If the image of a map from a Lie group is open, then is its differential onto?

Question

Let $G$ a simply connected Lie group, $E$ a finite-dimensional real vector space and $\pi:G\to E $ a smooth map such that its image $\pi(G)$ is open in $E$. Can I assure that the differential at the identity $\pi_{*,\epsilon}$ is onto? In the other way, if I know that the differential is onto, can I prove that $\pi(G)$ must be open?

Answer 1

Neither implication is necessarily true.

Consider the special case $G = \mathbb{R}$ and $E = \mathbb{R}$.

Let $\pi_1 : \mathbb{R} \to \mathbb{R}$ be the map $x \mapsto x^3$. Then the differential of $\pi_1$ at the identity is not onto, but the image of $\pi_1$ is $\mathbb{R}$, which is open.

Let $\pi_2 : \mathbb{R} \to \mathbb{R}$ be the map $x \mapsto \sin(x)$. Then the differential of $\pi_2$ at the identity is onto, but the image of $\pi_1$ is $[-1,1]$, which is not open.