TITLE: If a complex Lie group has the structure of an algebraic group, is this structure unique?
QUESTION [10 upvotes]: If $G$ and $H$ are algebraic groups over $\mathbb{C}$, and $f : G \rightarrow H$ is an isomorphism of complex Lie groups (i.e. a biholomorphic group isomorphism), then must $f$ be algebraic?  If not, are there additional hypotheses that make this true?
If $G$ and $H$ are projective then general GAGA machinery answers the question in the affirmative, but this is obviously rather restrictive (in particular, it forces $G$ and $H$ to be abelian).  But if $f$ is not an isomorphism, and instead is just a holomorphic group homomorphism, then it need not be algebraic, for example the map $\mathbb{C} \rightarrow \mathbb{C}^*$ given by $\lambda \mapsto e^\lambda$.

REPLY [6 votes]: An isomorphism of complex Lie groups between algebraic groups need not be algebraic in general. A counterexample is
$$
\begin{align*}
f:\mathbb{C}\times\mathbb{C}^\times&\xrightarrow{\sim} \mathbb{C}\times\mathbb{C}^\times,\\
(x,y)&\mapsto (x,y\cdot e^x).
\end{align*}
$$