Lef $f$ be a module homomorphism , $f$ is one-one if and only if $\ker f={0}$ I am following Herstein's Topics in Algebra and according to that book a module may be defined on a ring without 1 (https://i.stack.imgur.com/FMeIZ.png).
So if we consider a module $M$ over a ring $R$ without $1$, then will be the statement in the question true ? If yes , then how to prove it ?

Suppose that$Ker f=0$, $f(x)=f(y)$ implies that $f(x-y)=0$ and $x-y\in Ker f$ we deduce that $x-y=0$ and $x=y$.
$f$ is one to one, $f(x)=0=f(0)$ implies $x=0$.
