3.1. Algebraic tensor products 63(i.e., cpj(xi) = Oi,j) and let ·'lj; E Y* be arbitrary. Then for each 1 :S j :S n
we have
0 = !.pj 0 'lj; ( t Xi ®Yi) = 'lj;(yj ).
i=l
Since this holds for all 'lj;, it follows that each Yj is the zero vector.
Corollary 3.1.9 (Bases). If {xi}iEI C X and {yj}jEJ CY are bases, then
{Xi® Yj }(i,j)EJxJ
is a basis of X 0 Y. In particular, dim(X 0 Y) = dim(X)dim(Y).Since the set of elementary tensors spans X 0 Y, it is easy to see that
{xi® Yj}ci,j)EixJ also spans. That they are linearly independent follows
from Proposition 3.1.8 together with an exercise in "collecting like terms".^3
Though bases are important, it turns out that the following fact is often
more convenient. The proof uses the previous two results.
Corollary 3.1.10 (Unique representation). If {xihEI C Xis a basis, then
for each vector v EX 0 Y there is a unique set {YihEI CY such that
v = LXi ®Yi·^4
iEJ
We now consider inclusions and exact sequences of algebraic tensor prod-
ucts. The results are easy, but the corresponding statements in the C* -
setting are false.
Proposition 3.1.11 (Inclusions). If W C Y and X C Z are subspaces,
then there is a natural inclusion
W0XcY0Z.
Understanding the ambiguity of the previous proposition is 90 percent
of the proof. The "natural inclusion" is just the mapping
iw0ix: W0X-+ Y0Zinduced by the identity inclusions iw: W '--+ Y and ix: X <-+ Z. Hence we
are asserting that iw 0 ix is injective -which is a special case of something
more general.
Proposition 3.1.12 (Injectivity of tensor product maps). If cp: W ---+ Y
and 'lj;: X ---+ Z are injective linear maps, then
cp0'1j;: W0X-+Y0Z
is also injective.
(^3) 2=i,j Ai,j(Xi 181Yj)=2=i Xi 181 (2=j Ai,jYj)·
4of course, only finitely rnany yi's are nonzero.