I suspect there is a very simple argument for this question but I cannot get any further in this statement.
I am currently studying the connecting homomorphism between $H_0$ and $H_1$ (of group cohomology) and I need the following fact (prove or disprove)
Let $G$ be a group and $B$ and $C$ being $G$-group (groups with group action by $G$, the group operation is compatible with the group action).
Let $f:B\rightarrow C$ be a homomorphism. Prove or disprove that $\ker f$ (which is in $B$) is a subset of $B^G$. In other words, $\forall g\in G,k\in\ker f,g\cdot k=k$. (I mean $g$ acting on $k$ is $k$ itself, or $k$ is fixed by any $g\in G$)
I can provide more proof of work I have done (like I have reduced the problem of "well-definedness of connecting homomorphism" to this specific question. If the statement above is not generally true, then probably my deduction goes wrong.
Thank you very much for helping me.
You can see that $Ker f$ is stable by $G$ equivalently, for every $x\in B$, $g\in G$, $f(x)=1$ implies that $f(g.x)=g.f(x)=g.1=1$. But you do not have necessarily $g.x=x$ if $x\in Ker f$, consider the example: $G=G$ endowed with the adjoint action, $g.g'=gg'g^{-1}$, $C=\{1\}$ the trivial group endowed with the trivial action of $G$, and $f:G\rightarrow \{1\}$, $ker f=G$, but $G\neq G^G$ if $G$ is not commutative.