I want to know if any Lie group extension
$$1\to \mathbb C^*\to A\xrightarrow{\pi} \mathbb C\to 0 \tag{1}\label{1}$$
is trivial, where the group structure is multiplicative on $\mathbb C^*$ and additive on $\mathbb C$.
I guess the answer is Yes and it suffices to find a section $s:\mathbb C\to A$ as Lie group morphism such that $\pi\circ s=Id$. From the point of view of topology, $A$ is a principal $\mathbb C^*$-bundle over the contractible base $\mathbb C$. This implies $A$ is trivial as principal $\mathbb C^*$-bundle, so there is a section $s:\mathbb C\to A$. However, the section need not preserve the group structure. Can I modify the section $s$ so as preserve the group structure?
Edited remark: As @Roland's points out, considering the extension $(1)$ purely as group extension which lives in $Ext_{Ab}^1(\mathbb C,\mathbb C^*)$ is different from an extension as Lie groups (or algebraic groups), because a section in the category of groups does not need to be even continuous.
Second Edition: Here is another idea: If we pull back $A$ to its universal covering $\require{AMScd}$ \begin{CD} \mathbb C @>>> \tilde{A}@>\tilde{\pi}>>\mathbb C\\ @VVV @VVV @VV=V \\ \mathbb C^* @>>> A @>\pi>> \mathbb C \end{CD} If I can show the sequence on the first row is trivial, then a section $\tilde{s}:\mathbb C\to \tilde{A}$ will descend to a section $s:\mathbb C\to A$, but how do we show $Ext^1_{Lie-gr}(\mathbb C,\mathbb C)=0$ (although it sounds trivial)?