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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 $\pi(x) =u$. If we drop the assumption $\sigma(u) = \{ \lambda \in \mathbb{T}: Re(\lambda) \geq 0\}$, can we still get some $x$ such that $\pi(x) = u$? For example, if $\pi: B(H) \rightarrow B(H)/K(H)$, can we still get a unitary $x$ such that $\pi(x) =u$?

Since $\pi $ is onto, we know there there exists some $a \in A$ such that $\pi(a) =u = u^{-1}$.

So $uu\pi(a) = \pi(uu) \pi(a) =\pi(1)\pi(a)= uuu =u$. I'm not sure how to proceed to construct the unitary $x$ based on $a$.

Any help will be appreciated!

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Note you should assume $A$ is unital or $A$ doesn't even contain unitaries! A hint for the first part is to write $u$ as $e^{ith}$ where $h$ is a self-adjoint element of $A/I$. A hint for the second part is to continue looking at the example $A=B(H)$, $I=K(H)$ and to review anything you've learned about Fredholm operators/the Fredholm index.

If I'm honest, I don't know what you have in mind with this part of your post:

Since $\pi $ is onto, we know there there exists some $a \in A$ such that $\pi(a) =u = u^{-1}$.

So $uu\pi(a) = \pi(uu) \pi(a) =\pi(1)\pi(a)= uuu =u$. I'm not sure how to proceed to construct the unitary $x$ based on $a$.

Why are you assuming $u=u^{-1}$?

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