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Question. Do you know a specific example which demonstrates that the tensor product of monoids (as defined below) is not associative?


Let $C$ be the category of algebraic structures of a fixed type, and let us denote by $|~|$ the underlying functor $C \to \mathsf{Set}$. For $M,N \in C$ we have a functor $\mathrm{BiHom}(M,N;-) : C \to \mathsf{Set}$ which sends an object $K \in C$ to the set of bihomomorphisms $M \times N \to K$, i.e. maps $|M| \times |N| \to |K|$ which are homomorphisms in each variable when the other one is fixed. Then one can show as usual that $\mathrm{BiHom}(M,N;-)$ is representable and call the universal bihomomorphism $M \times N \to M \otimes N$ the tensor product of $M,N$. This is a straight forward generalization of the well-known case $C=\mathsf{Mod}(R)$ for a commutative ring $R$.

Actually, this is a special case of a more general tensor product in concrete categories, studied in the paper "Tensor products and bimorphisms", Canad. Math. Bull. 19 (1976) 385-401, by B. Banaschewski and E. Nelson.

Here are some examples: For $C=\mathsf{Set}$, the tensor product equals the usual cartesian product. This is also true for $C=\mathsf{Set}_*$. For $C=\mathsf{Grp}$, we get $G \otimes H \cong G^{\mathsf{ab}} \otimes_{\mathbb{Z}} H^{\mathsf{ab}}$, using the Eckmann-Hilton argument. (This differs from the "tensor product of groups" studied in the literature). The case $C=\mathsf{CMon}$ is very similar to the well-known case $C=\mathsf{Ab}$ and is spelled out here; namely, we have internal homs and therefore a hom-tensor-adjunction. The same is true for $C=\mathsf{Mod}(\Lambda)$ for a commutative algebraic monad $\Lambda$, see here, Section 5.3.

Note that the tensor product is commutative, and that it commutes with filtered colimits in each variable. However, the case $C=\mathsf{Grp}$ shows that it does not have to commute with coproducts. In particular, tensoring with some object is no left adjoint. Also, the free object on one generator is not a unit in general:

Let us consider $C=\mathsf{Mon}$. Then, we have

$\mathbb{N} \otimes M = M / \{ (mn)^p = m^p n^p \}_{m,n \in M, p \in \mathbb{N}}$

The usual proof of the associativity of the tensor product breaks down: There is a map $\beta : M \times (N \otimes K) \to (M \otimes N) \otimes K$ mapping $(m, n \otimes k) \mapsto (m \otimes n) \otimes k$, which is a homomorphism in the second variable. But what about the first variable? The equation $\beta(mm',t) = \beta(m,t) \beta(m',t)$ is clear if $t \in N \otimes K$ is a pure tensor. But for $t=(n \otimes k) (n' \otimes k')$ we end up with the unlikely equation

$((m \otimes n) \otimes k) ((m' \otimes n) \otimes k) ((m \otimes n') \otimes k') ((m' \otimes n') \otimes k')$ $=((m \otimes n) \otimes k) ((m \otimes n') \otimes k') ((m' \otimes n) \otimes k) ((m' \otimes n') \otimes k')$

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    Regarding associativity: have you thought about comparing $(M \otimes N) \otimes P$ with the object (which presumably exists) that represents the functor $\operatorname{TriHom}(M,N,P;-)$ (defined in the "obvious" way)? I usually think of tensor products as being associative due to the existence of an isomorphism with a canonical object $M \otimes N \otimes P$.2012-10-02
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    @Manny: There are surjective homomorphisms $M \otimes (N \otimes P) \leftarrow M \otimes N \otimes P \rightarrow (M \otimes N) \otimes P$, but I doubt that they are invertible.2012-10-02
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    By the way, the fact that $G \otimes H \cong G^\textrm{ab} \otimes_\mathbb{Z} H^\textrm{ab}$ for groups proves that $\otimes$ has _no_ unit at all – which is much stronger than saying that $\mathbb{Z}$ is not the unit.2012-10-02
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    I don't think your description of the tensor product for groups is correct--for instance, in $\mathbb{Z}^2\otimes\mathbb{Z}^2$, why should $(0,1)\otimes (0,1)$ commute with $(1,0)\otimes (1,0)$?2016-01-24
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    @EricWofsey: In $G \otimes H$ we may expand $gg' \otimes hh'$ in two ways (either starting with $gg'$ and then do $hh'$, or the other way round). Cancelling $g \otimes h$ on the left and $g' \otimes h'$ on the right (this is possible since these elements are invertible) gives you that $g \otimes h'$ and $g' \otimes h$ commute with each other.2016-01-24

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