Let $A$ be a $C^*$-algebra.
Let $\pi: A \rightarrow A/I$ be the canonical *- homomorphism, where $I$ is a closed ideal of A
Show that if $k$ is self-adjoint in $A/I$, then there exists a self-adjoint element $a$ in $A$ such that $\pi(a)=k$.
We have that $k= h + I$ for some $h \in A$. Now we just have to show that $h$ is self-adjoint. Since $k$ is self-adjoint, $(h+I)^*= h+I = h^* +I$. Hence, $h=h^*$.
Could someone let me know if my proof is correct? I feel like I didn't really use the *- homomorphism here.
If not, how should I proceed?
Thank you!
The proof you gave is not correct because having $h+I=h^*+I $ does not imply $h=h^*$. For instance, for any $h$ in $I$, you have $h+I=I$.
Instead you could proceed like this: Given a self-adjoint element $k$ in $A/I$, use surjectivity of $\pi$ to produce $x$ in $A$ with $\pi (x)=k$. Then let $h=(x+x^*)/2$ and check that $h$ is self-adjoint. Because $\pi$ is $*$-preserving, you have $\pi(h)=(k+k^*)/2=k$.