Suppose that $C$ is a complete closed monoidal category and $I$ is any small category. Then the functor category $Fun(I,C)$ is again a closed monoidal category with the pointwise tensor product $F \otimes G (x) = F(x) \otimes G(x)$. See here

I am wondering about an analog of this for strict 2-categories. I just care about the cartesian setting. So, specifically let $C$ be a complete strict 2-category whose underlying 1-category is cartesian closed. Let $I$ be a small strict 2-category (though I just care about the case that $I = (0 \to 1)$ is the arrow 1-category). Let $Lax(I,C)$ denote the lax functor 2-category of strict functors, lax natural transformations, and modifications.

Is it the case that the underlying category of $Lax(I,C)$ is Cartesian closed?

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    $\begingroup$ Consider the case when the underlying category of $C$ is trivial, so 2-endomorphisms of the identity form a commutative monoid. Then the underlying category of $\mathrm{Lax}(0\to1,C)$ will have single object with that endomorphism monoid, and it hardly ever has binary products. $\endgroup$ Apr 23, 2015 at 13:21
  • $\begingroup$ So the monoidal structure on C must be compatible with the 2-category structure on C. go ahead and assume that. $\endgroup$ Apr 23, 2015 at 13:25
  • $\begingroup$ Well in certain sense it is in that example. Could you add what kind of compatibility do you mean? $\endgroup$ Apr 23, 2015 at 13:32
  • $\begingroup$ For simplicity we might suppose that C has strict 2-products, and that this gives the monoidal structure. Your example doesn't have that unless C is trivial. $\endgroup$ Apr 23, 2015 at 13:39
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    $\begingroup$ On the otherhand your example does give a monoidal 2-category, just not cartesian monoidal. In that case the lax functor category does indeed happen to be monoidal, just not Cartesian monoidal, as expected. $\endgroup$ Apr 23, 2015 at 13:40

1 Answer 1


At the generality of closed monoidal categories, I think I have a counterexample that will go into my paper with Claudia Scheimbauer --- I need to go over it one more time, so let me not do it here.

But you care most about the Cartesian case. Let $\mathcal C$ be a strict 2-category, by which I mean a 1-category enriched in $\mathrm{Cat}$, the 1-category of categories. For objects $X,Y \in \mathcal C$, I will write $\mathcal C(X,Y) \in \mathrm{Cat}$ for the enriched hom, and $\mathcal C_0(X,Y) \in \mathrm{Set}$ for the uninriched hom. It is worth distinguishing two notions of "Cartesian". Let's say that an enriched product of $X$ and $Y$ is an object $X \times Y$ and a natural-in-$A$ isomorphism of categories $\mathcal C(A,X\times Y) \cong \mathcal C(A,X) \times \mathcal C(A,Y)$, and an unenriched product is an object $X \times_0 Y$ and a natural-in-$A$ isomorphism of SETS $\mathcal C_0(A,X\times_0 Y) \cong \mathcal C_0(A,X) \times \mathcal C_0(A,Y)$. Then any enriched product is also unenriched, but the converse I think can fail. (If, say, $\mathcal C$ is tensored over finite categories, then unenriched products are also enriched products.) So I will assume you mean that $\mathcal C$ has enriched products.

Let $X \overset x \to X'$ and $Y \overset y \to Y'$ be objects of the lax arrow category, since you only care about the case $I = (\bullet \to \bullet)$. I want to prove that their product is $(X \times Y) \overset{x\times y}\to (X' \times Y')$. Let's test this by mapping in with some other object $A \overset a \to A'$. Such a map consists of:

  • A map $(f,g) : A \to (X \times Y)$. There are $\mathcal C(A,X\times Y) = \mathcal C(A,X) \times \mathcal C(A,Y)$ of these.
  • A map $(f',g'): A' \to (X' \times Y')$. There are $\mathcal C(A',X' \times Y') = \mathcal C(A',X') \times \mathcal C(A',Y')$ of these.
  • A 1-morphism in $\mathcal C(A,X' \times Y') = \mathcal C(A,X') \times \mathcal C(A,Y')$ between certain compositions of the above data with $a,x,y$.

Now, I think that naturality and functoriality of $\mathcal C, \times$ tell you that the 1-morphism you need, after writing everything as a product, is something with source $(f'a,g'a)$ and target $(xf,yg)$, which is exactly what you need it to be.

So I think in this case the answer is Yes, the lax functor category has products, and they are computed pointwise.

  • $\begingroup$ Well, I misread, and didn't see you wanted cartesian closed categories, rather than just cartesian categories. But I think via a "functor of points" approach the same coordinate-full calculation works. $\endgroup$ Apr 23, 2015 at 17:09
  • $\begingroup$ Yes, the point is whether or not there are inner homs. $\endgroup$ Apr 24, 2015 at 8:57

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