Localization of Module Homomorphism is Multiplicative

Theorem
Let $A$ be a commutative ring with unity.

Let $S \subseteq A$ be a multiplicatively closed subset.

Let $S^{-1}A$ be the localization of $A$ at $S$.

Let $f_1$ and $f_2$ be $A$-homomorphisms:
 * $M_1 \stackrel {f_1} \longrightarrow M_2 \stackrel {f_2} \longrightarrow M_3$

For $i=1,2,3$, let $\struct { S^{-1}M_i, \iota_i}$ be the localization of $M_i$ at $S$.

Let $S^{-1}f_1$ and $S^{-1}f_i$ be the unique $S^{-1}A$-homomorphisms:


 * $S^{-1}M_1 \stackrel {S^{-1}f_1} \longrightarrow S^{-1}M_2 \stackrel {S^{-1}f_2} \longrightarrow S^{-1}M_3$

such that:
 * $\iota_{i+1} \circ f_i = \paren {S^{-1}f_i} \circ \iota_i$

Let:
 * $S^{-1}M_1 \stackrel {S^{-1}\paren {f_2 \circ f_1}} \longrightarrow S^{-1}M_3$

be the unique $S^{-1}A$-homomorphism such that:
 * $\iota_3 \circ \paren {f_2 \circ f_1} = \paren {S^{-1} \paren {f_2 \circ f_1} } \circ \iota_1$

Then we have:
 * $S^{-1} \paren {f_2 \circ f_1} = \paren {S^{-1} f_2} \circ \paren {S^{-1} f_1}$

Proof
Let $g : S^{-1}M_1 \to S^{-1}M_3$ be defined by:
 * $g:= \paren {S^{-1} f_2} \circ \paren {S^{-1} f_1}$

Then:

Thus, by, we have:
 * $g = S^{-1} \paren {f_2 \circ f_1}$