Lie algebra: ideal

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Homework Statement



Let [itex]\mathfrak{g}[/itex] be any lie algebra and [itex]\mathfrak{h}[/itex] be any ideal of [itex]\mathfrak{g}[/itex].

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

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

The Attempt at a Solution



See below
 
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Answers and Replies

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This is my attempt:

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

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

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

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

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

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

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