# 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 $$\pi(x) =u$$. 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!

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.
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 $$u=u^{-1}$$?