A Quandle is a set $Q$ with two binary operations $(x,y)\mapsto x\triangleright y$ and $(x,y)\mapsto x\triangleright^{-1} y$ which satisfies three axioms for all $x,y,z\in Q$:
- $x\triangleright x=x$;
- $(x\triangleright y)\triangleright^{-1}y=x=(x\triangleright^{-1}y)\triangleright y$;
- $(x\triangleright y)\triangleright z=(x \triangleright z)\triangleright(y\triangleright z)$.
My question is:
For each $y\in Q$, does there exist an element $y^{-1}\in Q$ such that $$x\triangleright^{-1}y=x\triangleright (y^{-1})$$ holds true for all $x\in Q$? If yes, is $y^{-1}$ unique for all $y$?
And no, such an element $y^{-1}$ need not exist. In the special case that $\rhd$ is the same operation as $\rhd^{-1}$ (in which case the quandle is called involutory), then of course you can take $y^{-1} = y$. So let's look at the smallest non-involutory quandle. It's the quandle generated by the rotations of the tetrahedron acting on its vertices, and also the only connected quandle of order 4, but you can simply look at its operation table if we take its elements to be ${1,2,3,4}$, namely: $$\begin{matrix}1 & 4 & 2 & 3 \\ 3 & 2 & 4 & 1 \\ 4 & 1 & 3 & 2 \\ 2 & 3 & 1 & 4\end{matrix}$$ Now notice that $2 \rhd^{-1} 1 = 4$ and 3 is the only element such that $2 \rhd 3 = 4$, whereas $ 3 \rhd^{-1} 1 = 2$ but 4 is the only element such that $3 \rhd 4 = 2$. Hence, there is no element that can serve as $1^{-1}$ in the sense you describe above.