This is actually P. Aluffi's "Algebra: Chapter $0$", exercise ${\rm III}.6.3$
Let $R$ be a ring, $M$ an $R$-module and $p\colon M\to M$ an $R$-module homomorphism such that $p^2=p$. (Such a map is called a projection). Prove that $M\cong\ker p~\oplus~{\rm im}~p$.
The standard approach is more or less the following: let $x\in M$, then $x-p(x)\in M$ and by the given property we deduce that $p(x-p(x))=p(x)-p(x)=0$ and therefore $x-p(x)\in\ker p$. Now we can write every element as $x=(x-p(x))+p(x)$ and the rest are routine verifications. (here is a reference regarding "standard approach").
Anyway, in the spirit of Aluffi's book, I was wondering if there is a different approach. More precisely, is it possible to deduce $M\cong\ker p~\oplus~{\rm im}~p$ only from $p^2=p$ not using the elementwise approach. I was thinking about constructing morphisms $\alpha,\beta$ (somehow related to $p$) such that $0\rightarrow\ker p\overset\alpha\rightarrow M\overset\beta\rightarrow{\rm im}~p\rightarrow0$ splits. An evident choice would be the inclusion arrow $\iota\colon\ker p\to M$ and $p$ itself, respectively, but I was not able to deduce anything else than trivialities from hereon.
Is this doable? Does anyone has a reference, where this approach is followed? Or do I miss something (maybe obvious) which prevents one from doing it without elements?
Thanks in advance!
The canonical short exact sequence $$0 \to \ker p \xrightarrow{i} M \xrightarrow{p} \text{im}~p \to 0$$ is split, because $\sigma : M \to \ker p$ defined by $\sigma = 1_M-p$ (is well defined because $p^2=p$) is a section for the sequence (this is, $\sigma i = 1_{\ker p}$) and thus by the splitting lemma, the sequence is split exact