Proposition 2

Let $A, B$ be two rings, $S$ a subset of $A$, $T$ a subset of $B$ and $f$ a homomorphism from $A$ to $B$ such that $f(S) \subset T$. There exists a unique homomorphism $f'$ from $A[S^{-1}]$ to $B[T^{-1}]$ such that $f'(a/1) = f(a)/1$ for all $a \in A$.

Suppose further that $T$ is contained in the multiplicative subset of $B$ generated by $f(S)$. Then, iff is surjective (resp. injective) so is $f'$.

The first assertion amounts to saying that there exists a unique homomorphism $f': A[S^{-1}] \to B[T^{-1}]$ giving a commutative diagram:

Now the relation $f(S) \subset T$ implies that $i_B^T(f(s))$ is invertible in $B[T^{-1}]$ for all $s \in S$ and it is sufficient to apply Proposition 1 to $i_B^T \circ f$. It follows easily from (2) that, for all $a \in A$ and $s \in \overline{S}$ (multiplicative subset of $A$ generated by $S$),

Suppose that $T$ is contained in the multiplicative subset generated by $f(S)$, which is precisely $f(\overline{S})$. Then it follows from (3) that, iff is surjective, so is $f'$.

Suppose now that $f$ is injective. Let $a/s$ be an element of the kernel off'. As the multiplicative subset generated by $T$ is $f(\overline{S})$, there is an element $s_1 \in \overline{S}$ such that $f(s_1)f(a) = 0$, whence $f(s_1a) = 0$ and consequently $s_1a = 0$ since $f$ is injective; then $a/s = 0$, which proves that $f'$ is injective.

Remark (6). If the elements of $T$ are invertible in $B$, $B[T^{-1}]$ is identified with $B$ by means of the isomorphism $i_B^T$ and $f'$ then becomes identical with the unique homomorphism $u'$ of $A[S^{-1}]$ to $B$ such that $u' \circ i_A^S = f$.