I know of the following monoidal structures over $\mathbf{Set}$ (taken from here):

  • The Cartesian product $\otimes=\times$ (categorical product)
  • The disjoint union $\otimes=+$ (categorial coproduct)
  • $A\otimes B = A\times S\times B + A + B$ for some given set $S$
  • $A\otimes B = (A\times S\times B\times S)^* \times (A + A\times S \times B) + (B\times S\times A\times S)^* \times (B + B\times S \times A)$ where $*$ is the Kleene star (free monoid construction)
  • If $A=\emptyset$ then $A\otimes B=B$, if $B=\emptyset$ then $A\otimes B=A$, otherwise $A\otimes B=I$

I want to know if there are any more that are known or, conversely, if there are any demonstrations that additional examples don't exist.

Also, what if we restrict to, say, braided or symmetric monoidal categories, or traced monoidal categories (or some example), are there any results/examples regarding the existence/non-existence of monoidal structures over $\mathbf{Set}$ in these cases?

  • $\begingroup$ What's + of sets? $\endgroup$ Commented Jan 28, 2014 at 10:56
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    $\begingroup$ @FernandoMuro It's the disjoint union, I've edited the original to make this explicit. $\endgroup$
    – JMAA
    Commented Jan 28, 2014 at 12:17
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    $\begingroup$ Welcome to MO ! $\endgroup$ Commented Jan 28, 2014 at 14:11
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    $\begingroup$ One simple observation: the unit object must be empty or a singleton. Proof: Call it $U$. Let $M$ be the monoid of set maps $U\to U$. Then the obvious action of $M\times M$ on $U\times U\cong U$must be such that both $M\times 1$ and $1\times M$ are acting by the original action of $U$. Therefore $M$ is commutative. $\endgroup$ Commented Jan 29, 2014 at 2:07
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    $\begingroup$ @Tom: right, more generally in any monoidal category $\text{End}(1)$ must be commutative by the Eckmann-Hilton argument. $\endgroup$ Commented Jan 29, 2014 at 2:11

4 Answers 4


Plausibly we can construct examples from the power series expansion of formal group laws. We can try to write down such examples by writing down nice functions $q(x)$ and hoping that the formal group law

$$q^{-1} (q(x) + q(y)) = \sum s_{i, j} x^i y^j$$

has positive integer coefficients so it can be upgraded to an "analytic bifunctor"

$$\text{Set} \times \text{Set} \ni (A, B) \mapsto \bigsqcup_{i, j} S_{i, j} \times A^i \times B^j \in \text{Set}$$

which is a candidate to be a monoidal structure. For example,

  • the disjoint union corresponds to choosing $q(x) = x$,
  • the cartesian product corresponds to choosing $q(x) = \log x$,
  • the third example corresponds to choosing $q(x) = \log (sx + 1)$.

The fourth example should also arise in this way with $q(x)$ related to the arctangent; the relevant formal group law is $\frac{x + y + 2s xy}{1 - s^2 xy}$ which resembles the tangent addition formula $\frac{x + y}{1 - xy}$. Speaking of which, this should give another family of examples (with $q(x) = \arctan sx$), namely

$$A \otimes B = (S^2 \times A \times B)^{\ast} \times (A + B).$$

Edit: Apparently we are already pretty close to exhausting all of the formal group laws that can be expressed as rational functions! See this blog post. It seems plausible, but I haven't checked, that

$$A \otimes B = (S^2 \times A \times B)^{\ast} \times (A + B + T \times A \times B)$$


Edit #2: Hmm. I can only find an associator in the case already mentioned in the problem, where $T = S + S$. In this case, as mentioned in the link, $A \otimes B$ can be interpreted as the set of all words which start with an element of $A$ or $B$ and then alternate between $A$ and $B$, with the alternations delimited by elements of $S$, and ending with an element of $A$ or $B$. Then $(A \otimes B) \otimes C$ and $A \otimes (B \otimes C)$ both describe the corresponding construction where we don't repeat among the choices $A, B, C$ consecutively.

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    $\begingroup$ Actually one ought to be able to construct an example with infinitely many parameters using the universal formal group law, which apparently (my reference here is definition 5.2.5 of Hazewinkel's Formal Groups and Applications) can be written with positive rational coefficients over a polynomial ring in countably many variables, so by replacing those variables with suitable multiples we should be able to get positive integer coefficients. $\endgroup$ Commented Jan 28, 2014 at 21:30
  • $\begingroup$ This sounds very promising, thank you. Unfortunately it rather stretches my understanding beyond breaking point. I am sure I can read up on formal group laws (thank you for the reference), but I don't see where the link is to monoidal structures over $\mathbf{Set}$. In particular, googling "analytic bifunctor" (with quotes) only returns this page. $\endgroup$
    – JMAA
    Commented Jan 28, 2014 at 22:49
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    $\begingroup$ @JMAA: by an analytic bifunctor I mean a bifunctor $\text{Set} \times \text{Set} \to \text{Set}$ (not working in full generality here) of the form $(A, B) \mapsto \bigsqcup_{i, j} S_{i, j} \times A^i \times B^j$ where $S_{i, j}$ is a family of sets. These can be thought of as formal power series in two variables over a polynomial ring in countably many variables, and if the corresponding formal power series $f(x, y)$ satisfies the axioms of a formal group law then that at least suggests that the bifunctor itself might give a monoidal structure, e.g. if $f(x, f(y, z)) = f(f(x, y), z)$ then... $\endgroup$ Commented Jan 28, 2014 at 22:52
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    $\begingroup$ ...we can at least hope to upgrade this formal power series identity to an associator. I put the term "analytic bifunctor" in quotes because I made it up but it's a natural extension of the notion of analytic functor, which you will find google hits for. $\endgroup$ Commented Jan 28, 2014 at 22:53
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    $\begingroup$ It seems like albany.edu/~lenart/articles/thesis.pdf is attacking the problem of categorifying the universal formal group law, but I can't extract a concrete answer from it on a quick read. $\endgroup$ Commented Jan 29, 2014 at 0:51

The following family of examples can be extracted from Lenart and Ray, Some applications of incidence Hopf algebras to formal group theory and algebraic topology (Postscript). Let $B_2$, $B_3$, $B_4$, … be any sequence of sets. An element of $X \otimes Y$ is a plane tree $T$, where every non-leaf has at least $2$ children, where every leaf is labeled with an element of $X \sqcup Y$, and every non-leaf with $i$ children is labeled with an element of $B_i$, subject to the condition:

If $v$ is a non-leaf but all the children of $v$ are leaves, then the children of $v$ do not come entirely from $X$ nor entirely from $Y$.

To see associativity, note that both $(X \otimes Y) \otimes Z$ and $X \otimes (Y \otimes Z)$ can be canonically bisected with plane trees whose non-leaves are labeled with $X \sqcup Y \sqcup Z$, subject to the analogous conditions. (The analogue of the boxed condition is that the children of $v$ do not come entirely from $X$, nor entirely from $Y$, nor entirely from $Z$.)

I think the logarithm of the corresponding group law is $t - \sum_{i \geq 2} |B_i| t^i$. Since this method can only categorify group laws whose logarithm has integer coefficients, this clearly cannot include many of the most obvious examples.

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    $\begingroup$ Nice! You can get rational coefficients by working in groupoids instead of sets and allowing yourself to take homotopy quotients (decategorification involves taking groupoid cardinality). But I guess this is outside the scope of the original question. $\endgroup$ Commented Jan 29, 2014 at 18:09
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    $\begingroup$ Right, and Lenart does that in some of his papers. But there are lots of examples in the category of sets that can't be obtained from this construction. For example, with this construction, the coefficient of $xy$ in the group law $x \oplus y$ is always even. $\endgroup$ Commented Jan 29, 2014 at 18:12

Here is another possible source of tensor products on $\mathbf{Set}$: transferring tensor products from categories $\mathbf{Set}$ is coreflective in. Unfortunately I have no good example.

