C*-algebra generated by a non-invertible element

Question

Let $x$ be a non-invertible element, and put $A:=C^*(x)$. Let $I$ be a closed ideal of $A$, and $\pi: A\to A/I$ be a natural quotient map. Is it possible that there is an invertible element $y\in A$ such that $\pi(x)=\pi(y)$?

Answer 1

For this question to make sense, you need to take $C^*(x) $ to be the unital C$^*$-algebra generated by $x $ (otherwise, "invertible" makes no sense).

1. Let $A=\mathbb C\oplus\mathbb C $, $x=(1,0) $, $y=(1,1) $, $I=\{(0,t):\ t\in\mathbb C \} $. Then $x-y=(0,1)\in I $, i.e, $\pi (x)=\pi (y) $.

2. Now let $A=C[0,1]$, $I=\{f:\ f(0)=0\}$, $x$ the function given by $x(t)=t$. For any $y\in C[0,1]$ is invertible, there exists $\delta>0$ with $|y(t)|\geq\delta$ for all $t$. In particular, $$ |(x-y)(0)|=|x(0)-y(0)|=|y(0)|\geq\delta, $$ so $x-y\not\in I$, i.e. $\pi(x)\ne\pi(y)$.

The first example shows that an invertible element with $\pi(y)=\pi(x)$ will exist when the ideal $I$ is complemented. But I don't know if this condition is necessary.