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Let $\mathcal{T}$ be a commutative algebraic theory (for example sets, abelian groups, commutative monoids, but not groups etc.). References include the nlab and Borceux' Handbook of Categorical Algebra 2, section 3.10. Then $\mathsf{Mod}(\mathcal{T})$ is a monoidal category with internal homs.

Question 1. (Answered: Yes) Can we find a property of concrete categories which holds for $\mathsf{Mod}(\mathcal{T})$ if and only if $\mathcal{T}$ is commutative? In other words, does commutativity of an algebraic category not depend on the presentation?

Question 2. (Answered: No) Let $\mathcal{T}$ be a commutative algebraic theory and $C=\mathsf{Mod}(\mathcal{T})$. Assume that $X \in C$ is a Co-$C$-algebra, i.e. we have a factorization of $\hom(X,-) : C \to \mathsf{Set}$ over $C$. Does this have to coincide with the usual factorization? This is well-known to be true in the examples I have mentioned above, for example for every abelian group $A$ there is only one natural abelian group structure on the hom-sets $\mathrm{hom}(A,B)$. This should be all well-known, but I don't know a reference.

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To me a "presentation" of an algebraic theory is a choice of operations and axioms, while Lawvere theories and monads embody the extensional essence of an algebraic theory. In any case, I think the answer to question 1 is negative: being a commutative algebraic theory is not stable under Morita equivalence, since e.g. the theory of $k$-modules for a field $k$ is commutative, but the theory of $\textrm{Mat}_n(k)$-modules is not for $n \ge 2$. –  Zhen Lin Nov 20 '12 at 8:22
    
Yes but these categories are not equivalent as concrete categories. –  Martin Brandenburg Nov 20 '12 at 8:32
    
Indeed, which is why I said Morita equivalence. Once you have fixed a forgetful functor, then the category entirely determines the algebraic theory. (To construct the symmetric monoidal closed structure you only need to know that the monad admits a commutative structure.) –  Zhen Lin Nov 20 '12 at 9:29
    
Why can't we consider the commutativity condition on the monad $Ran_U U$ as a condition on the concrete category $U: Mod(T) \to Set$? –  Todd Trimble Nov 20 '12 at 11:53
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Um, Martin, as a courtesy I was trying to clarify whether you were expecting something else as regards question 1, since the comment seems to be something of a simple observation. Also, I wanted to think more about your question 2. But, I'm happy to put down my comment as an answer, as long as you're happy. (Later, since I have to dash off to do something.) –  Todd Trimble Nov 20 '12 at 13:29

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up vote 6 down vote accepted

As for question 1: commutativity doesn't depend on the presentation of $T$. If $M = Mod(T)$ and $U: M \to Set$ is the forgetful functor, then commutativity can be formulated as saying that the monad $Ran_U U = U \circ Ran_U 1_M$ is commutative (or monoidal) in the sense of the nLab article here. Perhaps the most interesting aspect of this is that commutativity is a property, not an extra structure on a monad (where the structure of a strength constraint on an endofunctor on $Set$ is canonically given because every such endofunctor is canonically $Set$-enriched). These observations also lift to the enriched setting, provided of course that the functors involved are given as enriched functors (with respect to a base of enrichment $V$).

(Note: $Ran_U 1_M$, which invariably exists, is just the left adjoint $F$ of $U$ if $U$ has a left adjoint. Some related discussion on the codensity monad of a general functor $U: M \to Set$ can be found in this post by Tom Leinster.)

Edit: I had responded to question 2 earlier, but I am now editing that response out as it is superseded by Martin's second comment below, which makes the situation quite clear. Apologies for the noise.

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Hm, I think that 2) is wrong: Note that 2) is equivalent to the assertion that the enriched Yoneda embedding $C^{op} \to \hom_c(C,C)$ is an equivalence (here $c$ stands for continuous functors). Now take $C=$left $k$-modules for some ring $k$. Then it is well-known (special case of Eilenberg-Watts) that the right hand side equals the category of $k$-bimodules. Even when $k$ is commutative, this is not just $k$-modules. –  Martin Brandenburg Nov 21 '12 at 7:31
    
Martin, you're absolutely right, and that's a very clear way to look at the situation. I'm now going to edit my answer, and delete some comments which are now obsolete. –  Todd Trimble Nov 21 '12 at 14:08

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