Proposition 6

Let $f$ be a homomorphism from a ring A to a ring B and S a multiplicative subset of A. There exists a unique mapping j from $(f(S))^{-1}B$ to $S^{-1}B$ (where B is considered as an A-module by means of $f$) such that $j(b|f(s)) = b|s$ for all $b \in B, s \in S$. If $f' : S^{-1}A \to (f(S))^{-1}B$ is the ring homomorphism associated with $f$ (no. 1, Proposition 2), then $j \circ f' = S^{-1}f$. The mapping j is an isomorphism of the $S^{-1}A$-module structure on $(f(S))^{-1}B$ defined by $f'$ onto that on $S^{-1}B$ and also of the B-module structure on $(f(S))^{-1}B$ onto that on $S^{-1}B$ (resulting from the definition $S^{-1}B = (S^{-1}A) \otimes_A B$).

If $b, b'$ are in B, $s, s'$ in S, the conditions $b|s = b'|s'$ and $b|f(s) = b'|f(s')$ are equivalent, as follows from Corollary 1 to Proposition 4, which establishes the existence of j and shows that j is bijective; the uniqueness of j is obvious. Clearly j is an additive group isomorphism. If $a \in A, b \in B, s \in S, t \in S$, then

from which its follows that j is $(S^{-1}A)$-linear. Clearly $j \circ f^1 = S^{-1}f$. Finally, if $b \in B, b' \in B, s \in S$, then $j(b.(b'|f(s))) = j(bb'|f(s)) = bb'|s = b.(b'|s)$, which proves the last assertion.

The mapping j of Proposition 6 is called the canonical isomorphism of $(fS)^{-1}B$ onto $S^{-1}B$. These two sets are in general identified by means of $f'$; then $f' = S^{-1}f, i_B^S = i_B^{f(S)}$.