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For my question, let us consider the following scenario.

We have a quasi-variety $\mathcal{A}$ generated by a finite algebra $\mathbf{M}$ (i.e. $\mathcal{A} = \mathbb{ISP}(\mathbf{M})$). Now, let $\mathbf{A}$ be another finite algebra in $\mathcal{A}$. Assume that there exists an integer $k$ such that, for each $n \in \mathbb{N}$, every homomorphism $f \colon \mathbf{A}^n \rightarrow \mathbf{M}$ is essentially at most $k$-ary.

My question is: Is it guaranteed that there exists a finite algebra $\mathbf{M'} \in \mathcal{A}$ with $\mathbf{A} \in \mathbb{ISP}(\textbf{M'})$ such that, for each $n \in \mathbb{N}$, every homomorphism $f \colon \mathbf{A}^n \rightarrow \mathbf{M'}$ is essentially at most unary?

To illustrate this question, I will now give an example in which this is true. Let $\textbf{M}$ and $\textbf{A}$ be two finite distributive lattices. Now, there exists an integer $k$ such that the essential artiy of every $f \colon \mathbf{A}^n \rightarrow \mathbf{M}$ is at most $k$-ary. We can define $\mathbf{M'}$ to be the (up to isomorphism unique) two-element distributive lattice.

Note that $\mathbb{ISP}(\mathbf{M'}) = \mathbb{ISP}(\mathbf{M})$ is not required (although it is true in the example). It is only required to have $\mathbf{A} \in \mathbb{ISP}(\mathbf{M'})$.

Is this always possible? I believe it is not, but I do not know any counterexample up to this point in time. Does anybody know a counterexample, a proof or has a feeling of whether this statement is true or false?

Note: If the answer to my question would be positive, this would have some very nice consequences for centralizer clones (which is my main research interest at the time). In fact, this would mean (and I can explain later, why that is) that every centralizer clones that has bounded essential arity cannot contain many types of functions. Among them: Near-unanimity operations, minority operations, semiprojections etc.. Furthermore, it would lead to a full characterization of all minimal clones in centralizer clones with bounded essential arity.

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Exercise more care. Many would see your homomorphism f as having arity 1. You should instead say f is a map of n-variables from A to M, such that it is essentially k-ary when viewed as such a map but also that the associated map g from A^n to M given by g(veca) = f(a_1,..a_n) is a homomorphism. Gerhard "Ask Me About System Design" Paseman, 2010.08.18 –  Gerhard Paseman Aug 18 '10 at 16:23
    
Also, this condition on f can be viewed as a condition on the (is it pseudoquasivariety? I forget the terminology) subclass of the quasivariety of finite structures, such that all the congruences of powers are 'like' congruences induced by projections. This might mean that the quasivariety is congruence-modular, for example, or satisfies some Mal'cev condition. This might give you information to help find M' . Gerhard "Ask Me About System Design" Paseman, 2010.08.18 –  Gerhard Paseman Aug 18 '10 at 16:32
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Also, here is a feeling towards your problem. One way to have your conjecture be true is to have M' be much smaller than M, so that homomorphisms to M' do not have as much room to play. So consider an algebra M with small subalgebra lattice. M might be an n-valued Post algebra or something in an arithmetic variety in which M has no subalgebras. Then homomorphisms from powers to M might be essentially n-ary, but there may be no way to find M' to reduce the arity since there are no smaller candidates. Gerhard "Ask Me About System Design" Paseman, 2010.08.18 –  Gerhard Paseman Aug 18 '10 at 16:40
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The "feeling" turns out to be right. Thank you.

One can take the unique four-element boolean algebra and make each of the four elements a constant (instead of only the lowest and greatest alement). This can be defined to be $A$ and $M$.

However, this does not answer the question that motivated me to ask for this here. In the above scenarion, one can take $M$ to be the unique two-element boolean algebra and then treat the homomorphisms from $A^n$ to $A$ (where $A$ is defined as above) as a subset of the homomorphisms from $(M^2)^n$ to $M^2$ and then we are back to the situation that we wanted to avoid.

I would like to obtain an example in which such a trick is not possible.

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