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 $q \circ s= id_A$?


1 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .