Eric Forgy Boundary of a 2-Morphism and the Interchange Law

Contents

Under Construction

Contents

Attempt #2

Introduction

Thank you to Toby for helping me think this through. We had a discussion on some of these issues on the n-Forum.

He helped convince me that the boundary of a 2-morphism might not be exactly what I’m looking for. However, the ideas I presented here seem to be somehow neat and I can’t help but think there is something interesting to be uncovered here.

Source and Target Endomorphisms

So now, in my second attempt, instead of a single boundary endomorphism

α:yy,\partial\alpha: y\to y,

I want to consider two endomorphisms

s α:xxs_\alpha: x\to x

and

t α:yyt_\alpha: y\to y

satisfying

t αf=gs α.t_\alpha\circ f = g\circ s_\alpha.

Natural Transformations

The motivation for the above is the archetypal example of a strict 2-category Cat. In this case the 2-morphisms are natural transformations and the source and target endofunctors map to the components via

s α(ab)=α as_\alpha(a\to b) = \alpha_a

and

t α(ab)=α b.t_\alpha(a\to b) = \alpha_b.

Attempt #1

Introduction

In this note, we examine a boundary operation on 2-morphisms and derive its resulting calculus. Using this calculus, we derive the interchange law.

Boundary

To begin, consider two morphisms f,g:xyf,g:x\to y and a 2-morphism α:fg\alpha:f\Rightarrow g as illustrated below

y f x 1 y α 1 x y g x\begin{aligned} y &\stackrel{f}{\leftarrow} & x \\ \mathllap{1_y}{\downarrow} & {}^\alpha\Downarrow & \mathrlap{\,\,\,\,1_x}{\downarrow} \\ y &\underset{g}{\leftarrow} & x \end{aligned}

We would like to conceptually think of α\alpha as measuring the failure of this diagram to commute.

The approach we take here to make this explicit is to define a new endomorphism

α:yy\partial\alpha: y\to y

such that when we insert it into the diagram, the composites become equal, i.e.

{}

αf=g.\partial\alpha\circ f = g.

{}

In a way, we can think of α\partial\alpha as one-sided coequalizer that (co)equalizes the two parallel paths.

Vertical Composition

Next, consider three morphisms f,f,f:xyf,f',f'':x\to y and two 2-morphisms α:ff\alpha:f\to f' and α:ff\alpha':f'\to f''. The two 2-morphisms can be vertically composed giving a third 2-morphism αα:ff\alpha'\alpha:f\to f''.

y f x 1 y α 1 x y f x 1 y α 1 x y f x\begin{aligned} y &\stackrel{f}{\leftarrow} & x \\ \mathllap{1_y}{\downarrow} & {}^\alpha\Downarrow & \mathrlap{\,\,\,\,1_x}{\downarrow} \\ y &\underset{f'}{\leftarrow} & x \\ \mathllap{1_y}{\downarrow} & {}^{\alpha'}\Downarrow & \mathrlap{\,\,\,\,1_x}{\downarrow} \\ y &\underset{f''}{\leftarrow} & x \end{aligned}

From the definition of the boundary of α\alpha we have the following relations

αf=f,\partial\alpha\circ f = f',

{}

αf=f,\partial\alpha'\circ f' = f'',

and

(αα)f=f.\partial(\alpha'\alpha)\circ f = f''.

Putting this together indicates the following rule for vertical composition:

{}

(αα)f=ααf.\partial(\alpha'\alpha)\circ f = \partial\alpha'\circ\partial\alpha\circ f.

{}

Horizontal Composition

Now, consider four morphisms f 1,f 1:yzf_1,f_1:y\to z and f 2,f 2:xyf_2,f_2':x\to y together with two 2-morphisms α 1:f 1f 1\alpha_1:f_1\to f_1' and α 2:f 2f 2\alpha_2:f_2\to f_2'. The two 2-morphisms can be horizontally composed giving a third 2-morphism α 1α 2:(f 1f 2)(f 1f 2)\alpha_1\circ\alpha_2:(f_1\circ f_2) \to (f_1'\circ f_2').

z f 1 y f 2 x 1 z α 1 1 y α 2 1 x z f 1 y f 2 x\begin{aligned} z &\stackrel{f_1}{\leftarrow} & y &\stackrel{f_2}{\leftarrow} & x \\ \mathllap{1_z}{\downarrow} & \mathllap{{}^{\alpha_1}}{\Downarrow} & \mathllap{1_y}{\downarrow} & \mathllap{{}^{\alpha_2}}{\Downarrow} & \mathrlap{\,\,\,\,1_x}{\downarrow} \\ z &\underset{f_1'}{\leftarrow} & y &\underset{f_2'}{\leftarrow} & x \end{aligned}

Again, from the definition of the boundary of α\alpha we have the following relations

α 1f 1=f 1,\partial\alpha_1\circ f_1 = f_1',

{}

α 2f 2=f 2,\partial\alpha_2\circ f_2 = f_2',

and

(α 1α 2)f 1f 2=f 1f 2.\partial(\alpha_1\circ\alpha_2)\circ f_1\circ f_2 = f_1'\circ f_2'.

