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There are many $2$-categories, which are first specified by certain categories with extra structure; then the $1$- and $2$-morphisms are functors and natural transformations that preserve the extra structure. I want to understand the general procedure in finding the "correct" definitions of these $2$-morphisms, if there is any.

Example 1: Objects are tensor categories. Then $1$-morphisms should be tensor functors (some allow them to be lax) and $2$-morphisms are natural transformations $\eta$ which are compatible with the tensor structure. This means that $\eta(1)$ is an isomorphism and that for every pair of objects $x,y$ we have a commutative diagram which identifies $\eta_{x \otimes y}$ with $\eta_x \otimes \eta_y$.

Example 2: Take as objects cocomplete categories. Then $1$-morphisms are cocontinuous functors and $2$-morphisms are natural transformations $\eta$ which preserve colimits. The latter means that that for every colimit $\colim_i x_i$ the morphism $\eta(x)$ is the colimit of the morphisms $\eta(x_i)$. But wait, this is automatically true! This follows easily from the cocontinuity of the functors and the naturality of $\eta$. In how far is this "coincidence"?

So far I have never seen this definition of a "cocontinuous natural transformation", but actually this property is used very often when dealing with natural transformations in this situation. So perhaps it should be included in the definition? For example the "correct" definition of a homomorphism $f : G \to H$ of groups includes that $f$ preserves the unit, inversion and multiplication, although everyone knows that multiplication is enough and unfortunately some authors then take the "wrong" definition and get the correct one by a lemma. I hope it's clear that I don't want to offend anyone here and there is no "correct" definition, but perhaps the one which fits best into general patters of category theory.

Example 3: Objects are symmetric tensor categories. Then $1$-morphisms are tensor functors which preserve the symmetry (the functor $F$ maps the symmetry $x \otimes y \cong y \otimes x$ to the symmetry $F(x) \otimes F(y) \cong F(y) \otimes F(x)$; again this is a commutative diagram) and $2$-morphisms are natural transformations $\eta$ which are compatible with the tensor structure as in Example 1 and also are compatible with the symmetry. But what should this compatibility mean? Actually I have not been able to write down a diagram which connects $\eta$ with the symmetry and does not directly follow from the naturality. So perhaps we cannot even formulate a compatibiltiy condition here? Again I'm interested in how far this is "coincidence".

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  • $\begingroup$ these aren't sub 2-categories $\endgroup$ Commented Feb 6, 2011 at 16:28

3 Answers 3

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In all of your examples, the "correct" definition of 2-morphism can be obtained from the fact that there is a 2-monad whose algebras and morphisms are the structured categories and functors you describe, so that the 2-morphisms are the 2-morphisms of algebras over that monad. In Example 2, the vacuity of the condition to be a "cocontinuous natural transformation" follows from the fact that the 2-monad is "fully property-like," which in turn follows from its being "lax-idempotent," i.e. every functor between algebras is uniquely a lax structured functor. Some references on 2-monads can be found at http://nlab.mathforge.org/nlab/show/2-monad .

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A. This is really just an aspect of Mike Shulman's answer, but could be of some use in particular cases.

There's a 2-categorical limit called the power (or cotensor) of an object $B$ by the arrow-category $2$. This is an object $B^2$ with the property that morphisms from $A$ to $B^2$ are in bijection with pairs of morphism from A to B with a 2-cell between them. For example if B is a category then $B^2$ is the functor category $[2,B]$. If $B$ is a monoidal category then $B^2$ is $[2,B]$ with the evident (pointwise) monoidal structure.

In each of your examples, and more generally in Mike's setting, this limit exists in the structured 2-category, and is preserved by the forgetful 2-functor into Cat. Normally you would prove this given the definition of 2-cell. But you can also turn this around. Given a structure on B, if you know how to make $B^2$ into a structured object, then you can use this to define the structured 2-cells.

