# Background

Let $$R$$ be a ring or a semigroup. We say that $$x\in R$$ is a von Neumann regular element of $$R$$ if there exists $$y\in R$$ such that

$$xyx=x.$$

Any $$y\in R$$ satisfying the above equation is called a pseudoinverse of $$x$$. (It is usually not unique.) It is a useful notion which extends outside ring and semigroup theory. For example, it can be proven that for any $$(m\times n)$$-matrix $$A$$ over a field $$K$$ there exists (and can be effectively found) a $$(n\times m)$$-matrix $$X$$ over $$K$$ such that

$$AXA=A.$$

This pseudoinverse of $$A$$ can be used to solve the linear system

$$Ax=b.$$

This equation has solutions iff $$Xb$$ is a solution, which is easy to verify. It isn't much more difficult to check that if $$Xb$$ is a solution, then

$$x=Xb+(I-XA)y,$$

where $$y$$ is arbitrary, gives all solutions $$x$$ of the system.

Let's say that $$y$$ is a 2-pseudoinverse of $$x$$ if

$$yxy=y.$$

It is easy to see that if $$x$$ has a pseudoinverse $$y$$, then it has a 2-pseudoinverse too. Indeed, in this case, we have

$$(yxy)x(yxy)=yxyxy=yxy$$

so $$yxy$$ is a 2-pseudoinverse of $$x.$$

# Question

I would like to know whether the existence of a 2-pseudoinverse implies the existence of a pseudoinverse. That is, if I have $$y$$ such that

$$yxy=y,$$

will I always find $$y'$$ such that

$$xy'x=x?$$

In yet other words, if $$y$$ is a regular element with a pseudoinverse $$x,$$ must $$x$$ be a regular element? (Is it true in semigroups? Is it true in rings?)

The problem is that nothing in the equation $$yxy=y$$ allows me to reduce some expression to $$x,$$ which is essentially what I would have to do in order to prove that the answer is "yes". On the other hand, to prove it is "no", I would have to find a counter-example and I have no idea where to look.

It is clear that I can't look for counter-examples in semigroups or rings in which all elements are regular. (Which are called regular semigroups and von Neumann regular rings respectively). Unfortunately, I know next to nothing about regular elements in semigroups which are not regular and in rings which are not von Neumann regular.

Let $S$ be a non-regular semigroup with a zero element $0$. Then $0$ is regular, and every element of $S$ is a pseudoinverse of $0$ (i.e. $0x0 = 0$ for all $x\in S$). But not every element of $S$ is regular.