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
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.