Lie Algebra: Ideal Homework Proving \pi ^{-1} (\mathfrak{f}) is an Ideal

[Ted123]

Homework Statement

Let \mathfrak{g} be any lie algebra and \mathfrak{h} be any ideal of \mathfrak{g}.

The canonical homomorphism \pi : \mathfrak{g} \to \mathfrak{g/h} is defined \pi (x) = x + \mathfrak{h} for all x\in\mathfrak{g}.

For any ideal \mathfrak{f} of the quotient lie algebra \mathfrak{g/h}, consider the inverse image of \mathfrak{f} in \mathfrak{g} relative to \pi, that is: \pi ^{-1} (\mathfrak{f}) = \{X\in\mathfrak{g} : \pi (X)\in \mathfrak{f} \} .
Prove that \pi ^{-1} (\mathfrak{f}) is an ideal of the lie algebra \mathfrak{g}.

The Attempt at a Solution

See below

 

[Ted123]

This is my attempt:

Let x\in\mathfrak{g} and y\in\pi ^{-1}(\mathfrak{f}).

We want to show that [x,y]\in\pi ^{-1}(\mathfrak{f}). That is, if \pi ([x,y]) \in \mathfrak{f}.

Now \pi ([x,y]) = [\pi (x) , \pi (y)] since \pi is a homomorphism.

And since y\in \pi ^{-1}(\mathfrak{f}), \pi (y) \in \mathfrak{f}.

But [\pi (x) , \pi (y)] \in \mathfrak{f} for all \pi (x) \in \mathfrak{g/h} and \pi (y) \in \mathfrak{f} since \mathfrak{f} is an ideal of \mathfrak{g/h} and \pi (x) = x+\mathfrak{h} \in \mathfrak{g/h}.

Therefore \pi ([x,y])\in\mathfrak{f} and \pi ^{-1}(\mathfrak{f}) is an ideal of \mathfrak{g}.

Can anyone spot any mistakes or anything I've done wrong?