Attaining the norm of a C*-algebra quotient by an ideal The following statement is Exercise I.26 in K. Davidson's book C-Algebras by Example*:

Let $\mathfrak{A}$ be a C*-algebra, $\mathfrak{J}$ be an (two-sided
  and closed) ideal of $\mathfrak{A}$ and $A \in \mathfrak{A}$. Then
  there exists $J \in \mathfrak{J}$ such that $\|A - J\| = \|A + \mathfrak{J}\|$.

The book gives a hint of applying the Jordan decomposition (Corollary I.4.2 in the book) of the element $|A|-\|A+\mathfrak{J}\|I$. But I don't see how to proceed from here. Any help would be appreciated. Thank you!
 A: Here is a proof without using the hint.
It suffices to prove the following statement: If $\pi:A \to B$ is a surjective *-homomorphism between $C^*$-algebras and $y \in B$, then there exists $x \in A$ such that $\pi(x)=y$ and $\|x\|=\|y\|$. Also, we just need to prove the statement when $\|y\|=1$.
Special Case: $\underline{y=y^* \quad and \quad \|y\|=1}\\$
Since $\pi$ is surjective, there exists $a \in A$ such that $\pi(a)=y$. Replacing $a$ by $\frac{a+a^*}{2}$, we may assume in addition that $a=a^*$. 
Let $f: \mathbb{R} \to \mathbb{R}$ be the continuous function defined by
\begin{equation*}
f(t)=
\begin{cases}
t &\text{ , if  } \quad  |t| \leq 1 \\
\frac{t}{|t|} &\text{ , if} \quad  |t| >1
\end{cases}
\end{equation*}
Let $x=f(a)$. Then $\|x\| \leq \|f\|_\infty \leq 1$ and $\pi(x)=\pi(f(a))=f(\pi(a))=f(y)=y$. The last equation holds because $\sigma(y) \subseteq [-1,1]$ and $f(t)=t  $ for all $t \in [-1,1]$.
General Case:$\underline{\|y\|=1} \\$
Let
\begin{gather}
 Y=
\begin{pmatrix}
0 & y\\
y^* & 0 \\
\end{pmatrix}.
\end{gather}
Then $\|Y\|=\max\{\|y\|,\|y^*\| \}=1$ and $Y=Y^*$. Apply the special case to the surjective *-homomorphism  $\pi_2:M_2( A) \to M_2(B)$ given by 
\begin{gather}
\pi_2
\begin{pmatrix}
x_{11} & x_{12}\\
x_{21} & x_{22} \\
\end{pmatrix}
=
\begin{pmatrix}
\pi(x_{11}) & \pi(x_{12})\\
\pi(x_{21}) & \pi(x_{22}) \\
\end{pmatrix}. 
\end{gather}
and $Y$, there exists $X \in M_2(A)$ such that $\|X\| \leq 1$ and $\pi_2(X)=Y$. Let $x \in A$ be the $(1,2)$-entry of X. Then $\|x\| \leq \|X\| \leq 1$ and $\pi(x)=y$.
Since $\pi$ is norm-decreasing, $1=\|y\| =\|\pi(x)|\leq \|x\|$. This completes the proof.
