Csdrhrt

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!

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.