How to go from $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ to $0 \rightarrow A^G \rightarrow B^G \rightarrow C^G$

Question

Let $G$ be a profinite group and $A, B,C$ abstract $G$-modules. If I have an exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$, how does one get an exact sequence $0 \rightarrow A^G \rightarrow B^G \rightarrow C^G$ where the exponent $G$ means all the elements fixed by the action of $G$. In the book I am reading all it says is we pass from the former to the latter, but I have no idea how to get the second exact sequence. Any explanation would be appreciated.

Answer 1

Consider the following exact sequence of $G$-modules: $$0\longrightarrow A \stackrel{f}\longrightarrow B \stackrel{g}\longrightarrow C \longrightarrow0$$ Using the $G$-equivariance of $f$ and $g$, we see that any $a\in A^G$ is taken to some $f(a)\in B^g$. I.e. one can observe that $f(a)=f(g\cdot a)=g\cdot f(a)$. This clearly also holds for $g$ for the same reason. Now, we want to show that the following sequence is exact: $$0\longrightarrow A^G \stackrel{f|_{A^G}}\longrightarrow B^G \stackrel{g|_{B^G}}\longrightarrow C^G.$$ It is clear that each of $f|_{A^G}$ and $g|_{B^G}$ are $G$-equivariant (these are all trivial $G$-modules). It is also clear that $f|_{A^G}$ is injective given that it is the restriction of an injective map.

To see that this is exact about $B^G$ we take $a\in A^G$. Then $g(f(a))=0=g|_{B^G}(f|_{A^G}(a))$, i.e. $\text{im}(f|_{A^G})\subset \text{ker}(g|_{B^G})$. Next, say that $b\in B^G$ satisfies $g|_{B^G}(b)=0$, then by the exactness of the original sequence it follows that we can find some $a\in A$ such that $f(a)=b$. But by $G$-equivariance of $f$, we know that for any $g\in G$, $$g\cdot f(a)=g\cdot b = b=f(g\cdot a).$$ But $f$ is injective, so we must have that $g\cdot a=a$ for every $g\in G$, i.e. $a\in A^G$ satisfies $f(a)=b$, showing $\text{ker}(g|_{B^G})\subset \text{im}(f|_{A^G})$, giving exactness as desired.

Why is exactness at $C^G$ lost?