Question

(1) Let L, L' be Lie Algebra over C. Prove that: (a) If \(\phi: L \to L'\) is a homomorphism of Lie Algebra then \(L/ker\phi \cong Im\phi\). (b) If I is any ideal of L included in \(ker\phi\) there exists a unique homomorphism. \(\psi: L/I \to L'\) making the following diagram commute (\(\pi = \) canonical map) \(\begin{array}{ccc} L & \stackrel{\phi}{\to} & L' \\ \downarrow{\pi} & & \uparrow{\psi} \\ L/I & & \end{array}\)

          (1) Let L, L' be Lie Algebra over C. Prove that:
(a) If \(\phi: L \to L'\) is a homomorphism of Lie Algebra then \(L/ker\phi \cong Im\phi\).
(b) If I is any ideal of L included in \(ker\phi\) there exists a unique homomorphism.
\(\psi: L/I \to L'\) making the following diagram commute (\(\pi = \) canonical map)

\(\begin{array}{ccc}
L & \stackrel{\phi}{\to} & L' \\
\downarrow{\pi} & & \uparrow{\psi} \\
L/I & & 
\end{array}\)

(1) Let L, L' be Lie Algebra over C. Prove that:
(a) If ϕ: L → L' is a homomorphism of Lie Algebra then L/kerϕ≅ Imϕ.
(b) If I is any ideal of L included in kerϕ there exists a unique homomorphism.
ψ: L/I → L' making the following diagram commute (π = canonical map)

L ϕ→ L'
↓π ↑ψ

L/I

Added by Luc-A Q.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

08/17/2023

Step 1

Step 1: Recall that the kernel of a homomorphism φ: L → L' is defined as ker(φ) = {x in L | φ(x) = 0}. Show more…

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1) Let L, L' be Lie algebras over C. Prove that: (a) If φ: L → L' is a homomorphism of Lie algebras, then L/ker(φ) is isomorphic to Im(φ). (b) If I is any ideal of L included in ker(φ), then there exists a unique homomorphism π: L/I → L' making the following diagram commute (canonical map): L → L' T L/I