description/proof of that bijective Lie algebra homomorphism is 'Lie algebras - homomorphisms' isomorphism
Topics
About: Lie algebra
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
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.
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.
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.
Main Body
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.
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.
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.
