The rack kink map is bijective

Self-distributivity gives $(w\ast x)\ast x=(w\ast x)\ast (x\ast x)$ for every $w,x$. Thus $R_x(R_x(w))=R_{x\ast x}(R_x(w))$. Since $R_x$ is surjective, the two permutations agree: $R_{x\ast x}=R_x$, or $R_{\pi (x)}=R_x$.

Fix $z\in R$ and put $x=R_z^{-1}(z)$, which exists by [F1]. Then $x\ast z=z$. Self-distributivity says $R_z\circ R_x=R_{x\ast z}\circ R_z=R_z\circ R_z$. Cancel the bijection $R_z$ on the left to obtain $R_x=R_z$. Therefore $\pi (x)=R_x(x)=R_z(x)=x\ast z=z$, so $\pi$ is surjective and the displayed formula gives a preimage of every $z$.

If $\pi (x)=\pi (y)$, then step 1.1 gives $R_x=R_{\pi (x)}=R_{\pi (y)}=R_y$. Hence $R_y(x)=R_x(x)=\pi (x)=\pi (y)=R_y(y)$. Since $R_y$ is injective, $x=y$. Thus $\pi$ is injective as well as surjective, and step 1.2's formula is its inverse. If $R$ is empty, the unique map $R\to R$ is bijective and the conclusion is vacuous.