Two varieties of universal algebras are categorically equivalent iff their respective full subcategories of finitely generated free algebras are equivalent. Roughly speaking, this follows because they have the same equational theory in a generalized sense.


Now suppose each variety $\mathcal{V}\_i$ has cogenerator $K\_i$, and that the full subcategory of finite powers $K\_1^n$ is equivalent to the full subcategory of finite powers $K\_2^n$. Then it does not follow that $\mathcal{V}\_1 \cong \mathcal{V}\_2$.


**Examples**


1. Let $\mathcal{V}\_1 = \mathsf{BA}$ where $K\_1 = 2$, and $\mathcal{V}\_2 = \mathsf{DL}\_{01}$ where $K\_2 = 2$. They are cogenerators because each $A \in \mathcal{V}\_i$ can be embedded into a product $K\_i^J$. In both cases the homomorphisms $K\_i^n \to K\_i$ are the projections $\pi\_i : 2^n \to 2$. But clearly $\mathsf{BA} \ncong \mathsf{DL}\_{01}$, also because epis are not extremal in the latter via $3 \hookrightarrow 2 \times 2$.
2. Let $\mathcal{V}\_1$ be sets with a single constant, and $\mathcal{V}\_2$ be sets with a single unary operation satisfying $u(x) \approx u(y)$. In both cases let $K\_i$ be the two element algebra. The full subcategories of finite powers are the same functions, yet $\mathcal{V}\_1 \ncong \mathcal{V}\_2$ because epis are not extremal in $\mathcal{V}\_2$ via $\emptyset \hookrightarrow 1$.


In example (1) each $K\_i$ is an injective cogenerator. In example (2), each $\mathcal{V}\_i$ is a rather degenerate example of a commutative variety i.e. the interpretation of any $n$-ary operation $\sigma$ in any algebra $A \in \mathcal{V}$ defines an algebra homomorphism $\sigma\_A : A^n \to A$. A more natural example is abelian groups with cogenerator $\mathbb{Q}/\mathbb{Z}$.


Let me ask a rather specific question:



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Here is an alternative formulation:



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> > Under the conditions above, does it follow that the clone of operations on $K$ contains precisely those functions which are preserved by all homomorphisms $K^n \to K$?
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There are many positive examples. For vector spaces with $K = \mathbb{F}$ then its operations = its homomorphisms i.e. the linear maps. For abelian groups with $K = \mathbb{Q}/\mathbb{Z}$ then the only operations preserved by all homomorphisms are the $\mathbb{Z}$-linear maps. For sets with $K = 2$ the functions preserved by all boolean functions are the projections (the minimal clone). For sets with an involutive unary operation the smallest injective cogenerator has two self-loops and one two-loop, and its powers determine this variety.



When does a cogenerator determine a variety?
I understand the question to be: Let $\mathcal V$ be a commutative variety with an injective cogenerator $K$ and assume that $\mathcal V$-epis are surjective. Does it follow that the clone of operations on $K$ contains precisely those functions which are preserved by all homomorphisms $K^n\to K$? (I.e., does the clone of $K$ equal its double centralizer?)


Answer: no.


Sketch:
If $\mathcal V$ is the variety of all modules over a commutative ring $R$, then $\mathcal V$ is a commutative variety, has an injective cogenerator, and epis are surjective, so all hypotheses are met. The question is whether the double centralizer clone of an injective cogenerator equals the clone itself.


Let $k$ be a field and let $V$ be a $k$-vector space. Let $R$ be the commutative ring $k\oplus V$ with multiplication defined by $(a,u)(b,v)=(ab,av+bu)$. Let $\mathcal V$ be the variety of $R$-modules. If $V^\*$ denotes the dual space of $V$, then an injective cogenerator for $\mathcal V$ may be constructed with universe $k\oplus V^\*$ and scalar multiplication specified by 
$$
(a,u)(b,f) = (ab+f(u),af)
$$
for $(a,u)\in R$ and $(b,f)\in K$.


One can compute the double centralizer clone of $K$ and find it to be the clone of the $S$-module structure on $K$, where $S = k\oplus V^{\*\*}$ (constructed as $R$ was constructed, but using the double dual of $V$ rather than $V$ itself). The action of $S$ on $K$ is given by
$$
(a,\varphi)(b,f) = (ab+\varphi(f),af).
$$
So, if $V$ is infinite dimensional, this provides a negative answer.


