# Is this condition sufficient for a variety to be reversible?

A variety $$V$$ is said to be reversible, if for each $$n>0$$ and fundamental operation $$f$$ there are $$m\geq n$$ and $$r$$ along with terms $$T_{2},\dots,T_{r}$$ and $$S_{1},\dots,S_{m}$$ such that if $$G,H$$ are the following the functions, then $$G,H$$ are inverses.

1. $$G(x_{1},\dots,x_{m})=(f(x_{1},\dots,x_{n}),T_{2}(x_{1},\dots,x_{m}),\dots,T_{r}(x_{1},\dots,x_{m})$$.

2. $$H(x_{1},\dots,x_{r})=(S_{1}(x_{1},\dots,x_{r}),\dots,S_{m}(x_{1},\dots,x_{r})).$$

If $$V$$ is a reversible variety, then for each fundamental operation of arity greater than $$0$$, we have $$|f^{-1}[\{a\}]|=|f^{-1}[\{b\}]|$$.

Suppose that $$V$$ is a variety whose theory is axiomatized by finitely many identities and where $$V$$ is generated by its finite members.

Furthermore, assume that whenever $$X\in V$$ and $$X$$ is finite and $$f$$ is an $$n$$-ary fundamental operation, then $$\{(x_{1},\dots,x_{n}):f(x_{1},\dots,x_{n})=a\}=|X|^{n-1}.$$ Then is the variety $$V$$ reversible?

• Do your assumptions imply that the algebras are congruence regular or Hamiltonian? That might give you a leg up. Gerhard "Not After Hamilton The Rapper" Paseman, 2019.01.28. Jan 28, 2019 at 16:41
• If $G$ is a group that contains a non-normal subgroup $H$, then $G$ is not Hamitonian, but $G$ is still reversible and the variety of groups is reversible. Regularity is out of the question since the variety of all quandles is reversible but if $(X,*,*^{-1})$ is a quandle with more than 3 elements and $x*y=x*^{-1}y=y$ for all $x,y\in X$, then every equivalence relation on $(X,*,*^{-1})$ is a congruence on $(X,*,*^{-1})$. In fact, this means reversibility does not imply any non-trivial property characterized by Mal'cev conditions. Jan 28, 2019 at 21:23

I don't know the answer to this question, but will make an extended remark.

Let $$X$$ be a finite set and let $$f:X^n\to X$$ be any $$n$$-ary operation on $$X$$, $$n>0$$.

Claim. The following conditions are equivalent.

(i) $$f$$ is surjective with uniform kernel.
(Equivalently, for each $$a\in X$$ the set $$f^{-1}(a)=\{(x_{1},\dots,x_{n}):f(x_{1},\dots,x_{n})=a\}$$ has size $$|X|^{n-1}.$$)

(ii) There exist $$n$$-ary operations on $$X$$, $$T_2,\ldots, T_n$$ and $$S_1,\ldots, S_n$$ such that, if $$G,H$$ are $$G(\bar{x}) = (f(\bar{x}), T_2(\bar{x}), \ldots, T_n(\bar{x}))$$ and $$H(\bar{x}) = (S_1(\bar{x}), S_2(\bar{x}), \ldots, S_n(\bar{x})),$$ then $$G$$ and $$H$$ are inverse bijections between $$X^n$$ and $$X^n$$.

The question asks, if $$V$$ is a variety satisfying:

I. $$V$$ is finitely axiomatizable.
II. $$V$$ is generated by its finite members.
III. Item (i) above holds for the interpretation of any fundamental operation of arity at least $$1$$ on each finite member of $$V$$,

then must Item (ii) above hold in the strong sense that the $$S$$'s and $$T$$'s are term operations, but in the weak sense that we allow other parameters $$m$$ and $$r$$ in place of some instances of $$n$$?

Roughly, this asks if having Item (i) hold throughout the finite part of $$V$$ implies that Item (ii) is enforced by the equational theory of $$V$$.

This seems plausible to me, but it also seems that there are some extraneous elements in the question. I don't think that $$V$$ being finitely axiomatizable is relevant. I don't think the additional flexibility of introducing parameters $$m$$ and $$r$$ possibly different from $$n$$ helps, but I haven't tried to check any examples. (It is clear at least that $$m$$ must equal $$r$$ if $$V$$ is generated by its finite members.) I also think the result, if true, is not a property of varieties; that is, the question can be asked for a single (fundamental) operation of $$V$$: if $$V$$ is generated by its finite members and $$f$$ is a fundamental operation of positive arity satisfying Item (i) above, then must Item (ii) above hold?

Here is a sketch of a proof of the claim.

(ii) implies (i): Let $$\pi_1: X^n\to X$$ be the first projection map. It is surjective with uniform kernel. Since $$G: X^n\to X^n$$ is a bijection, $$\pi_1\circ G$$ ( = $$f$$) is also surjective with uniform kernel.

(i) implies (ii): For each $$a\in X$$, choose a bijection $$\beta_a: f^{-1}(a)\to X^{n-1}$$. (Item (i) is the statement that such a bijection exists.) If $$G: X^n\to X^n: \bar{x}\mapsto (f(\bar{x}),\beta_{f(\bar{x})}(\bar{x})),$$ then $$G$$ is a bijection, and the appropriate component functions exist for both $$G$$ and its inverse $$H$$.