Corollary 2

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

Let $\mathfrak{a}=f^{-1}(\mathfrak{h}')$. By Cor. $1,\mathfrak{a}$ has a Cartan subalgebra $\mathfrak{h}$. By Cor. 2 of Prop. $4,f(\mathfrak{h}) =\mathfrak{h}'$. We show that $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$. Let $\mathfrak{n}$ be the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$. It is enough to prove that $\mathfrak{h}=\mathfrak{n}$. If $x\in \mathfrak{n},f(x)$ belongs to the normalizer of $\mathfrak{h}'$ in $\mathfrak{g}'$, so $f(x)\in \mathfrak{h}'$ and $x\in \mathfrak{a}$; but $\mathfrak{h}$ is its own normalizer in $\mathfrak{a}$, so $x\in \mathfrak{h}$.