I have completed a proof of a theorem characterizing von Neumann regularity of endomorphisms but I think my proof is imperfect. The theorem is stated in the language of modules but they could be groups just as well, and probably many other things.
My question is:
The theorem below states the equivalence of three conditions. How can I prove $(3)\Longrightarrow (1)$ without proving $(2)$ in between?
The theorem says what follows.
Let $M$ be an $R$-module. Let $\mathrm{E}:=\operatorname{End}(M),$ and $\rho\in\mathrm E$. The following conditions are equivalent.
$(1)$ $(\exists \sigma\in\mathrm E) \;\rho\sigma\rho=\rho;$
$(2)$ $\ker\rho$ and $\operatorname{im}\rho$ are direct summands in $M.$
$(3)$ There exist idempotent $\pi_1,\pi_2\in\mathrm E$ such that $\ker \pi_1=\ker\rho$ and $\operatorname{im}\pi_2=\operatorname{im}\rho.$
Proof.
$(1)\Longrightarrow(2).$ Let $\rho\sigma\rho=\rho$ for some $\sigma\in\mathrm E$.
Suppose $m\in\ker\rho\cap\operatorname{im}\rho\sigma.$ Then $m\rho=0$ and $m=x\rho\sigma.$ Therefore $x\rho=x\rho\sigma\rho=0$ and $m=x\rho\sigma=0.$ Thus
$\ker\rho\cap\operatorname{im}\rho\sigma=0.$
Now let $m\in M.$ We have $(m-m\rho\sigma)\rho=m\rho-m\rho\sigma\rho=m\rho-m\rho=0,$ so $(m-m\rho\sigma)\in\ker\rho$ and since $m\rho\sigma\in\operatorname{im}\rho\sigma,$ we have $M=\ker\rho+\operatorname{im}\rho\sigma.$ Thus $M=\ker\rho\oplus\operatorname{im}\rho\sigma.$
Suppose $m\in\operatorname{im}\rho\cap\ker\sigma\rho.$ Then $m\sigma\rho=0$ and $m=x\rho.$ Therefore, $m=x\rho=x\rho\sigma\rho=0.$ Thus
$\operatorname{im}\rho\cap\ker\sigma\rho=0.$
Now let $m\in M.$ We have $(m-m\sigma\rho)\sigma\rho=m\sigma\rho-m\sigma\rho\sigma\rho=m\sigma\rho-m\sigma\rho=0,$ and since $m\sigma\rho\in\operatorname{im}\rho,$ we have $M=\operatorname{im}\rho+\ker\sigma\rho.$ Thus $M=\operatorname{im}\rho\oplus\ker\sigma\rho.$
$(2)\Longrightarrow (1).$ Suppose $M=\ker\rho\oplus K=\operatorname{im}\rho\oplus L.$ From the first isomorphism theorem, we have an isomorphism $i :(\ker\rho\oplus K)/\ker\rho \longrightarrow\operatorname{im}\rho$ such that for any $m\in M$
$(m+\ker\rho)i=m\rho.$
We can write uniquely any $m\in M$ as a sum $m=a_{\ker\rho}+a_K.$ Let $m,n\in M$ with decompositions $m=a_1+a_2,$ $n=b_1+b_2$ where $a_1,b_1\in\ker\rho$ and $a_2,b_2\in K.$ Then the residue classes of $m,n$ in $(\ker\rho\oplus K)/\ker\rho$ are equal iff $a_2=b_2.$ Therefore, the map \sigma'':(\ker\rho\oplus K)/\ker\rho\longrightarrow \ker\rho\oplus K such that
$(a_{\ker\rho}+a_K+\ker\rho)\longmapsto (0+a_K)$
is well-defined and it is easily seen to be a homomorphism. We define \sigma':\operatorname{im}\rho\longrightarrow M by the formula \sigma'=i^{-1}\sigma''. We can define $\sigma$ by putting \sigma|_{\operatorname{im}\rho}=\sigma' and $\sigma|_L=0.$ Now $\rho=\rho\sigma\rho.$
$(2)\Longrightarrow (3).$ Suppose $M=\ker\rho\oplus K=\operatorname{im}\rho\oplus L.$ We define $\pi_1,\pi_2$ by $\pi_1|_{\ker\rho}=0 \text{ and }\pi_1|_K=\operatorname{id}_K,$ and $\pi_2|_{\operatorname{im}\rho}=\operatorname{id}_{\operatorname{im}\rho}\text{ and } \pi_2|_L=0.$
$(3)\Longrightarrow (2)$. It is easy to see that for an idempotent map $\pi:M\longrightarrow M,$ we have $M=\operatorname{im}\pi\oplus\ker \pi.$ The implication follows.
This ends the proof. However, I'm not satisfied with it. I have proven four implications, whereas only three are usually proven in such cases. I am able to fix it by proving $(1)\Longrightarrow (3),$ thus obtaining a cycle $(1)\Longrightarrow (3)\Longrightarrow (2)\Longrightarrow (1).$ But this is not good enough for me. I need $(3)\Longrightarrow (1)$ because I want to be able to see whether it is possible to generalize this theorem to structures in which the kernel cannot be defined as a substructure but has to be defined as a congruence. It is easy to see that the theorem generalizes to groups and perhaps it generalizes to rings too (although I haven't thought about it). That's because there is nothing specific to modules in the proof. The only important thing is that we can meaningfully say that the kernel is a direct summand. That's impossible I think when the kernel is just a relation, as in the plethora of various "semi-structures". If I were able to discard $(2)$ in the theorem, I would have a theorem that translated easily to the language of such structures. Could you please help me find a proof of $(3)\Longrightarrow (1)$ that doesn't use direct sums in any way and uses the kernels as congruences? Is it possible?
I suspect this must have a lot to do with category theory, but I know very little about it. Please, if you have to use it, do it bearing in mind that I will not understand anything other than the most elementary explanation.