This is a follow up question on About unitary element in a $C^*$-algebra
Let $A$ be a $C^*$-algebra. Let $\pi: A \rightarrow A/I$ be the canonical *- homomorphism, where $I$ is a closed ideal of A Let $u \in A/ I$ be unitary and $\sigma(u) = \{ \lambda \in \mathbb{T}: Re(\lambda) \geq 0\}$, where $\mathbb{T}$ is the unit circle. Show that there exists some unitary $x \in A$ such that $\pi(x) =u$.
For any general unitary $u$, if $\pi: B(H) \rightarrow B(H)/K(H)$, can we still get a unitary $x$ such that $\pi(x) =u$? ''
I tried to do it following Mike's hints.
I know that by Theorem 2.12 in Murphy, since $\sigma(u) \ne \mathbb{T}$, there exists a self-adjoint $a$ such that $e^{ia}$. Since the $\pi$ is surjective, there exists $x \in A$ such that $\pi(x) = e^{ia}$. Now below is that part I'm not sure about.
Since $a$ is self-adjoint, we have that there exists some self-adjoint $k \in A $ such that $\pi(e^{ik}) = e^{ia}$. So we've found a unitary $e^{ik}$.
Could someone please let me know if the above claim is correct? If so, how is that true? If not, how do I continue with this proof?
For the second part, I know that by Atkinson's Theorem of $\pi$ we have that $\pi(u)$ is invertible. By a Fredholm argument, we have that $\pi(u) \ne \pi(w)$ for any invertible $w$. From here I'm not sure how to continue. I might be completely off...
Thank you for your attention!
Once you have that $u=e^{ia}$, find some $k\in A$ such that $\pi(k)=a$. Then it follows from continuity of $\pi$ that $\pi(e^{ik})=u$.
The issue with your second question is indeed an issue of Fredholm index (or more generally, an issue arising from K-theory). If $y\in B(H)$ is a Fredholm operator with index $\neq 0$, and $u=\pi(y)$ is a unitary, then any lift of $u$ is a Fredholm operator with the same index as $y$, hence cannot be a unitary, as all unitary (invertible) operators have index $0$.