In examples where the structure is given by a 2-monad, and in particular in examples which involve structure described by operations $B^n\to B$, natural transformations between these, and equations, then you can always do this in a "pointwise way". (But if you choose a strange way to make $B^2$ into a structured object you will get a strange notion of 2-cell.)

Suppose, for example, that $B$ is a monoidal category. Once you agree to make $[2,B]$ monoidal in the pointwise way, then you can define a monoidal transformation to be a monoidal functor with codomain $[2,B]$, and this will agree with the standard definition which you referred to.

In the case of a cocomplete category $B$, you don't need to choose how to make $[2,B]$ cocomplete, it just is. And then you can consider cocontinuous functors with codomain $[2,B]$; once again this will give no extra condition to be satisfied by a natural transformation between cocontinuous functors

The case of symmetric monoidal categories can be treated in the same way.

B. Regarding the case of symmetric monoidal categories, there is a general phenomenon here. As you add structure to your objects in the form of operations $B^n\to B$ (like a tensor product) you generally introduce preservation conditions on both morphisms and 2-cells (although there are special cases, as in your Example 2, where the 2-cell part is automatic). But if you introduce structure in the form of natural transformations between the operations $B^n\to B$ (such as a symmetry), this results in new preservation conditions for the morphisms but not for the 2-cells.

C. Despite all this, there can be more than one choice for the 2-cells. The general principles described by Mike (and by me) would suggest that if our structure is categories with pullback, so that our morphisms are pullback-preserving functors, the 2-cells should be all natural transformations between these. But sometimes it's good to consider only those natural transformations for which the naturality squares are pullbacks. (These are sometimes called cartesian natural transformations.) See this paper for example.

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Let me focus in on your third example, but I'll also wave at your other two. Here's everything in full:

A symmetric strong monoidal category consists of:

  • 0-morphisms: a category $C$.
  • 1-morphisms: a functor $\otimes: C \times C \to C$; and a functor $1: \{\ast\} \to C$.
  • 2-morphisms: a natural isomorphism $\alpha: \otimes \circ (\otimes,\operatorname{id}) \to \otimes \circ (\operatorname{id},\otimes)$ of functors $C^{\times 3} \to C$; natural isomorphisms $\lambda: \otimes \circ (1,\operatorname{id}) \to \operatorname{id}$ and $\rho: \otimes \circ (\operatorname{id},1) \to \operatorname{id}$ of functors $C \to C$; and a natural isomorphisms $\sigma: \operatorname{flip}\circ \otimes \to \otimes$ of functors $C^{\times 2} \to C$.
  • 3-morphisms: a pentagon, a triangle, two hexagons, and $\sigma^2 = 1$.
  • (4-morphisms and higher: trivially satisfied, because we've reached the highest category number.)

A strong morphisms of symmetric strong monoidal categories $(C,\otimes_C,\dots) \to (D,\otimes_D,\dots)$ consists of:

  • 1-morphisms: a functor $f: C \to D$.
  • 2-morphisms: a natural isomorphism $\phi: f\circ \otimes_C \to \otimes_D \circ (f,f)$ of functors $C^{\times 2} \to D$; and a natural isomorphism $\varphi: f\circ 1_C \to 1_D$.
  • 3-morphisms: $\phi,\varphi$ should commute with $\alpha,\lambda,\rho,\sigma$
  • (4-morphisms, which would intertwine the data $\phi,\varphi$ with the pentagon, etc., are trivially satisfied.)

(You can, of course, replace the two words "strong" by "lax" or "oplax" by allowing some of the natural isomorphisms to be simply natural transformations, but then you have to decide which direction you want them to go.)

