Let $R$ be a unital ring and $f:M\to N$ be an epimorphism of modules. I know how to prove that a morphism of modules is a monomorphism iff it is injective, and that a surjective morphism is an epimorphism. But how can I prove that an epimorphism is surjective?
If $f$ is an epimorphism, then for each $R$-module $P$ and morphisms $\varphi,\psi:N\to P$ with $\varphi\circ f=\psi\circ f$, we have that $\varphi=\psi$. We want to show that for each $y\in N$ there exists $x\in M$ with $y=f(x)$. I'm guessing it involves a clever choice of $P$, $\varphi$ and $\psi$ but I can't think what it should be. Any tips will be greatly appreciated.
I am particularly interested in the corresponding result for morphisms of graded modules, but I should be able to work this bit out from the ungraded case 🙂
Best Answer
Let $\pi$ be the projection $\pi:N\to N/\operatorname{Im} f$ and $g:N\to N/\operatorname{Im}f$ the zero map. Then $\pi(f(x)) = 0$ and $g(f(x)) = 0$ for all $x\in M$. As such, since $f$ is an epimorphism, $\pi=g$, implying $\pi(x)$ is zero for all $x\in N$. It must then be true then that $\operatorname{Im}f = N$, so $f$ is a surjection.