Inverse image of invertible element under surjective C*-homomorphism

Question

Suppose that $A$ and $B$ are unital C*-algebras and that $\varphi:A \to B$ is a surjective *-homomorphism. If $b \in B$ is invertible, must there be some invertible $a \in A$ with $\varphi(a)=b$?

I wondered about this problem and it has been bothering me ever since. I think that using functional calculus I was able to show that it does if $b$ has a normal pre-image, but this can only happen if $b$ is normal. I have been unable to generalize it to arbitrary $b$, which leads me to believe that in general it will be false. However, I have been unable to think of a counterexample.

Answer 1

This is false in general, even if $b$ is normal. Consider $A=C(\mathbb D)$, $B=C(\mathbb T)$, and $\varphi$ the restriction map. Take $b$ to be the identity function. If $\varphi(a)=b$, then $a:\mathbb D\to \mathbb C$ is such that $a|_{\mathbb T}$ is the identity. Consider $g(z)=z$ on $\mathbb D$ and $g(z)=z/|z|$ for $|z|\geq1$. Then $\varphi(g(a))=g(\varphi(a))=g(b)=b$, that is $g\circ a:\mathbb D\to\mathbb D$ is continuous and is the identity on the circle. By topological considerations, such $g\circ a$ has $0$ in its range. This implies that $a$ has $0$ in its range (for if $g(a(z))=0$ then $a(z)=0$), so $a$ is not invertible.