Extension of a finite group by a connected group necessarily splits? Suppose that $G$ is a compact abelian group. I denote by $G_0$ the connected component of the identity in $G$.
If $G_0$ is open in $G$ (equivalently $G/G_0$ is finite) is it true that $G\cong G_0\times G/G_0$?
I also assume that $G/G_0$ is equipped with the quotient topology and that everything is Hausdorff.
*If the title confuses you, the statement above is equivalent to whether the short exact sequence $0\rightarrow G_0\rightarrow G \rightarrow G/G_0\rightarrow 0$ splits.
 A: Yes, this is true. The case for Lie groups is pretty easy (and I explained it here). For the general case, one can prove this using the fact compact groups are inverse limits of Lie groups.
Let $G$ be a compact abelian group, and denote $A=G/G^0$. By a corollary of Peter-Weyl, every identity neighbourhood contains a subgroup which is co-Lie (i.e. the quotient by it is a Lie group). Thus we can form a net of subgroups $N_\alpha$, ordered by reverse inclusion (so $\alpha \geq \beta$ if and only if $N_\beta \subseteq N_\alpha$), such that $N_\alpha\subseteq G^0$ for all $\alpha$ (sine $G^0$ is open) and such that $\bigcap N_\alpha=\{1\}$. Denote $G_\alpha=G/N_\alpha$ and denote by $p_\alpha:G\to G_\alpha$ the natural quotient map.
Since each $p_\alpha$ is open, continuous and surjective, the subgroup $p_\alpha(G^0)$ is open and connected, and hence is the connected component of $G_\alpha$. Since $N_\alpha\subseteq G^0$, we have by the third isomorphism theorem (for topological groups) that $G_\alpha/G_\alpha^0=(G/N_\alpha)/(G^0/N_\alpha)\cong G/G^0 = A$. Thus, by the case for Lie groups, we know $G_\alpha \cong G_\alpha^0\times A_\alpha$ for $A_\alpha\cong A$.
Denote $A'_\alpha = p_\alpha^{-1}(A_\alpha)$ and $A'=\bigcap A'_\alpha$. I will show that $A'$ is mapped isomorphically onto $A$ via the quotient $q:G\to G/G^0$, which shows the sequence does indeed split.
First let us see that it is mapped injectively into $A$ via $q$. We have, for every $\alpha$, $$p_\alpha(A'_\alpha\cap G^0)\subseteq p_\alpha(A'_\alpha)\cap p_\alpha(G^0)=A_\alpha\cap G_\alpha^0=\{1\}.$$ Thus $A'_\alpha\cap G^0\subseteq N_\alpha$ for every $\alpha$, so $\bigcap A'_\alpha\cap G^0\subseteq\bigcap N_\alpha=\{1\}$. This precisely mean that the intersection of $A'=\bigcap A'_\alpha$ with $\ker q=G^0$ is trivial, i.e. $q$ mapps $A'$ injectively into $A$.
It remains to show $q(A')=A$. I think this is pretty clear. For every $\alpha$ we know $q(A'_\alpha)=A$. To see this, first note that $p_\alpha(A'_\alpha\cdot G^0)$ contains both $p_\alpha(A'_\alpha)=A_\alpha$ and $p_\alpha(G^0)=G_\alpha^0$ and hence is all of $G_\alpha=G_\alpha^0\times A_\alpha$, and therefore $A'_\alpha\cdot G^0\cdot \ker p_\alpha = G$, but $\ker p_\alpha = N_\alpha \subseteq G^0$, so $A'_\alpha\cdot G^0=G$. Since $\ker q= G^0$ we have $q(A'_\alpha)=A$. But by compactness this implies $q(A')=A$ (since if $a\in A$ then for every $\alpha$ there is $a_\alpha\in A'_\alpha$ such that $q(a_\alpha)=a$; this net has some converging subnet which converges to some $a'\in A$, and by continuity $q(a')=a$).
This answer turned out a lot longer than I expected... Using the language of inverse limits it would have been a lot shorter, I think, if you know how to prove $G\cong \varprojlim G_\alpha$ in a natural way and not just that every neighbourhood contains a co-Lie subgroup.
