108 3. Tensor Products
What one must check is that this isomorphism takes
a® (bn)n E A ® (fJ Bn) C IB\ ( K ® ( E9 'Hn))
n nto the element
(a® bn)n E II A® Bn C IB\ ( Ejj(K ® 'Hn)).
n nThus we have constructed a *-homomorphism A® (I1n Bn) ___,. ITn A® Bn,
but it won't be surjective if A is infinite dimensional.
However, everything is fine for E. Indeed, if we fix an algebraic basis
{x1, ... , Xm} C E, then an arbitrary sequence (dn)n E ITn E ® Bn has a
unique representation as
m m
(dn)n = (L Xj ® bj,n)n = L(Xj ® bj,n)n,
j=l j=lwhere we used the fact that (bj,n) is a bounded sequence, for every j. But
each of the sequences (xj ® bj,n)n comes from E ® (I1n Bn), so the proof is
complete. D
With the previous identification in hand, the following is immediate from
the definition of ®-exactness.
Lemma 3.9.5. If E is a finite-dimensional operator system with algebraic
basis { x1, ... , Xm} and Bn are unital C* -algebras, then there is a natural
contractive linear mapping
ITn(E®Bn) --tE® (I1nBn)
EBn(E ® Bn) EBn Bn
such that the image of an element (L,"J!= 1 Xj ® bj,n)n E I1n(E ® Bn) under
the mapping
is
If E is ®-exact, then this map is an isometric isomorphism.
In the proposition below, both A and the Hilbert space Hare separable.