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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...


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

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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.

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