Let $R$ be a commutative ring with unity and $S$ be a submonoid of $(R,\cdot)$ and $M,N$ be $R$-modules. Let $\phi:M\rightarrow N$ be an $R$-module homomorphism. Assume that for any $d\in S$, the left multiplication map $N\rightarrow N:x\mapsto dx$ is a bijection.
How do I prove that there exists a unique $R$-module homomorphism $\Phi:S^{-1}M\rightarrow N$ such that $\Phi(\frac{x}{1})=\phi(x)$?
To construct such $\Phi$, we must try to figure out how $\Phi$ looks like.
By the assumption, we can prove that if $\frac{x}{s}=\frac{x'}{s'}$ then $s\phi(x')=s'\phi(x)$. So I tried to use this to figure out how $\Phi$ looks like, but I am stuck. How do I prove the existence of such $\Phi$?
Best Answer
There is only one possible way to define $\Phi$. Namely, since $\Phi(x/1)=\phi(x)$, $\Phi(x/s)$ must be some element $y\in N$ such that $sy=\phi(x)$. Since multiplication by $s$ is a bijection on $N$, there exists exactly one such $y$.
So define $\Phi(x/s)$ to be the unique element $y\in N$ such that $sy=\phi(x)$. You then just have to check that this is well-defined and a homomorphism. These are straightforward verifications. For instance, here is how you show it is well-defined. If $x/s=x'/s'$, then there exists $t\in S$ such that $ts'x=tsx'$. Let $y\in N$ be such that $sy=\phi(x)$ and $y'\in N$ be such that $s'y'=\phi(x')$. We then have $$tss'y=ts'\phi(x)=\phi(ts'x)=\phi(tsx')=ts\phi(x')=tss'y'.$$ Since multiplication by $tss'$ is a bijection on $N$, this implies $y=y'$.