Corollary 2

Let $f:\mathfrak{g}\rightarrow \mathfrak{g}'$ be a surjective homomorphism of Lie algebras. If $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g},f(\mathfrak{h})$ is a Cartan subalgebra of $\mathfrak{g}'$.

Indeed, $f(\mathfrak{h})$ is a nilpotent subalgebra of $\mathfrak{g}'$. On the other hand, consider the representation $x \rightarrow$ ad $f(x)$ of $\mathfrak{h}$ on $\mathfrak{g}'$. By Prop. 9 (iv) of §1, no. 3, $f(\mathfrak{g}^0(\mathfrak{h})) ={\mathfrak{g}'}^0(\mathfrak{h})$. Now $\mathfrak{g}^0(\mathfrak{h}) =\mathfrak{h}$, and on the other hand it is clear that ${\mathfrak{g}'}^0(\mathfrak{h}) ={\mathfrak{g}'}^0(f(\mathfrak{h}))$. Hence, $f(\mathfrak{h}) ={\mathfrak{g}'}^0(f(\mathfrak{h}))$ and it suffices to apply Prop. 4.