For example, here's the compatibility between $\sigma$ and $\phi$. Let me continue to use $\circ$ for composition of 1-morphisms, and if $f:C\to D$ is a functor and $\xi$ a morphism in $C$, I'll write $f(\xi)$ for the corresponding morphisms in $D$. I'll write composition of 2-morphisms of functors (= morphisms in a category) as $\bullet$. Then the axiom is that $\sigma_D \bullet \phi = \phi \bullet f(\sigma_C)$ as natural isomorphisms $f\circ \otimes_C \to \otimes_D \circ \operatorname{flip} \circ (f,f)$ of functors $C^{\times 2} \to D$. You absolutely do need this axiom. For example, up to isomorphism there is a unique monoidal functor from super vector spaces to $\mathbb Z/2$-representations, but it is not symmetric monoidal.

But you're interested in the next level --- you know this much. You want:

A natural transformation of strong morphisms of symmetric monoidal categories $(f,\phi_f,\varphi_f),(g,\phi_g,\varphi_g) : (C,\dots) \to (D,\dots)$ consists of:

  • 2-morphisms: a natural transformation $\eta: f \to g$.
  • 3-morphisms: $\eta$ should intertwine the $(\phi,\varphi)$s.
  • (4-morphisms and higher are trivially satisfied)

Now for your actual question: why doesn't $\eta$ care about the symmetry $\sigma$? The answer is that you should demand a compatibility between $\eta$ and $\sigma$, but this demand is a certain 4-morphism, which automatically exists.

More generally, your "coincidences" begin to occur at codimension $-2$, because there is a unique $(-2)$-category. We're working at categorical dimension $2$ (objects are categories, which is to say 0-morphisms of a 2-category), so the $(-2)$-codimensional things are 4-morphisms. At codimension $-1$, a condition either holds or it doesn't --- it's a property. At the level of this discussion, the $(-1)$-codimensional morphisms are commutative diagrams of natural transformations. At codimension $0$ and above, it's actual data.

Something similar happens in your second example. For a functor to preserve colimits is a property. So a colimit preserving functor is actually two pieces of data: a 1-morphism (the functor) and a (hell of a lot of) 3-morphism(s). A natural transformation of this should be a 2-morphism and a (bunch of) 4-morphism(s), except that 4-morphisms always canonically exist.


Here is a very good exercise, to test your 2-category-fu.

  1. You know what is the 2-category of strong monoidal categories. Fix a strong monoidal category $(C,\otimes,1,\alpha,\lambda,\rho)$. Write down the 2-category of strong $C$-categories, which is to say categories $X$ with a $C$-action. I'll start. A $C$-category is: (0-morphisms) a category $X$; (1-morphisms) a functor $\otimes: C \times X \to X$; (2-morphisms) ...; (3-morphisms) .... I'll let you figure out what the rest is. I'll also let you figure out what 1-morphisms of $C$-categories are and what are the 2-morphisms.

  2. Repeat the exercise, but this time fix a symmetric structure on $C$, and figure out which are the symmetric $C$-categories. Hint: if $X$ is a symmetric $C$-category, then it shouldn't matter whether $C$ acts from the left or from the right.

  3. Just like not every monoidal functor is symmetric, not every $C$-category is symmetric. How much extra is being symmetric? At which levels of morphisms of $C$-categories do you need to demand more data/properties/...?

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  • $\begingroup$ I'm not sure I agree entirely with your description of the second example. A colimit-preserving functor has a 2-morphism as well: the comparison isomorphism between the image of the colimit and the colimit of the image. It just so happens that that isomorphism is uniquely determined if it exists. $\endgroup$ Commented Feb 7, 2011 at 6:04
  • $\begingroup$ Hrm. No, I disagree. In part, I disagree because I don't think "the colimit" is as well-defined as your comment implies (although I think you'll agree with me). Namely, a colimit of a diagram is any initial cocone over the diagram. So when I say that a functor "preserves colimits", what I'm saying is that any colimit cocone you feed into it comes out as some colimit cocone. In particular, I disagree that the comparison isomorphism should need to exist. $\endgroup$ Commented Feb 7, 2011 at 18:24

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