We have exact sequence of groups $0\rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$ and an injective resolution of group $G$: $0\rightarrow G\rightarrow I^0\rightarrow I^1\rightarrow 0$. Since $I^*$ are injective we get $$0\rightarrow Hom(A'',I)\rightarrow Hom(A,I)\rightarrow Hom(A',I)\rightarrow 0$$ and hence we have a short exact sequence of cochain complexes $$0\rightarrow Hom(A'',I^*)\rightarrow Hom(A,I^*)\rightarrow Hom(A',I^*)\rightarrow 0.$$ Now I want to get (not very) long exact sequence in homology: $$0\rightarrow Hom(A'',G)\rightarrow Hom(A,G)\rightarrow Hom(A',G)\rightarrow Ext(A'',G)\rightarrow Ext(A,G)\rightarrow Ext(A',G)\rightarrow 0$$
When I draw a diagram, I think I found the way (a map) how elements from $Hom(A',G)$ go in $Ext(A'',G)$, but I don't know how to show that the exactness holds. Any idea?
I thaught that last two $Ext$'s will be easier, but I don't see now how to show the exatness too.