Localization of Module Homomorphism is Multiplicative

Theorem

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:

$S^{-1} \paren {f_2 \circ f_1} = \paren {S^{-1} f_2} \circ \paren {S^{-1} f_1}$

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:

$\ds g \circ \iota_1$

$=$

$\ds \paren {S^{-1}f_2} \circ \paren {S^{-1}f_1 } \circ \iota_1$

$\ds $

$=$

$\ds \paren {S^{-1}f_2} \circ \iota_2 \circ f_1$

Definition of $S^{-1} f_1$

$\ds $

$=$

$\ds \iota_3 \circ \paren {f_2 \circ f_1}$

Definition of $S^{-1} f_2$

Thus, by Definition of $S^{-1} \paren {f_2 \circ f_1}$, we have:

$g = S^{-1} \paren {f_2 \circ f_1}$

$\blacksquare$