I understand the definition of the kernel of a morphism, $f:A\to B$, say, in an additive category. It is a morphism $i: K \to A$ such that $fi=0$ and is universal with the property that if $j:K' \to A$ is another morphism such that $fj=0$, then there is a unique morphism $g:K'\to K$ such that $j=ig$.
I want to know how do you show the morphism $i$ always exists for any $f$ in this category?
It is written here that the inclusion $i : ker(f) \to A$ is the kernel of $f$ in the sense of definition above in this category, however, I don't know how to show this. I'm confused about whether one simply begins by saying $f(i(x)) = f(0) = 0$? But then doesn't this presupposes $i$ is the kernel of $f$ before we've proven it is a kernel? Wouldn't this work in any other category also then?
Category of modules is concrete category: we have sets with additional structure. It makes sense to write down $\ker f=\{x\in M\,|\, f(x) = 0\}$ in ${\bf{Mod}}_R$ for some $f\colon M\to N$, but obviously not in categories that are not concrete. You want to confirm that $i\colon \ker f\to M$ is kernel in categorical sense. Note that $i$ is defined by $i(x) = x$.
First of all, $f(i(x)) = f(x) = 0$, for all $x\in\ker f$, by defintion of $\ker f$ and thus $f\circ i = 0$. Furthermore, take $g\colon M'\to M$ such that $f\circ g = 0$. That means that for all $x\in M'$ we have $f(g(x)) = 0$ which implies that $g(x)\in\ker f$, for all $x\in M'$. Thus, $g$ factors as $M'\to\ker f\stackrel{i}{\to}M$. It is trivial to confirm that this factorization is unique.