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Is the following proof correct?

Claim: Let $g$ be a semi-simple Lie algebra, and $f: g\rightarrow h$ be a homomorphism of Lie algebras. Then $Im f \leq h$ is a semi-simple Lie algebra.

Proof attempt It is a standard result that $Im f$ is Lie subalgebra of $h$, and that $f$ induces an isomorphism of Lie algebras $\bar{f}: g/ker{f} \rightarrow im{f}$. Furthermore, as $g$ is assumed semi-simple so is its quotient $g/ker{f}$. Suppose for contradiction that there is a non-zero solvable ideal $I \lhd Im f$, and let $J$ be its preimage in $g/ker(f)$. It is an easy check that $J \lhd g/ker(f)$, and that $\bar{f}([J,J]) = [I,I]$. Hence by induction $\bar{f}(D^n J) = D^n I$. Now let $n$ be such that $D^n I = 0$. Then $\bar{f}(D^n J) = D^n I = 0$, hence as $\bar{f}$ is an isomorphism $D^n J = 0$, which gives as the required contradiction.

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  • $\begingroup$ See this duplicate. If $f=0$, then $im(f)=0$. But is $0$ a semisimple Lie algebra? $\endgroup$ Commented May 5, 2019 at 18:28

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It looks fine, but there is a simpler way. Since $\mathfrak g$ is semisimple, its ajoint representation is simisimple too and therefore there is an ideal $\mathfrak j$ of $\mathfrak g$ such that $\mathfrak g=\ker f\oplus\mathfrak j$. But then $\mathfrak g/\mathfrak\ker f\simeq\mathfrak j$, which is semisimple.

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  • $\begingroup$ Yes this is much better, thanks for posting $\endgroup$
    – gen
    Commented May 5, 2019 at 17:20

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