1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

94 3. Tensor Products

In the previous section we saw that for certain inclusions A c B, we

have inclusions A @max C C B @max C for every C. Our next result is in this

same spirit, but we should first recall some terminology.

Definition 3. 7 .5. If A is unital, an extension 0 -----> J -----> A ~ A/ J -----> 0 is

called locally split if for each finite-dimensional operator system E c A/ J

there exists a u.c.p. map O": E-----> A such that 1f o O" = idE.^19

Proposition 3. 7.6.^20 Assume that 0 -----> J -----> A ~ A/ J -----> 0 is locally

split. Then for every B the sequence

0-----> J Q9 B-----> A Q9 B-----> (A/ J) Q9 B-----> 0

is exact.

Proof. We know that

A@B

J Q9 B ~ (A/ J) @a B

for some C*-norm and our job is to show that II· Ila is the spatial norm. By

Takesaki's Theorem (Theorem 3.4.8) it suffices to show that if y E (A/ J)GB,

then

llYlla '.S llYllmin·

Since y E (A/ J) 8 B is a finite sum of elementary tensors, we can find

a finite-dimensional operator system X C A/ J such that y E X 8 B C

(A/ J) 8 B. By assumption, we can find a u.c.p. map e: X-----> A which lifts

X and this induces a u.c.p. (hence contractive) map

e Q9 idB: X Q9 B-----> A Q9 B.

The remainder of the proof is contained in the following diagram:

(A/J) Q9 B

isometric r inclusion

X Q9 B ~ A Q9 B -----+ 1~~ ~ (A/ J) @a B.

Indeed, both maps on the bottom are contractions and since e was a lifting,
it follows that the element y E X 8 B C X Q9 B gets mapped to the element
YE X 8 BC (A/ J) @a B. Thus llYlla :S llYllmin as desired. D

Now let us reverse the roles and ask if some property of B will ensure
that 0 -----> J Q9 B -----> A Q9 B -----> (A/ J) Q9 B -----> 0 is exact for all choices of A
and ideals J <I A. We'll need a lemma.

(^19) The nonunital definition replaces operator systems with operator spaces and u.c.p. maps
with c.c. maps.
(^20) This is the easy half of the Effros-Haagerup lifting theorem; see Theorem C.4.