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.

as concrete categories. – Martin Brandenburg Nov 20 '12 at 8:32Moritaequivalence. 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 9:29you'rehappy. (Later, since I have to dash off to do something.) – Todd Trimble♦ Nov 20 '12 at 13:29