Proposition 11

Let

be a diagram of B-modules and B-linear mappings, and let

be the corresponding diagram of A-modules. Then $(\mathscr{E})$ is an exact sequence if and only if $(P\otimes \mathscr{E})$ is an exact sequence.

Suppose that the sequence $(\mathscr{E})$ is exact. Since the right B-module P is projective, the sequence $(P\otimes \mathscr{E})$ is exact (II, §3, No. 6, p. 251, Proposition 5 and II, §3, No. 7, p. 257, Corollary 6).

Conversely, suppose that the sequence $(P\otimes \mathscr{E})$ is exact. Let Q be a $(B,A)_k$-bimodule, inverse to P, and $\theta : Q\otimes_AP\rightarrow_sB_d$ an isomorphism. Consider the commutative diagram

$Q\otimes_AP\otimes_B{V'}^{1_Q\otimes 1_P\otimes f}/$/ $Q\otimes_AP\otimes_BV^{1_Q\otimes 1_P\otimes g}/$/ $Q\otimes_AP\otimes_BV''$

$\theta \otimes 1_{V'}\theta \otimes 1_V\theta \otimes 1_{V''}$

$V'f$ // V $g$ // $V''$.

Since Q is a projective A-module and the sequence $(P\otimes \mathscr{E})$ is exact, the first line of this diagram is an exact sequence. Since the vertical arrows are isomorphisms, the second line is also exact.