Proving inverse image of ideal is an ideal

Question

Let $L_1,L_2$ two Lie algebras.

For a Lie morphism $\varphi : L_1 \to L_2$, let $I\subset L_2$ an ideal.

How we prove $$\varphi^{-1}(I) \subset L_1$$ is an ideal?

My try:

$$[\varphi^{−1}(I),L_1]=[\varphi^{−1}(I),\varphi^{−1}(L_2)]=\varphi^{−1}([I,L_2])⊂\varphi^{−1}(I)$$

Additional doubt: If $S\subset L_1$ is a subalgebra, do you have $\varphi(S) \subset L_2$ is a subalgebra?

Thanks!

Answer 1

Let $I$ any ideal in $L_2$. Of course $\varphi^{-1}(I)\neq\emptyset$ since $\varphi(0)=0\in I$. Now for all $x,y\in\varphi^{-1}(I)$ we have $\varphi(x), \varphi(y)\in I$ and, in particular, $\varphi(x+y)=\varphi(x)+\varphi(y)\in I$. Therefore $x+y\in\varphi^{-1}(I)$.

Finally for every $r\in L_1$ and any $x\in\varphi^{-1}(I)$ we have $\varphi(r)\in L_2$ and $\varphi(x)\in I$; hence $\varphi([r,x])=[\varphi(r),\varphi(x)]\in I$. Therefore $[r,x]\in\varphi^{-1}(I)$.

Answer to your additional doubt: YES!