What does $(B+I)/I\sim B/(B\cap I)$ tell us? Let $A$ be a $C^*$-algebra in which $B$ is a $C^*$-subalgebra and $I$ is a closed ideal. In several books on $C^*$-algebras I have encountered the following:

$(B+I)/I$ is $*$-isomorphic to $B/(B\cap I)$.

It seems important, but none of the books I read gives a hint why this is important.
So what does this isomorphism say actually?
Thanks!
 A: Here's a general algebraic interpretation that I've always found compelling.
Within a fixed algebra $A$, we're given a subalgebra $B$ and an ideal $I$ (something you can quotient by, I'm skipping the specifics). We are interestered in the quotient $B/I$, but this does not make sense in general, since we don't necessarily have $I \subset B$.
There are 2 different ways to go about this:


*

*either extend the subalgebra $B$ so that $I$ lies in this extension (the smallest such subalgebra is $B+I$)

*or restrict $I$ so that this restriction lies in $B$ (the largest such ideal is $B \cap I$).
The isomorphism $(B+I)/I \cong B/(B\cap I)$ tells us that both approaches yield the same result.
A: As azarel mentioned, the form of this 'second' isomorphism should be familiar from elementary group, ring, or module theory.
A useful application is the following:
If $I,J$ are closed ideals of the $C^*$-algebra $A$, then $I+J$ is a closed ideal as well.
(proof: the natural 'second isomorphism' map $I/(I\cap J)\to A/J$ is a $*$-(iso)morphism onto $(I+J)/J$, hence the latter is closed in $A/J$. Now $(I+J)/J$ and $J$ are Banach, hence $I+J$ as well.)
In turn, closed ideals are nice because their quotients yield complete spaces, e.g. the quotient of a C*-algebra by a closed ideal is again a C*-algebra.
