Let \(R\) be a ring and \(a\in R\). Consider any solutions \(\chi: R\to R\) of the f.eq \(\chi(ax)=1+\chi(x)\). This is an abel function of multiplication.
The set of all these solutions, denote it with \({\mathcal A}_a\), is a very big set. Note that for every \(r\in R\) we have the inclusion \({\mathcal A}_a +r\subseteq {\mathcal A}_a\) and, let \(\Theta\) be the set of solutions \(\theta: R\to R\) of the f.eq. \(\theta(ax)=\theta(x)\), then \({\mathcal A}_a+\Theta\subseteq {\mathcal A}_a\). This means we have many solutions.
Let \(h\in R\) be a root of the polynomial \(p\in R[x]\) where \(p(x)=2x-1\). Assume that \(\chi\in {\mathcal A}_a\) has a right inverse \(\phi\) (a retraction), i.e. \(\chi\) is surjective/ \(\chi\phi={\rm id}_R\), then define \(f(x):=\phi(\chi(x)+h)\). It satisfies
\[\begin{align}
f^2(x)& =\phi( \chi( \phi(\chi(x)+h) )+h)&&&\\
& =\phi( \chi(x)+2h) &&&\\
& =\phi( \chi(x)+1) &&&\\
& =ax &&&
\end{align}\]
In short: let \({\mathcal A}_a^{\rm surj}\) the set of surjective abel functions of multiplication by \(a\), and \(V(2x-1)\) be the set of roots of the previous polynomial, let \(\mathcal F_a\) be set of solutions of the f.eq \(f^2(x)=ax\) in \(R\), then we have a map
\[{\rm pt}\times_{R^R}({\mathcal A}_a^{\rm surj}\times R^R)\times V(2x-1) \to \mathcal F_a\]
Sending triples \((\chi,\phi,h)\), where \(\chi\phi={\rm id}_R\) to functions \(f(x):=\phi(\chi(x)+h)\). A you can see the size of \(\mathcal F_a\) can be bounded from below by studying if this assignment is injective.
The set of all these solutions, denote it with \({\mathcal A}_a\), is a very big set. Note that for every \(r\in R\) we have the inclusion \({\mathcal A}_a +r\subseteq {\mathcal A}_a\) and, let \(\Theta\) be the set of solutions \(\theta: R\to R\) of the f.eq. \(\theta(ax)=\theta(x)\), then \({\mathcal A}_a+\Theta\subseteq {\mathcal A}_a\). This means we have many solutions.
Let \(h\in R\) be a root of the polynomial \(p\in R[x]\) where \(p(x)=2x-1\). Assume that \(\chi\in {\mathcal A}_a\) has a right inverse \(\phi\) (a retraction), i.e. \(\chi\) is surjective/ \(\chi\phi={\rm id}_R\), then define \(f(x):=\phi(\chi(x)+h)\). It satisfies
\[\begin{align}
f^2(x)& =\phi( \chi( \phi(\chi(x)+h) )+h)&&&\\
& =\phi( \chi(x)+2h) &&&\\
& =\phi( \chi(x)+1) &&&\\
& =ax &&&
\end{align}\]
In short: let \({\mathcal A}_a^{\rm surj}\) the set of surjective abel functions of multiplication by \(a\), and \(V(2x-1)\) be the set of roots of the previous polynomial, let \(\mathcal F_a\) be set of solutions of the f.eq \(f^2(x)=ax\) in \(R\), then we have a map
\[{\rm pt}\times_{R^R}({\mathcal A}_a^{\rm surj}\times R^R)\times V(2x-1) \to \mathcal F_a\]
Sending triples \((\chi,\phi,h)\), where \(\chi\phi={\rm id}_R\) to functions \(f(x):=\phi(\chi(x)+h)\). A you can see the size of \(\mathcal F_a\) can be bounded from below by studying if this assignment is injective.
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)