Why are cochains in group cohomology exact as a functor of the coefficients? I am stuck with exercise $1$ of section $3$ of chapter $1$ in the book Cohomology of number fields by Neukirch. The exercise is to show that the functor from $A \rightarrow C^n(G,A)$ is exact, where $G$ is a profinite group and $A$ is a $G$ module. 
By definition $C^n(G,A)=X^n(G,A)^G$. Here $X^n(G,A)$ is defined to be continuous map from $G^{n+1}$ to $A$ with discrete topology on $A$. Now from page $32$ of the same book  using  proposition $1.3.7$, I deduce that $X^n(G,A)=X^n(G,Ind_G(A))^G$ that is I get that $C^n(G,A)=({X^n(G,Ind_G(A))^G})^G$. But $(-)^G$ is left exact and I know that $A \rightarrow Ind_G(A)$ is exact. I am lost after that. Any help will be very favorable.   
 A: As Alex Youcis commented, you can understand this much more cleanly using inhomogeneous cochains.  Given a continuous map $F:G^n\to A$, you can define $f:G^{n+1}\to A$ by $$f(g_0,g_1,\dots,g_n)=g_0F(1,g_0^{-1}g_1,\dots,g_0^{-1}g_n)$$ and you can easily check that $f\in C^n(G,A)$.  Conversely, given $f\in C^n(G,A)$, you can define $$F(g_1,\dots,g_n)=f(1,g_1,\dots,g_n)$$ and this is inverse to the previous construction.  This gives a natural isomorphism of abelian groups between $C^n(G,A)$ and the set of all continuous maps $F:G^n\to A$.  But this latter construction is trivially seen to be exact (for right-exactness, note that if $p:B\to C$ is a surjection of $G$-modules and $F:G^n\to C$ is continuous, we can compose $F$ with any set-theoretic section of $p$ to get a continuous map $G^n\to B$).
Another way to think about what's going on here is that the topological $G$-set $G^{n+1}$ is actually freely generated by the subset $\{1\}\times G^n$.  So, $G$-equivariant continuous maps out of $G^{n+1}$ are the same thing as just any continuous maps out of $\{1\}\times G^n$.
