174 5. Exact Groups and Related TopicsProof. Assume first that r is KW-exact and let 0 -----+ I -----+ A -----+ A/ I -----+ 0
be an arbitrary short exact sequence. We must show that
0-----+l®CHI')-----+A®C~(r)-----+ (A/I) ®C~(I')-----+O
is also exact (Theorem 3.9.1). This, however, is trivial because 0 -----+ I -----+
A-----+ A/I-----+ 0 can be thought of as a short exact sequence of r-algebras by
equipping A with the trivial action (and recalling that A® C~(r) ~A ><Irr
in this case - see Exercise 4.1.2).
For the converse, assume that r is exact and let 0 -----+ I -----+ A -----+ A/ I -----+
0 be an arbitrary short exact sequence of r-algebras. To prove that the
sequence of reduced crossed products is exact, we will exploit the fact that
full crossed products preserve exactness.
By Fell's absorption principle (Proposition 4.1.7), there is an embedding
7rA: A ><Irr'--> (A ><1 I')® C~(r) such that
7rA(aAs) =as® As,
for all a E A and s E r (just apply Fell's principle to any faithful represen-
tation of the full crossed product). One easily checks that
o _______, I ><1 r ® c~(r) _______, A)<! r ® c~(r) _______, (A/I))<! r ® c~(r) _______, o~Ir ~Ar
0 ------+ A ><1rI'~A/IrA/I ><Irr ------+^0
is a commutative diagram. Note that the top row is exact since we assumed
r to be exact.
It turns out that there is also a c.c.p. map <PA: (A ><1 r)®C~(r)-----+ A ><lrI'
such that <PA o 7rA = idAXlrI' and we have a commutative diagram
0 ------+ I)<! r ® C~(r) ------+ A ><Ir® C~(r) ------+ (A/I) ><Ir® C~(r) ------+ 0Wr l WA l
0 ------+ ------+ 0.We will construct <PA below, but first one should chase through the two
diagrams above to see that the bottom row is also exact; hence the proof is
complete once we know that <PA exists.
The construction of <PA, though elementary, is somewhat tedious. It is
sufficient to find a c.c.p. map (A><1rI')®CHr)-----+ A><lrI' such that aA 8 ®Ag 1--+
0 ifs "I-g and aA 8 ®Ag 1--+ aA 8 ifs = g. Indeed, if successful, then composing
with the canonical quotient map (A ><1 r) ® C~(r)-----+ (A ><Irr)® C~(r) will
give the desired map <PA.
Starting from a faithful representation A c JIB(H), we represent A ><Irr
faithfully on 1i ® f^2 (r) via the induced regular representation (Definition