I am working on commutative Lie-groups and I am not so sure about the following problem. Suppose we have an exact sequence of commutative Lie-groups $1 \rightarrow A \rightarrow B \rightarrow C \rightarrow 1$ and denote with $(-)^0$ the connected component to the identity of the Lie-groups. Suppose that $A$, $B$ and $C$ have finely many connected components. Hence by a theorem, which is called "Open Mapping Theorem" (11.1.8) in Neeb, Hilgert "Structure and geometry of Lie-Groups" the induced morphism of the Lie-algebras of $B$ and $C$ is surjective. Hence we have a surjective Lie-group homomorphism $B^0 \rightarrow C^0\rightarrow 1$. Now I would like to look at this kernel and hopefully it is connected. Does anybody know, if this is true? Do we maybe have the identity $im(A^0)=B^0$?
Thanks for any comment
Added version:
What I really want to show is the following:
Let $k=\mathbb{C}$ or $\mathbb{R}$. Let. $X/k$ be a smooth commutative group scheme of finite type and $A/k$ an abelian variety. Suppose we have an exact sequence of commutative Lie-groups $0 \rightarrow k^\star \rightarrow X(k) \overset{\beta}{\rightarrow} A(k) \rightarrow 0$.
I want to show, that (if we denote again with $(-)^0$ the identity) component, that $\beta(X(k)_{max.komp.})\subset A(k)^0$ holds. I tried that using a theorem, which states if we have an exact sequence of commutative connected Lie-groups, we obtain an exact sequence of the maximal compact subgroups, which uses the representation of a connected compact Lie-group. Thats why I wanted to reduce to the case of the identity components, but this sequence is not exact anymore I think.
Again: Edited :D
Sorry, I realize I should really learn more about that :D So I reduced the problem to the following:
Assume we have a surjective Lie-grouphomomorphism of abelian Lie-groups $G \rightarrow H$ and the corresponding Lie-algebra homomorphism $L(G)\rightarrow L(H)$ is surjective too. What I want to show is, that then the Lie-algebra homomorphism $L(G_{max.compact})\rightarrow L(H)$ is surjective too, again $G_{max.compact}$ is the maximal compact subgroup of $G$. Maybe this is because $G_{max.compact}$ is the maximal compact subgroup? Does anyone know, why if is true?
Thanks for you're help!