A question on exact sequences and group cohomology Question:
If $0\longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0,$ is an exact sequence of $G$-modules, then show that $$0 \longrightarrow A^G \longrightarrow B^G \longrightarrow C^G \longrightarrow H^1(G,A) \longrightarrow  H^1(G,B) \longrightarrow H^1(G,C)$$ is exact, where if A is a G-module, then $A^G$ is the set of $G$-invariants of $A.$
Attempt:
The map $A^G \longrightarrow B^G$ is the restriction of $A \longrightarrow B$ to $A^G.$ (and similarly, the map $B^G \longrightarrow C^G$ is defined). The map $\eta:H^1(G,A) \longrightarrow H^1(G,B)$ is defined as follows. If $j:A\longrightarrow B$ and $h\in Z^1(G,A),$ then $\eta(\bar{h})=\overline{j\circ h}$ where bar denotes the quotients from $B^1(G,A)$ and $B^1(G,B).$
I have proved everyting except for the fact that ${\rm im}(H^1(G,A) \longrightarrow H^1(G,B))=\ker(H^1(G,B) \longrightarrow H^1(G,C)).$ Could someone help me with this please? Thank you!!
 A: Let $f: A \to B, g:B \to C$ be the corresponding maps. Then the map $H^1(G,A) \to H^1(G,B)$ is defined by $\phi \mapsto f \circ \phi$, and similarly for $H^1(G,B) \to H^1(G,C)$.
It follows that $\text{Im}(H^1(G,A) \to H^1(G,B)) \subseteq \text{Ker}(H^1(G,B) \to H^1(G,C))$. For the hard part let $\psi \in \text{Ker}(H^1(G,B) \to H^1(G,C))$. Then $g \circ \psi=0$ in $H^1(G,C)$, this means that there exists some $x \in C$ such that $g \circ \psi(\sigma)=\sigma x - x$. Since $g$ is surjective, $x=g(y)$ for some $y \in B$. Therefore
$$g(\psi(\sigma))=\sigma g(y)-g(y)=g(\sigma y- y).$$
Let $\psi'$ be defined by $\psi'(\sigma)=\psi(\sigma)-(\sigma y- y)$. It represents the same element as $\psi$ in $H^1(G,B)$, and $g \circ \psi'(\sigma)=0$ for all $\sigma$. Thus for each $\sigma \in G$,  $\psi'(\sigma) \in \text{Ker}(g)=\text{Im}(f)$. Since $f$ is injective there exists a unique $a_\sigma$ such that $\psi'(\sigma)=f(a_\sigma)$. Let $\phi: G \to A$ be given by $\sigma \mapsto a_\sigma$. Now just check that $\phi \in Z^1(G,A)$ i.e $\phi(\sigma \tau)=\sigma \phi(\tau)+ \phi(\sigma)$ for all $\sigma, \tau$.
Of course the question is trivial if you know about derived functors and the fact that $H^{*}(G,-)=\text{Ext}_{\mathbb{Z}[G]}(\mathbb{Z},-)$ is the right derived functor of $\text{Hom}_{\mathbb{Z}[G]}(\mathbb{Z},-)=(-)^G$. You can read about that in a homological algebra book.
