776: Bijective Lie Algebra Homomorphism Is 'Lie Algebras - Homomorphisms' Isomorphism

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description/proof of that bijective Lie algebra homomorphism is 'Lie algebras - homomorphisms' isomorphism

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Topics

About: Lie algebra

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Note
• 4: Proof

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Starting Context

• The reader knows a definition of Lie algebra.
• The reader knows a definition of %structure kind name% homomorphism.
• The reader knows a definition of bijection.
• The reader admits the proposition that any bijective linear map is a 'vectors spaces - linear morphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that any bijective Lie algebra homomorphism is a 'Lie algebras - homomorphisms' isomorphism.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the not necessarily finite-dimensional }F\text{ Lie algebras }\}$
$V_2$: $\in \{\text{ the not necessarily finite-dimensional }F\text{ Lie algebras }\}$
$f$: $:V_1\to V_2$, $\in \{\text{ the Lie algebra homomorphisms }\}\cap \{\text{ the bijections }\}$
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Statements:
$f\in \{\text{ the 'Lie algebras - homomorphisms' isomorphisms }\}$
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2: Natural Language Description

For any field, $F$, any not necessarily finite-dimensional Lie algebras, $V_1,V_2$, and any bijective Lie algebra homomorphism, $f:V_1\to V_2$, $f$ is a 'Lie algebras - homomorphisms' isomorphism.

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3: Note

In general, a bijective morphism is not necessarily an isomorphism: for example, a bijective continuous map is not necessarily a 'topological spaces - continuous maps' isomorphism, because the inverse is not necessarily continuous, which is the reason why we need to specifically prove this proposition.

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4: Proof

Whole Strategy: Step 1: define the inverse, $f^{-1}:V_2\to V_1$; Step 2: see that $f^{-1}$ is Lie algebra homomorphic; Step 3: conclude the proposition.

Step 1:

As $f$ is bijective, there is the inverse, $f^{-1}:V_2\to V_1$.

Step 2:

Let us see that $f^{-1}$ is Lie algebra homomorphic.

$f$ is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any bijective linear map is a 'vectors spaces - linear morphisms' isomorphism.

Let $v_1,v_2\in V_2$ be any. $f^{-1}([v_1,v_2])=[f^{-1}(v_1),f^{-1}(v_2)]$? $f\circ f^{-1}([v_1,v_2])=[v_1,v_2]$. $f([f^{-1}(v_1),f^{-1}(v_2)])=[f\circ f^{-1}(v_1),f\circ f^{-1}(v_2)]=[v_1,v_2]$. As $f$ is injective, that implies that $f^{-1}([v_1,v_2])=[f^{-1}(v_1),f^{-1}(v_2)]$, so, yes.

Step 3:

So, $f$ is a 'Lie algebras - homomorphisms' isomorphism.

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References

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