If I assume the previous statement, would it then be true that $H^{i}(Y^{*}) \cong Im(H^{i}(X^{*})) \cong H^{i}(X^{*})$? i.e. would we be done?
It follows immediately from the long exact sequence of cohomology: since $H^{i}(Z)=0$ for all $i$, the long exact sequence is a collection of exact sequences of the form $0\to H^{i}(X)\to H^{i}(Y)\to 0$. But exactness here just means the two are isomorphic.