If f is surjective then f is not a right divisor of zero Let $R$ be a ring and $M$ a R-module. For $r\in R$ define $f:M\to M$ by $f(s)=sr$.
Show that $f$ is injective if and only if $r$ is not a right zero divisor.
I have done a similar problem to this in the past, involving surjectivity instead of injectivity and came up with:
Suppose f is surjective and gf=0. To prove f is not a right divisor of zero, we need to show that g=0, i.e. that g(m)=0 for all m. So let m be in M. Since f is surjective, m=f(n) for some n in M. Thus g(m)=g(f(n))=(gf)(n)=0(n)=0.
However I am stuck with how to show this problem using injectivity, and was wondering could anybody provide guidance on this? 

$f$ is injective if and only if 
$f(s)=0  \Rightarrow s=0$ if and only if
$sr=0  \Rightarrow s=0$  if and only if
$r$ is not a right zero divisor.
