Quotients and Unit Balls

Let $A$ be an C*-algebra, $I$ an closed ideal of $A$ and $\pi :A\to A/I$ the canonical projection. Then we have $\pi \left(A_1\right)={\pi \left(A\right)}_1$ ($\subset$ is obvious) as follows.

First we cosider the unital case. Let $x$ be an element whose image under $\pi$ is in the unit ball, i.e. ${\pi \left(x\right)}^{*}\pi \left(x\right)\le 1$. Consider a function $$f\left(x\right)=\left\{\begin{array}{cc}1& \left(1<t\right)\\ t& \left(-1\le t\le 1\right)\\ 1& \left(t<-1\right)\end{array}\right.$$ on the real line. Then we have $f\left(x^{*}x\right)\le 1$ and $\pi \left(f\right(x^{*}x\left)\right)=f\left(\pi \right(x^{*}x\left)\right)={\pi \left(x\right)}^{*}\pi \left(x\right)$. Put $k=x^{*}x-f\left(x^{*}x\right)$. This is in the kernel of $\pi$ and so is its positive part $k_{+}$ ($ker\pi$ is a sub C*-algebra of $A$). We have $x^{*}x=f\left(x^{*}x\right)+k\le 1+k_{+}$. $x{(1+k_{+})}^{-\frac{1}{2}}$ is in the unit ball and mapped to $\pi \left(x\right)$.

In the non unital case, consider the unitization $\mathrm{uA}$ of $A$. Then $I$ is identified to a closed ideal of $\mathrm{uA}$ and we have $u(A/I)\simeq \mathrm{uA}/I$. Any element of ${(A/I)}_1$ ($\subset \mathrm{uA}/I$) can be lifted up to ${\mathrm{uA}}_1$, and the lift must be in $A$, since $I$ is a subgroup of $A$.