Why must trivial extension of C*-algebra be split Short Exact Sequence?

Question

Background: Suppose $0 \to B \to E -q-> A \to 0$ is a short exact sequence of C*-algebras. Since B sits as an ideal of E, there is a natural *-homomorphism from E to the multiplier algebra of B (by the universal property of B) which extends the identity map on B. Composing this map with the quotient by B map, we obtain the Busby map: $A \to M(B)/B$

We say the extension is trivial if the Busby map $A \to M(B)/B$ lifts to a *-homomorphism $A \to M(B)$.

I am wondering why does this imply that there is a *-homomorphism section $s: A \to E$ such that $q \circ s= id_A$?

Answer 1

After some struggling, I figure out that the answer is actually quite straightforward. Here is the key observation that is used:

For $B$ an ideal of $E$, the natural *-homomorphism $\sigma: E \to M(B)$ is multiplicative and extends the identity map on $B$. Hence for any $a \in E, b \in B$, we have $\sigma(a)b = \sigma(a)\sigma(b) =\sigma(ab)=ab$ because $ab \in aB \subset B$ ($B$ is an ideal in $E$).

This fact is used to prove the section map constructed is actually multiplicative.