Coequalizer in the category of modules

Question

I am trying to prove that the category of modules is cocomplete. It suffices to show that it has all coequalizers and coproducts. It's relatively easy to show that all coproducts exist, and I am left to prove that all coequalizers exist in the category of modules. Thus, let $f, g : M \to N$ be $R$-module homomorphisms where $M$ and $N$ are $R$-modules. I am guessing that the coequalizer must be $N / \sim$, where $\sim$ is an equivalence relation of some sort. However, if we simply defined $\sim$ as the smallest equivalence relation on $N$ such that $f(n) \sim g(n)$ for all $n \in N$, I don't know how to show that the $R$-module structure on $N / \sim$ defined as $r \cdot [n] := [rn]$ is well defined. I am thinking that perhaps the $\sim$ should be defined slightly differently, in which case I would need to prove the universal property. However, I am not sure how. Can you help?

Answer 1

The relation $\sim$ should be defined as $f(m) \sim g(m)$ for all $m \in M$. This means that $\sim$ generates the submodule $N' = \{f(m)-g(m) | m\in M\}$.

To prove that the $R$-module structure on $N/\sim$ is well defined, $\forall n \in N$, $m \in M$, $r \in R$, $$ r(n + f(m) - g(m)) = rn + f(rm) - g(rm) $$ So $r(n + f(m) - g(m)) - rn$ is in the submodule $N'$.