Let $A$ be a $C^*$algebra and $I$ be an ideal. Suppose that $I$ admits an approximate identity consisting of projections in $I$. Let $u \in U(A/I)$. Then is there a partial isometry $v \in A$ such that $\pi(v)= u$, where $\pi$ is the quotient map?
I think if we use perturbation we can find such partial isometry but I couldn't. Could anybody help me out?
Given $u$ in $U(A/I)$, choose any $w$ in $A$ such that $\pi (w)=u$. One then has that $$ \pi (w^*w-1)=0, $$ so $w^*w-1\in I$. Given any $\varepsilon >0$, we then have by hypothesis that there exists a projection $p$ in $I$ (chosen e.g. from an approximate identity) such that $$ \varepsilon >\|(w^*w-1) - p(w^*w-1)\| = $$ $$ = \|(1-p)(w^*w-1)\| = \|p^\perp (w^*w-1)\|, $$ where $p^\perp = 1-p$. Letting $y=wp^\perp $, observe that $$ \pi (y) = \pi (wp^\perp ) = \pi (w-wp)=\pi (w) = u, $$ and moreover that $$ \|y^*y-p^\perp \| = \|p^\perp w^*wp^\perp -p^\perp \| = $$ $$ = \|p^\perp (w^*w-1)p^\perp \| \leq \|p^\perp (w^*w-1)\|\|p^\perp \| <\varepsilon . $$
Since $p^\perp$ is idempotent, anything nearby is approximately idempotent so, by choosing $\varepsilon $ sufficiently small, we may assume that $$ \|(y^*y)^2 - y^*y\| < \frac14, $$ which in turn implies that $\frac12$ cannot be in the spectrum of $y^*y$.
Let $$ f:\mathbb R \setminus \left\{\textstyle \frac12\right\} \to \mathbb R $$ be the continuous function vanishing identically on $(-\infty ,1/2)$, and such that $f(t)= t^{-1/2}$, for $t\in (1/2, \infty )$. Clearly $$ tf(t)^2 = 1_{(\frac12,\infty )}, $$ which is a real idempotent function, hence $$ f(y^*y)\ y^*y \ f(y^*y) $$ is a self-adjoint idempotent element, i.e. a projection. Therefore $$ v:= y f(y^*y) $$ is a partial isometry thanks to the fact that $v^*v$ is a projection. To conclude we claim that $\pi (v)=u$, and this is easy to see since $$ \pi (v) = \pi (y) \pi (f(y^*y)) = \pi (y) f(\pi (y^*y)) = $$ $$ = u f(u^*u) = u f(1) = u. $$