This is most likely trivial, but I don't get it. In Humphrey's Introduction to Lie algebras, page 82, he says:
It is clear that, if $\phi \colon L \to L'$ is epimorphism [of finite dim. Lie algebras], and $y \in L$, then $\phi (L_a (\operatorname{ad} y)) = L'_a (\operatorname{ad} \phi(y))$,
where $L_a(\operatorname{ad} y)$ is generalized eigenspace of operator $\operatorname{ad} y$, for eigenvalue $a \in K$, that is $$ L_a(\operatorname{ad} y) = \bigcup_{k \geq 1}(\operatorname{Ker}(\operatorname{ad}(y) - aI))^k,$$ an increasing union that stabilizes for $k$ large enough. Similary $L'_a (\operatorname{ad} \phi(y))$.
Question:
Inclusion $\phi (L_a (\operatorname{ad} y)) \subseteq L'_a (\operatorname{ad} \phi(y))$ is trivial even to me, but why does the other inclusion $\phi (L_a (\operatorname{ad} y)) \supseteq L'_a (\operatorname{ad} \phi(y))$ hold?
My attempt: Take $\phi(x) \in L'_a (\operatorname{ad} \phi(y))$. Then, for some $k$ we have $(\operatorname{ad}(\phi(y)) - aI)^k(\phi(x))=0$, from which we see that $(\operatorname{ad}(y) - aI)^k(x) \in \operatorname{Ker} \phi$. But I need to have $(\operatorname{ad}(y) - aI)^k(x) = 0$, or to find some $z$ such that $(\operatorname{ad}(y) - aI)^k(z) = 0$ and $x-z \in \operatorname{Ker} \phi$.
I found an answer in this thesis, on page 12. I will rewrite it here for the sake of completeness. It is trivial, but not obvious.
Take $z=\phi(x) \in L'_a (\operatorname{ad} \phi(y))$, and let $a_1, \ldots, a_n$ be all eigenvalues of $\operatorname{ad} x$. Using Jordan decomposition, we have direct sums $$ L_{a_1}(\operatorname{ad} y) \oplus \ldots \oplus L_{a_n}(\operatorname{ad} y)=L, \quad L'_{a_1}(\operatorname{ad} \phi(y)) \oplus \ldots \oplus L'_{a_n}(\operatorname{ad} \phi(y)) = L'.$$ Last equality follows because $\phi (L_a (\operatorname{ad} y)) \subseteq L'_a (\operatorname{ad} \phi(y))$ and $\phi$ being epimorphism.
If $a$ is not eigenvalue of $\operatorname{ad} y$, then there is nothing to prove. Now take $a=a_1$, decompose $x=x_1 + \ldots +x_n$, where $x_i \in L_{a_i}(\operatorname{ad} y)$. We have $$ L'_{a_1} (\operatorname{ad} \phi(y)) \ni z-\phi(x_1) = \phi(x)-\phi(x_1) = \phi(x_2) + \ldots + \phi(x_n) \in L'_{a_2}(\operatorname{ad} \phi(y)) \oplus \ldots \oplus L'_{a_n}(\operatorname{ad} \phi(y)). $$ Because the sum is direct, we have $z=\phi(x_1) \in \phi(L_{a_1}(\operatorname{ad} y))$.