Lemma 1

Let $f$ and $g$ be homomorphisms from B to A. The following conditions are equivalent:

(i) The $(B,A)$-bimodules $A^f$ and $A^g$ are isomorphic.

(ii) There exists an inner automorphism (I, §8, No. 4, p. 102, Example 2) $\theta$ of A such that $g=\theta \circ f$.

The automorphisms of the right A-module $A_d$ are the mappings $x\mapsto ax$, where $a$ is an invertible element of A. Such an automorphism is a B-linear mapping from $A^f$ to $A^g$ if and only if we have

for every $x$ in A and every $b$ in B. This relation is equivalent to $g(b) =af(b)a^{-1}$ for every $b$ in B, that is, to $g=\theta \circ f$, where $\theta$ is the inner automorphism $x\mapsto axa^{-1}$ of A.