1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

(jair2018) #1
306 10. Summary

the proof of Theorem 5.1.10). Since (B Q9 A) ><Irr~ B Q9 (A ><Ir I') in this
case (Exercise 4.1.3), we have a commutative diagram
O ~ (J@A) ><1rI' ~ (B@A) ><lrI' ~ ((B/J)@A) ><lrI') ~^0

~i ~i ~i
0 ~ J@(A><lrI') ~ B@(A><lrI') ~ (B/J)@(A><lrI') ~ 0.
Since the top row is exact, by Theorem 5.1.10, so is the bottom row. D

Free products. As in the nuclear case, full free products of exact C* -
algebras are almost never exact. But reduced free products behave well.
Theorem 10.2.10. The reduced free product {with arbitrary amalgamation)
of exact C*-algebras is exact {Corollary 4.8.3).

10.3. Quasidiagonal C*-algebras

Subalgebras. Quasidiagonality obviously passes to subalgebras.

Extensions. Extensions of QD C*-algebras need not be QD; the Toeplitz
algebra is an elementary counterexample. Even for extensions which have
a *-homomorphic splitting, it isn't known if quasidiagonality is preserved.
However, there are some useful partial results. First we need a simple fact
about multiplier algebras.
Proposition 10.3.1. Let I be residually finite-dimensional. Then the mul-
tiplier algebra M(I) is also RFD.

Proof. Let 1T'n: I-+ An be surjective *-homomorphisms such that each An
is unital and QD (e.g., finite-dimensional) and the map EB7rn is faithful.
Extend to *-homomorphisms ii'n: M(I)-+ An.^2
In general, if J <1 A is an essential ideal and we have a *-homomorphism
p: A-+ B with the property that pJJ is injective, then p must be injective
on all of A (since any nonzero ideal intersects J). Hence we see that the
*-homomorphism

is also injective. D

Proposition 10.3.2. Assume 0 -+ I -+ A ~ B -+ 0 is exact, where I is
RFD and B is QD. Then A is QD.

(^2) 0ne could quote the noncommutative Tietze extension theorem, but here it isn't necessary.
For each n let en EI be a lift of the unit of An, and simply define iTn(x) = 1rn(enxen) for all
x E M(J). This map might appear to be just u.c.p. but multiplicative domains suggest something
more.

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