Finally, putting this together indicates the following rule for horizontal composition:

{}

(α 1α 2)f 1f 2=α 1f 1α 2f 2.\partial(\alpha_1\circ\alpha_2)\circ f_1\circ f_2 = \partial\alpha_1\circ f_1\circ\partial\alpha_2\circ f_2.

{}

Interchange Law

The interchange law is a consistency requirement relating the vertical and horizontal compositions. In this section, we demonstrate that when there is a well-defined boundary map, the interchange law is satisfied automatically.

To fix the notation, we begin with 6 morphisms f 1,f 1,f 1:yzf_1,f_1',f_1'':y\to z and f 2,f 2,f 2:xyf_2,f_2',f_2'':x\to y and 4 2-morphisms α 1:f 1f 1\alpha_1:f_1\to f_1', α 1:f 1f 1\alpha_1':f_1'\to f_1'', α 2:f 2f 2\alpha_2:f_2\to f_2', and α 2:f 2f 2\alpha_2':f_2'\to f_2'' as illustrated below:

z f 1 y f 2 x 1 z α 1 1 y α 2 1 x z f 1 y f 2 x 1 z α 1 1 y α 2 1 x z f 1 y f 2 x\begin{aligned} z &\stackrel{f_1}{\leftarrow} & y &\stackrel{f_2}{\leftarrow} & x \\ \mathllap{1_z}{\downarrow} & \mathllap{{}^{\alpha_1}}{\Downarrow} & \mathllap{1_y}{\downarrow} & \mathllap{{}^{\alpha_2}}{\Downarrow} & \mathrlap{\,\,\,\,1_x}{\downarrow} \\ z &\underset{f_1'}{\leftarrow} & y &\underset{f_2'}{\leftarrow} & x \\ \mathllap{1_z}{\downarrow} & \mathllap{{}^{\alpha_1'}}{\Downarrow} & \mathllap{1_y}{\downarrow} & \mathllap{{}^{\alpha_2'}}{\Downarrow} & \mathrlap{\,\,\,\,1_x}{\downarrow} \\ z &\underset{f_1''}{\leftarrow} & y &\underset{f_2''}{\leftarrow} & x \end{aligned}

If we vertically compose first and then horizontally compose, we obtain the 2-morphism

(α 1α 1)(α 2α 2).(\alpha_1'\alpha_1)\circ(\alpha_2'\alpha_2).

Conversely, if we horizontally compose first and then vertically compose, we obtain the 2-morphism

(α 1α 2)(α 1α 2).(\alpha_1'\circ\alpha_2')(\alpha_1\circ\alpha_2).

The interchange law states the the resulting 2-morphism should not depend on the order of composition so that

(α 1α 1)(α 2α 2)=(α 1α 2)(α 1α 2).(\alpha_1'\alpha_1)\circ(\alpha_2'\alpha_2) = (\alpha_1'\circ\alpha_2')(\alpha_1\circ\alpha_2).

Let’s now consider the boundary

[(α 1α 1)(α 2α 2)]f 1f 2 =(α 1α 1)f 1(α 2α 2)f 2 =α 1α 1f 1α 2α 2f 2 =α 1f 1α 2f 2 =(α 1α 2)f 1f 2 =(α 1α 2)α 1f 1α 2f 2 =(α 1α 2)(α 1α 2)f 1f 2 =[(α 1α 2)(α 1α 2)]f 1f 2.\begin{aligned} \partial[(\alpha_1'\alpha_1)\circ(\alpha_2'\alpha_2)]\circ f_1\circ f_2 &= \partial(\alpha_1'\alpha_1)\circ f_1\circ\partial(\alpha_2'\alpha_2)\circ f_2 \\ &= \partial\alpha_1'\circ\partial\alpha_1\circ f_1\circ\partial\alpha_2'\circ\partial\alpha_2\circ f_2 \\ &= \partial\alpha_1' \circ f_1'\circ \partial\alpha_2'\circ f_2' \\ &= \partial(\alpha_1'\circ\alpha_2')\circ f_1'\circ f_2' \\ &= \partial(\alpha_1'\circ\alpha_2')\circ\partial\alpha_1\circ f_1\circ\partial\alpha_2\circ f_2 \\ &= \partial(\alpha_1'\circ\alpha_2')\circ\partial(\alpha_1\circ\alpha_2)\circ f_1\circ f_2 \\ &= \partial[(\alpha_1'\circ\alpha_2')(\alpha_1\circ\alpha_2)]\circ f_1\circ f_2. \end{aligned}

