Exactness of Inflation restriction sequence, Galois Cohomology I am trying to prove the following.
Let $K/k$ be a finite Galois extension, $G= G(K/k)$, $k \subset F \subset K$ with $K/k$ normal and $H=G(K/F)$. Then:
$ \rho : C^{2} (G,A) \rightarrow C^{2} (H,A) $ defined as $\rho (f) ( \sigma , \tau ) = f( \sigma, \tau)$ induces a homomorpshim $\rho^{*} : H^{2} (G,A) \rightarrow H^{2} (H,K^{\times})$, we call this restriction. And $ \lambda : C^{2} (G/H, F^{\times}) \rightarrow C^{2} (G,K^{\times}) $ given by $\lambda (f)( \sigma, \tau) = f (\sigma H , \tau H) $ induces a homomorphism $\lambda^{*} : H(G/H, F^{\times}) \rightarrow H^{2} (G,K^{\times})$, we call this inflation.
I already proved this part, you only have to see that it sends cochains on cochains and coboundaries on coboundaries as far as I understand.
Then, I have to prove that the following sequence is exact:
$$
1 \rightarrow H^{2} (G/H, F^{*} ) \overset{\lambda^{*}}{\longrightarrow} H^{2} (G,K^{*}) \overset{\rho^{*}}{\longrightarrow}  H^{2} (H,K^{*})
$$
As far as I understand, I have to prove that $\lambda^{*}$ is injective as the image of the trivial homomorphism is the identity element. Also, I have to prove that $Im(\lambda^{*})=Ker(\rho^{*})$. I have proved that $Im(\lambda^{*}) \subset Ker(\rho^{*})$ computing $\rho^{*}$ over a generic element of $Im(\lambda^{*})$ and seeing it's zero. But I wasn't able to prove the opposite implication.
Regarding the injectivity of $\lambda^{*}$, I tried to prove that if $\lambda (f) = \delta (g)$ with $g \in C^{1} (G,K^{*})$ then there exists $h \in C^{1} (G/K,F^{*})$ such that $f=h$, and that should mean that $\lambda^{*}$ is injective, as it would mean that $\lambda^{*} (\overline{f})=\overline{1}$ implies $\overline{f} = \overline{1}$.
The thing is, that I cannot find that element $h \in C^{1} (G/K,F^{*})$, I think need to prove that $g$ can be factored on to the quotient to define a $h=g'$ such that $g(\sigma) = g' (\sigma H)$ for all $\sigma \in G$.
I haven't used any Galois theory on this, so I guess here is where those hyphotesis comes into play. I know that $G/H \cong G(F/k)$ but I haven't been able to use that isomorphism.
Any hints?
 A: This property actually generalizes in group cohomology, but the condition isn't straightforward. For this to work, you need $H^1(H,K^*)$ to vanish. So any straightforward computation is bound to fail.
More generally, if $A$ is a $G$-module and $H$ is a normal subgroup of $G$, if $k \geq 1$, the inflation-restriction sequence $$1 \longrightarrow H^k(G/H,A^H) \longrightarrow H^k(G,A) \longrightarrow H^k(H,A)$$ is exact as long as all the $H^l(H,A)$ vanish for $1 \leq l < k$. This is proved rather cleanly with group cohomology tools (the long exact sequence, shifts).
I managed to find a (mostly) elementary proof of the injectivity: let $A=K^*$ be the $G$-module noted additively (for ease of notation). By Hilbert 90, $H^1(H,A)=0$, and $A^H=F^*$ (set of elements of $A$ that are invariant under the action of $H$). 
Let $f:(G/H)^2 \rightarrow A^H$ correspond to an element of $H^2(G/H,A^H)$, that is, for every $g,h,k \in G$, $[g] \cdot f([h],[k]) - f([gh],[k]) + f([g],[hk])-f([g],[h])=0$. Assume that $f$ vanishes in $H^2(G,A)$. Thus, there exists $t:G \rightarrow A$ such that for all $g,h \in G$, $f([g],[h])=g \cdot t(h) - t(gh) + t(g)$. We want to show that $f$ vanishes in $H^2(G/H,A^H)$. 
Taking $g \in H$, we find that $f(e_{G/H},\cdot)$ is constant equal to $l \in A^H$. Taking $h \in H$, we find that $[g] \cdot l = f([g],e_{G/H})$. Now, considering $t$, it follows that $t(e_G)=l$ (taking $h = e_G$). 
Restricting $t$ to $H$, we find that for all $g,h \in H$, $t(gh)=t(g)+g(t(h))-l$, so that $c=t-l$ is a cocycle $H \rightarrow A$. Since $H^1(H,A)=0$, there is some $p \in A$ such that $t(h) = l+h(p)-p$ for every $h \in H$. 
Now, take $g \in G,h \in H$: then $t(gh)-t(g)=g(t(h))-f([g],[h])=g(t(h)-l) = gh(p)-g(p)$. Thus $g \in G \longmapsto t(g)-g(p)$ is invariant under the right action of $H$, it corresponds to a function $f_1:G/H \rightarrow A$. Let $g,h \in G$: $g \cdot f_1([h])-f_1([gh])+f_1([g]) = g(t(h)-h(p))-(t(gh)-gh(p))+t(g)-g(p) = f([g],[h]) - g(p)$. It follows that the image of $f_1+p$ is invariant under $H$, so $f_1+p:G/H \rightarrow A^H$ satisfies $\partial (f_1+p)=f$ and $f$ represents the zero element in $H^2(G/H,A^H)$. 

I haven't been able to find a similar proof for the restriction map. I'm pretty sure it should exist, though. 
But using the toolbox from group cohomology, the statement is much easier to show:
First, consider the exact sequence of $G$-modules $0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}[G] \rightarrow J_G \rightarrow 0$, where the action on $\mathbb{Z}$ is trivial, the first map is $1 \longmapsto \sum_{g \in G}{g}$, and the action on the second term is the relevant permutation of the coordinates. 
This exact sequence is split (in abelian groups), so remains exact when tensoring with $A$: we get an exact long sequence in cohomology $H^1(S,A[G]) \rightarrow H^1(S,A \otimes J_G) \rightarrow H^2(S,A) \rightarrow H^2(S,A[G])$ with $S=H$ or $G$. The (first) trick is to notice that for $i \geq 1$, $H^i(S,A[G])=0$ -- this is proved using the definition. Thus we have isomorphisms $H^2(S,A) \cong H^1(S,A \otimes J_G)$.
Second, note that we also have an exact sequence $1 \rightarrow A^H \rightarrow A[G]^H \rightarrow (A \otimes J_G)^H \rightarrow H^1(H,A)$. By Hilbert 90, $H^1(H,A)=0$, and we can thus take the same exact sequence in cohomology of $G/H$-modules: $H^1(G/H,A[G]^H) \rightarrow H^1(G/H,(A \otimes J_G)^H) \rightarrow H^2(G/H,A^H) \rightarrow H^2(G/H,A[G]^H)$. But $A[G]^H \cong A[G/H]$ so its cohomology groups vanish. Thus we have an isomorphism $H^2(G/H,A^H) \cong H^1(G/H,(A \otimes J_G)^H)$. 
The third element is that these isomorphisms commute with the inflation and restriction morphisms. So it is enough to show the exactness of the sequence for $A \otimes J_G$ in degree $1$. But in this degree, the computations are manageable.  
