Abbreviation: RRng
A \emph{regular ring} is a rings with identity $\mathbf{R}=\langle R,+,-,0,\cdot,1 \rangle $ such that
every element has a pseudo-inverse: $\forall x\exists y(x\cdot y\cdot x=x)$
Let $\mathbf{R}$ and $\mathbf{S}$ be regular rings. A morphism from $\mathbf{R}$ to $\mathbf{S}$ is a function $h:R\rightarrow S$ that is a homomorphism:
$h(x+y)=h(x)+h(y)$, $h(x\cdot y)=h(x)\cdot h(y)$, $h(1)=1$
Remark: It follows that $h(0)=0$ and $h(-x)=-h(x)$.
\begin{examples} \end{examples}
$\begin{array}{lr}
f(1)= &1
f(2)= &
f(3)= &
f(4)= &
f(5)= &
f(6)= &
\end{array}$