1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

290 9. Local Reflexivity

which is surjective (by Kaplansky's density theorem [53, Theorem I.7.3])

if we start with a large enough index set I (e.g., the directed set of finite

subsets of B*).

The key observation is that idA 0 O": A 0 B1-+ A 0 B** C (A 0 B)** is

continuous with respect to the minimal norm since

n

ll(idA 0 O")(Lak@(xk(i))iEI)ll(A®B)**

k=l

n

= llstrong*-~i~ Lak 0 xk(i)ll(A®B)**

k=l

n

:S sup II L ak 0 Xk ( i) llA®B

iEJ k=l

n

= II L ak 0 (xk(i))iEillA@Br

k=l

for every ,E~=l ak 0 (xk(i))iEI E A 0 B1. (The last equality follows from

Lemma 3.9.4.) Letting J <J B1 be the kernel of the map B1 -+ B**, it follows

that A 0 J is contained in the kernel of

A 0 B1 -+ (A 0 B)**

and thus this *-homomorphism factors through

A A@ 0 B1 J =A 0 (B I /J) =A 0 B**

since we assumed A to be exact. The last step is to verify that the map we

have constructed,

A 0 B**-+ (A 0 B)**,

agrees on elementary tensors with the canonical map A 0 B** -+ (A 0 B)**.

But this is not hard, so we consider the proof complete. 0

Though rarely needed, it is worth mentioning that property C" is a local

property in the sense that one can always reduce to the separable case.

Lemma 9.2.8. A C* -algebra A is locally reflexive if and only if every sep-

arable C* -subalgebra of A is locally reflexive.

Proof. Local reflexivity passes to subalgebras thanks to Propositions 9.2.5
and 9.2.3. Hence we must show the converse. We may assume that A is
unit al.

Let a finite-dimensional operator system E C A*, a finite-dimensional
subspace F C A and c > 0 be given. It suffices to find a u.c.p. map