in my functional analysis class right now we are studying the basics of C* Algebras and I was recently asked this question about the second isomorphism theorem for C* Algebras, but first let me cite the definition of a C* Algebra ismorphism:
Abstract characterization of C* Algebras
Now we are asked this:
Let A be a C* Algebra and $ B \subset A $ a C* subalgebra of A and I an ideal. We are also reminded of this result: We know that B+I is a C* subalgebra of A. We are asked the following:
We are to show the following C* isomorphism: $ B/(B \cap I) \cong (B+I)/I $
I am stuck in part b as I do not really know how to show a bijective bounded (continuous) linear map which preserves multiplication and the * operation. I really need someone out there to show me how to find and prove a C*-isomorphism as desired. Also here is a related post here: sum of C*-subalgebra and ideal
Define $\phi:(B+I)/I\to B/B\cap I$ by $$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$ Of course we need to check that this is well-defined. If $b_1+j_1=b_2+j_2$, then $$ b_1-b_2=j_2-j_1\in B\cap I, $$ so $b_1+B\cap I=b_2+B\cap I$. The map is obviously linear, multiplicative, $*$-preserving, and onto.
As for injectivity, if $b_1+B\cap I=b_2+B\cap I$, then $j=b_1-b_2\in B\cap I$; so $$b_1+0+I=b_2+j+I.$$
Finally, $\phi$ is bounded because every $*$-homomorphism between C$^*$-algebras is. In fact, because it is a $*$-isomorphism between C$^*$-algebras, it is isometric.