# Lie algebra homomorphisms preserve semi-simplicity?

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.

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

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.