Consider a strict monoidal category $C$, and let $E := Func(C, C)$ be the endofunctor category. We already know that any endofunctor category can be "given" a strict monoidal structure, by setting the product to be the composition $F \circ G$ on objects and the "horizontal composition" $\alpha \underline{\circ} \beta$ on morphisms.

But if we "use" the fact that $C$ is a strict monoidal category, we can try to "give" $E$ another monoidal structure, what i call "object-wise" in the title:

Given two functors $F, G$ we define their product using the product on $C$:

$(F \otimes G)(a) = F(a) \otimes G(a)$.

In order to define the tensor product on morphisms (in $E$, that is natural transformations), consider two of them:

$\alpha: F_1 \to F_2$

$\beta: G_1 \to G_2$

we then have the following (commutative by naturality) diagram (where the product of natural transformations is defined as the product of the components at each object, as morphisms of $C$):


and we can mimic the definition of horizontal composition, and call this "horizontal tensor product":

$\alpha \underline{\otimes} \beta := (\alpha \otimes 1_{G_2}) \circ (1_{F_1} \otimes \beta) = (1_{F_2} \otimes \beta) \circ (\alpha \otimes 1_{G_1})$

Even though this is far from a proof that the above definition actually endows a monoidal structure, let alone strict, my (wild) guess is that such a proof would be quite straightforward: we should be able to lift associativity, left and right unitors and (eventually) braiding or commutation, as well as strictness, from the "base" category $C$.

I tried google, but couldn't find anything. Am i missing something, that makes this a special case of another well-known construction? Or is this just an uninteresting construction? Or maybe my wild guess is plain wrong, and this does not endow $E$ with a different monoidal structure?

  • $\begingroup$ Without working through the details, I'd imagine this construction should work for any functor category $C^J$; no reason to limit the domain to $C$. $\endgroup$
    – user14972
    May 17, 2017 at 19:26
  • $\begingroup$ More explicitly, if you can formulate everything right, everything follows by simply exponentiation by $J$. e.g. the operation itself can be constructed by noting any functor $C \times C \to C$ gives a functor $(C \times C)^J \to C^J$ by composition, and $(C \times C)^J \cong C^J \times C^J$. Alternatively, if various identities on 'elements' of $C$ hold, they should hold when you substitute functor values for those elements. $\endgroup$
    – user14972
    May 17, 2017 at 19:33
  • $\begingroup$ That is right, there is no need to take an endofunctor category: any functor category, with as codomain a monoidal category, should be perfectly good for this construction. $\endgroup$ May 17, 2017 at 20:01

1 Answer 1


$\require{AMScd}$As you was told in the comment, the monoidal structure is defined by two functors $\otimes \colon C\times C \to C$ and $I\colon * \to C$ (choice of the unit object) that fit in diagrams $$ \begin{CD} C\times C \times C @>id\times \otimes>> C\times C \\ @V\otimes \times idVV @VV\otimes V\\ C \times C @>>\otimes> C \end{CD} \qquad\qquad \begin{CD} C\times * @>C\times I>> C\times C \\ @A\cong AA @AA I\times C A\\ C @>>\cong> *\times C \end{CD} $$ where in the second one the antidiagonal is filled with $\otimes : C\times C \to C$, and the two resulting triangles are commutative (there's no way to draw diagonal arrows in amsCD, huh?), plus other conditions that you know very well. Now, exponentiation with any category $\cal J$ is a functor ${\bf Cat}\to{\bf Cat}$ that commutes with limits, in particular with finite products, so that if $\otimes^J : (C\times C)^J \cong C^J \times C^J \to C^J$ is your new candidate tensor product, the same diagrams will commute basically in the same way.

This is an easy and tedious exercise (I suspect that's the way mathematicians say "I didn't do it, and I never will because it's boring, but I'd know what to do if I had to").

Let me expand with an interesting detour. You are probably aware of the fact that whenever you have a group $G$ (a monoid is enough, but the definition is uglier) then the set of functions $f:G\to\mathbb C$ becomes a group with respect to the convolution product $$ f * g \colon x \mapsto \int_{\lambda\in G} f(\lambda)g(x\lambda^{-1})d\mu_G $$ where $d\mu_G$ is a measure that you canonically endow $G$ with.

You can do it with functors, too!

If $C$ is a monoidal category, then the category of presheaves $[C, {\bf Set}]$ is monoidal not only because it is cartesian, but also because it is endowed with a convolution product, called Day convolution, and defined by the following colimit: consider the diagram $$ \begin{CD} C\times C @>F\times G>> {\bf Set}\times{\bf Set} @>\times>> {\bf Set}\\ @V\otimes VV \\ C \end{CD} $$ for two given presheaves $F,G$. Send an object $c\in C$ into the colimit $$ \underset{a\otimes b\to c}{\mathrm{colim}}\; Fa\times Gb $$ where the subscript is a shorthand for "over all the triples $(a,b,\phi : a\otimes b\to c)$". A massive amount of computations shows, then, that this is a monoidal structure. Or, if you don't want to do this amount of computations, you can package the above colimit as a coend ;-) nudge nudge. $$ \int^{a,b}\hom(a\otimes b,c)\times Fa\times Gb $$ Of course, under suitable assumptions this construction survives to the case of $[C,C]$.

  • 2
    $\begingroup$ You're talking about two different monoidal structures here, without mentioning it: the Day convolution is dramatically different than the pointwise product, which corresponds to the ordinary product of complex-valued functions. If you want a Day convolution on endofunctors of a possibly large category, by the way, then you'd better only be considering the small endofunctors. I think it's non-tedious to prove the result about exponentiation of monoidal products by identifying strict monoidal categories with monoids in Cat. Exponentiation is a monoidal functor since it preserves products. $\endgroup$ May 17, 2017 at 22:50
  • $\begingroup$ > "the Day convolution is dramatically different than the pointwise product" it is, that's why I wanted to mention it :) $\endgroup$
    – fosco
    May 18, 2017 at 7:44

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