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Let $C$ be the category of algebraic structures of a certain 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 $M \times N \to K$, where this is defined to be a map $|M| \times |N| \to |K|$ which is a homomorphism 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 $R$.

Question. Has this tensor product of arbitrary algebraic structures already been studied or used in the literature?

Edit. Yes, see

B. Banaschewski and E. Nelson, Tensor products and bimorphisms, Canad. Math. Bull. 19 (1976) 385-401.

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 some Eckmann-Hilton argument. In particular this yields nothing new (and differs from the tensor product of groups which is known 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 generalized ring $\Lambda$ (i.e. a commutative algebraic monad) in the sense of Durov (see here, Section 5.3).

But what about, say, $C=\mathsf{Mon}$? Do you know other interesting special cases of this tensor product?

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, the tensor product is no left adjoint. Also, the free object on one generator is not a unit in general. For example, for $C=\mathsf{Mon}$ 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

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

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

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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 $M \otimes N \otimes P$. –  Manny Reyes Oct 2 '12 at 16:11
    
@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. –  Martin Brandenburg Oct 2 '12 at 16:26
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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. –  Zhen Lin Oct 2 '12 at 19:39
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2 Answers 2

This does not exactly answer your question, but it should be pointed out that in some situations such as groups, Lie algebras, ... one wants to consider other kinds of tensor products in which the key notion is that of a biderivation. An example of this is the commutator map $[\; ,\; ]: M \times N \to G$ where $M,N$ are normal subgroups of the group $G$. See a bibliography on this nonabelian tensor product with 120 items.

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This does not address my question, -1. –  Martin Brandenburg Oct 2 '12 at 10:39
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I understand your gripe; but sometimes a slightly different question can have more possibilities. Loday and I stumbled on the tensor product I mentioned coming from a different direction, when it embeds in a larger structure called a crossed square, relevant to homotopy theory. So I was hoping you might be interested. Not to worry. –  Ronnie Brown Oct 2 '12 at 20:38
    
I am interested, and many others probably too. But this is not the right place to advertise your work. Because it is not related to my question at all. –  Martin Brandenburg Oct 3 '12 at 9:43
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EDIT. This doesn't work: see the comments.

I'm not sure what you're thinking of when you say "usual proof of the associativity", but the one I have in mind doesn't use commutativity.

Define a multihomomorphism of algebras to be a function of finitely many variables that is a homomorphism in each variable separately, and then consider "trihomomorphisms" (terhomomorphisms?) $f : A \times B \times C \to D$. It is clear that we get a unique bihomomorphism $g : A \times (B \otimes C) \to D$ such that $g(a, b \otimes c) = f(a, b, c)$ – just think of $f$ as an $A$-indexed family of bihomomorphisms – and we also have a unique bihomomorphism $h : (A \otimes B) \times C \to D$ such that $h(a \otimes b, c) = f(a, b, c)$. Conversely, any bihomomorphism $A \times (B \otimes C) \to D$ or $(A \otimes B) \times C \to D$ gives rise to a unique trihomomorphism. Thus, we have natural bijections $$\textrm{Multi}(A \otimes B, C; D) \cong \textrm{Multi}(A, B, C; D) \cong \textrm{Multi}(A, B \otimes C; D)$$ and so the Yoneda lemma implies $$(A \otimes B) \otimes C \cong A \otimes (B \otimes C)$$ as required. The same argument using "quadrihomomorphisms" (quaterhomomorphisms?) should be enough to verify the hexagon axiom.

It's not so clear to me how to make this argument work in the general setting of strong monads over a symmetric monoidal closed category... but it probably can be done, since Kock [1971] proved it for commutative monads.

As for literature – Borceux mentions it very, very briefly in [Handbook of categorical algebra, Vol. 2, §3.10].

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Nicely said! This is exactly the kind of reasoning I had in mind with my comment above. –  Manny Reyes Oct 2 '12 at 23:48
    
@Zhen Lin: You should check the details more carefully! Namely, why is $g$ a homomorphism in the first variable? Write it down in the case for monoids and you will see a problem. –  Martin Brandenburg Oct 3 '12 at 6:43
    
Yes, I see the problem now. Hmmm. I'll have to think about this more... –  Zhen Lin Oct 3 '12 at 9:20
    
Borceux only treats the commutative case (which is very similar to the well-known case of modules, see also my remarks about Durov's generalized rings). As for the general case, they claim "But the tensor product optained in this way is often not interesting when the theory is not "commutative enough"" and indicate what happens for groups. Why is the tensor product not interesting when it imposes some sort of commutativity? See also the example $\mathbb{N} \otimes -$ for monoids: it forces the power maps to be homomorphisms. –  Martin Brandenburg Oct 3 '12 at 9:47
    
@Martin: As you asked for other special cases, I mention that I have used a number of special cases of bimorphisms: cubical sets, $\omega$-groupoids, crossed complexes, chain complexes, simplicial abelian groups. The last 2 cases are interesting because the 2 categories are equivalent (if chain complexes are $0$ in dim $<0$) but this equivalence does not preserve the usual tensor products. All these are of interest in obtaining monoidal closed structures. If you want detailed references, let me know. –  Ronnie Brown Oct 14 '12 at 16:43
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