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If $\phi: L_1 \rightarrow L_2"$ is a surjective Lie algebra homomorphism, is it true that $\phi (Z(L_1))=Z(L_2)$. I see that $\phi (Z(L_1))$ is in $Z(L_2)$, but if $\phi^{-1}(0)$ is not $0$, i.e $\phi$ not injective I think that the other inclusion is not true. Is that right? would a projection work as a counterexample?

inquisitor
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You are completely correct that $\varphi(Z(L_1))\subseteq Z(L_2)$ (as is easily seen by writing up what this means and using that the homomorphism is surjective).

You are also correct that in general, one need not get the entire center of $L_2$ this way. For a general example, one can take any nilpotent but not abelian Lie algebra (such as the Lie algebra consisting of $3\times 3$ strictly upper triangular matrices). As it is nilpotent, it has a non-trivial center. If one quotients by the center, one again gets a non-trivial nilpotent Lie algebra, which thus has a non-trivial center. But as the center of the original Lie algebra is sent to $0$ in this quotient, one cannot get the entire center of the quotient this way.

Tobias Kildetoft
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    It took me a while to see why if $L$ is nilpotent it has non-zero center (assuming is not abelian or $0$). I guess the argument is just: $[L^{k-1}L]=0$ so $L^{k-1}$ is in the center (where $k$ is the minimum st $L^k=0$) – inquisitor Feb 07 '13 at 18:58
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    @inquisitor Yes, exactly. – Tobias Kildetoft Feb 07 '13 at 19:07
  • If the map was an isomorphism rather than a surjective morphism, then the centers would be preserved obviously, right? – FireFenix777 Sep 19 '20 at 23:49
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    @FireFenix777 yes, $\phi(Z_1) = Z_2$. You can deduce it using the above result: use that $\phi(Z_1) \subset Z_2$ and apply $\phi$ to $\phi^{-1}(Z_2) \subset Z_1$ – inquisitor Sep 21 '20 at 09:37