Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\cal A$ be a Banach algebra and $I$ be a closed ideal on it. Let $\phi: \cal A/I\to \cal A$, $\phi(a+I)=a$. Is $\phi$ well defined?

if $a+I=b+I$ then $a-b\in I$, so $\phi(a-b+I)=\phi(I)=0$. Since $\phi(a-b+I)=a-b$, $a-b=0$ and so $a=b$ conclude that $\phi(a+I)=\phi(b+I)$. Thus $\phi$ is well defined.

But let $x\neq0$ and $x\in I$. Since $\phi(x+I)=\phi(I)=0$ and $\phi(x+I)=x$, $x=0$. But we let $x\neq0$, so $\phi$ is not well defined. I'm confused...

Thanks

share|improve this question

migrated from mathoverflow.net Nov 10 '13 at 7:43

This question came from our site for professional mathematicians.

    
Banach or not, your $\phi$ can only be well-defined if $I=0$... –  darij grinberg Nov 10 '13 at 5:24
    
To elaborate on darij's point: in your "proof" of well-definedness, you assume your conclusion: when you go from '$\phi(a-b+I)=a-b$, $a-b+I=I$, and $\phi(I)=0$' to '$a-b=0$,' the only way to conclude this is by already assuming that $\phi$ is well-defined. –  Noah S Nov 10 '13 at 7:16

1 Answer 1

Note that, \phi(a-b+I)=0 implies that a-b belongs to I not a-b=0. Because any element in the quotient algebra is an equivalence class.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.