Proposition 2

Let $G'$ be a finite dimensional Lie group over $k$ and $f$ an open morphism from G to $G'$. The image under $f$ of a regular element of G is a regular element of $G'$. If the kernel of $f$ is contained in the centre of G, an element $g\in G$ is regular if and only if $f(g)$ is regular.

Indeed, let $\mathfrak{g}'$ be the Lie algebra of $G'$ and $\mathfrak{h}$ the ideal in $\mathfrak{g}$ given by the kernel of $Tf|\mathfrak{g}$. Let $\rho$ be the linear representation of G on $\mathfrak{h}$ defined by $\rho (g) =$ Ad $g|\mathfrak{h}$ for all $g\in G$, and let Ad $\circ f$ be the linear representation of G on $\mathfrak{g}'$ given by the composite of $f$ with the adjoint representation of $G'$. These linear representations define an exact sequence of G-modules:

By Lemma $2,r_{Ad}=r_{\rho}+r_{Ad\circ f}$. Since $r_{Ad\circ f}=r_{Ad}\circ f$ and since $f$ is an open map, $r_{Ad\circ f}^0=r_{Ad}^0\circ f$. Consequently:

Thus, if $g$ is regular, $(r_{Ad}-r^0_{Ad})(f(g)) = 0$, which means that $f(g)$ is regular. If the kernel of $f$ is contained in the centre of G,

$r_{\rho}(g) =r_{\rho}^0(g) =$ dim$\mathfrak{h}$ for all $g\in G$. Consequently, if $f(g)$ is regular, $r_{Ad}(g) =r_{Ad}^0(g)$, in other words, $g$ is regular.