Let $\phi:A\rightarrow B$ be a surjective $*$-homomorphism between unitary C*-agebras (not necessarily commutative). The question is that if for any $b\in B$ self-adjoint and invertible, there has to be some $a\in A$ self-adjoint and invertible such that $\phi(a)=b.$
I know on one side that there is $a\in A_{sa}$ s.t. $\phi(a)=b$ and also that given $a\in A$ such that $\phi(a)=b$, there is $a'\in A$ s.t. $aa'-1\in \operatorname{Ker}\phi$. I also know that $A/\operatorname{Ker}\phi\cong B.$ But all this does not suggest any way of proving it or giving a counterexample... Any help is welcome.
It's not true in general. Let $A=C[-2,2]$, $B=C([-2,-1]\cup[1,2])$ and $\phi:A\to B$ the restriction map, that is $$ \phi(f)=f|_{[-2,-1]\cup[1,2]}. $$ Then $\phi$ is a surjective $*$-homomorphism.
Let $b\in B$ be the function $b(t)=t$. Then $b$ is selfadjoint and invertible (with inverse $b^{-1}(t)=1/t$ in $B$). Now suppose that $a\in A$ is selfadjoint (so, real valued) and $\phi(a)=b$. This means that $a(t)=t$ for $t\in[-2,-1]\cup[1,2]$. As $a$ is continuous, there exists $t_0\in [-2,2]$ with $a(t_0)=0$, and so $a$ is not invertible.