Suppose $(\mathbf{C},\otimes,I)$ is a monoidal category with a coreflection from $\mathbf{Set}$; that is, there are functors $F\colon \mathbf{Set} \to \mathbf{C}$ and $G \colon \mathbf{C} \to \mathbf{Set}$ such that $F$ is left adjoint to $G$ with isomorphic unit $\eta_X \colon X \to GF(X)$. Moreover, suppose that $FG \colon \mathbf{C} \to \mathbf{C}$ is monoidal. Then $X \otimes Y := G(FX \otimes FY)$ is a monoidal product on $\mathbf{Set}$. (See e.g. Theorem 5 here.)

For example, these conditions are automatically fulfilled if $\mathbf{C}$ is the category of topological spaces and $G$ is the forgetful functor. However, a quick browse gives results classifying the tensor products on $\mathbf{Top}$ (*), so this is probably not a good source of examples. Are there others?

(*) That are canonical - meaning they just give the tensor product $\times$ on $\mathbf{Set}$ - and closed; but we are precisely looking for noncanonical ones...


So this is a partial answer, using the ideas of Qiaochu Yuan's answer. Whilst this question is obviously of general interest, it turns out that for the specific purposes I have in mind only products where the unit is the singleton are useful (since, as Tom Goodwillie points out, the only possible units in $\mathbf{Set}$ are the unit and the singleton). I leave it here in the hope that it may be useful to other people.

This answer then says that the only monoidal structure over $\mathbf{Set}$ that has the singleton as the unit which can be derived from Qiaochu Yuan's construction is the cartesian product. This is because there are no possible functions $q(x):\mathbb{C}\rightarrow \mathbb{C}$ such that $$q^{-1}(q(x)+q(y)) = \sum_{i,j}s_{i,j}x^iy^j$$ for integer coefficients and that $q(1)=0$ (which would give the tensor unit as the singleton set).

For such a $q$ to exist, it would have to have $q^{-1}(0)=1$ and therefore $$1 = \sum_{i,j}s_{i,j}$$ so that $s_{\alpha,\beta}=1$ for some specific pair $(\alpha,\beta)$ and all other $s_{i\neq\alpha,j\neq\beta}=0$ since they're all positive integers. Moreover, $\alpha=\beta$ since the LHS of the first displayed equation above is symmetric under $x\leftrightarrow y$ interchange. So we have $$\log f(x)f(y) = \log f((xy)^\alpha)$$ where $\log f = q$. Requiring $f$ to be analytic and expanding them as power series, it is easily seen that the only possible choice is $\alpha=1$, which is simply $q(x)=\log x$ and gives rise to the cartesian product, as noted by Qiaochu Yuan.

The same argument and conclusion is found when using $$q^{-1}(q(x)q(y)) = \sum_{i,j}s_{i,j}x^i y^j$$ rather than the above formal group law.

What this means is that if it can be proved that the above construction gives every possible monoidal structure on $\mathbf{Set}$ then all except the cartesian product have the empty set as their unit.

  • $\begingroup$ @QiaochuYuan in which case the argument runs just as above and leads to the same conclusion, just that this time we find $q(x)q(y)=q((xy)^\alpha)$ rather than the same case with $f$. $\endgroup$
    – JMAA
    Commented Jan 29, 2014 at 20:15
  • $\begingroup$ Wait, I don't understand this argument at all. How can you conclude $1 = \sum s_{i, j} x^i y^j$? $\endgroup$ Commented Jan 29, 2014 at 22:47
  • $\begingroup$ @QiaochuYuan Sorry, my mistake (now corrected). From the first equation you set $x=y=1$ in order to find the second equation (which holds whether you're using the sum or multiplicative group laws), which gives you that only one such $s_{i,j}$ is non-zero (and thus equal to unity and must be such $i=j=\alpha$). Then returning to general $x$ and $y$ the only remaining question is which $\alpha$ are possible. $\endgroup$
    – JMAA
    Commented Jan 30, 2014 at 9:44
  • $\begingroup$ Oh, I see. $q$ doesn't even need to be mentioned in this argument; you can just start with an analytic bifunctor $\bigsqcup_{i, j} S_{i, j} \times A^i \times B^j$ and set $A, B$ to the singleton. For the singleton to be the identity I guess other ideas are required (I would be very surprised if this construction came close to exhausting monoidal structures on $\text{Set}$). $\endgroup$ Commented Jan 30, 2014 at 19:26

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