Is the Lie algebra morphism induced by surjective Lie group morphism also surjective? Let $G$ be a matrix lie group and   $\Pi : G\to \Pi(G)$ a surjective Lie group morphism. Let $\mathfrak g$ and $\mathfrak h$ be the respective Lie algebras of $G$ and $\Pi(G)$. Then there is a unique morphism of Lie algebras $\pi : \mathfrak g \to \mathfrak h$ which makes the following diagram commute:
$$\begin{matrix}
&& \Pi & \\
&G & \to & \Pi(G)\\
\exp &\uparrow & & \uparrow & \exp\\
&\mathfrak{g} & \to &\mathfrak{h}\\
&& \pi \\
\end{matrix}$$
Is $\pi$ surjective? $\quad$
ie. Can we write $\mathfrak h = \pi(\mathfrak g)$?

Using a dimension argument and Sard theorem (or anything to assure that the identity is a regular value of $\Pi$ ), the dimension of G is the sum of the dimensions of the kernel and image of $\Pi$ and the same holds for $\pi$ so it remains to show that the dimensions of the kenerls coincide, using that $\Pi ( e^X) = e^{\pi (X)} $ it's easy to show that the kenerl of $\pi$ is the lie algebra of $\ker \Pi$ and thus the dimensions coincide.
