Is a Mapping Between Lie Algebras an Isomorphism if it Takes a Basis to a Basis?

Ted123
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If a mapping between Lie algebras \varphi : \mathfrak{g} \to \mathfrak{h} takes a basis in \mathfrak{g} to a basis in \mathfrak{h} is it an isomorphism of vector spaces?
 
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Good question. What do you think?
 


Office_Shredder said:
Good question. What do you think?

I'm fairly sure it is. Is that right?
 


Assuming that by "takes a basis to a basis" you mean "one to one and onto", a Lie Algebra is completely determined by its basis, isn't it?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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