This demonstrates that the action of (α 1α 1)(α 2α 2)(\alpha_1'\alpha_1)\circ(\alpha_2'\alpha_2) on f 1f 2f_1\circ f_2 is the same as the action of (α 1α 2)(α 1α 2)(\alpha_1'\circ\alpha_2')(\alpha_1\circ\alpha_2) on f 1f 2f_1\circ f_2 when the boundary is defined. In other words, when the boundary map is defined, horizontal and vertical composition automatically satisfy the interchange law.

I don't agree with this conclusion. What you've shown is that, when a well-defined boundary operation exists, the interchange law is satisfied up to boundaries. But you haven't proved that (α 1α 1)(α 2α 2)=(α 1α 2)(α 1α 2)(\alpha_1'\alpha_1)\circ(\alpha_2'\alpha_2) = (\alpha_1'\circ\alpha_2')(\alpha_1\circ\alpha_2); you've only proved that [(α 1α 1)(α 2α 2)]=[(α 1α 2)(α 1α 2)]\partial[(\alpha_1'\alpha_1)\circ(\alpha_2'\alpha_2)] = \partial[(\alpha_1'\circ\alpha_2')(\alpha_1\circ\alpha_2)]. (Actually, you haven't even proved that, since you've also got those ffs hanging on the end.) So the theorem only holds when you have an injective well-defined boundary operation (and sometimes not even then).

I think that this is likely to be very rare. Consider the case of a strict 22-groupoid (or even some more general strict 22-category in which morphisms are invertible but 22-morphisms might not be). Then there is a well-defined boundary operation, with α=gf 1\partial\alpha = g \circ f^{-1} for α:fg\alpha\colon f \Rightarrow g (as you know). But this is far from injective, since there may be many 22-morphisms from ff to gg. So the interchange law that you've proved amounts to (f 1f 2f 2 1f 1 1=(f 1f 2)(f 1f 2) 1(f_1'' \circ f_2'' \circ f_2^{-1} \circ f_1^{-1} = (f_1'' \circ f_2'') \circ (f_1 \circ f_2)^{-1}, which is true enough, but this does nothing to prove the actual interchange law, which is an equation between 22-morphisms.

Toby Bartels

2-Groupoids

When dealing with groupoids, the boundary map takes on a more intuitive form, i.e.

α=gf 1.\partial\alpha = g\circ f^{-1}.

For vertical composition, we obtain

(αα)=αα\partial(\alpha'\alpha) = \partial\alpha'\circ\partial\alpha

and for horizontal composition

(αα)=αfαf 1.\partial(\alpha'\circ\alpha) = \partial\alpha'\circ f\circ\partial\alpha\circ f^{-1}.

Note that the boundary of horizontally composed 2-morphisms is reminiscent of the semidirect product.

Source and Target Morphisms

Given a 2-morphism α:fg\alpha:f\to g, in addition to the boundary morphism α\partial\alpha, we can consider source and target morphisms

s(α)=fs(\alpha) = f

and

t(α)=g.t(\alpha) = g.

With source and target morphisms, we can write the boundary endomorphism as

αs(α)=t(α).\partial\alpha\circ s(\alpha) = t(\alpha).

2-Morphisms out of Identities

When a 2-morphism has an identity morphism as its source, i.e. s(α)=1 xs(\alpha) = 1_x, then the boundary of the 2-morphism is equal to the target

α=t(α).\partial\alpha = t(\alpha).

Horizontally composing two such 2-morphisms results in

(αα)=αα=t(α)t(α)\partial(\alpha'\circ\alpha) = \partial\alpha'\circ\partial\alpha = t(\alpha')\circ t(\alpha)

so the boundary of horizontally composed 2-morphisms whose sources are identities is simply the composition of the respective target morphisms.

Boundary of a Natural Transformation

Categories, functors, and natural transformations comprise the archetypal example of a strict 2-category. Therefore, it is natural to consider the boundary of a natural transformation. So given categories CC and DD, functors F,G:CDF,G:C\to D, and a natural transformation α:FG\alpha:F\Rightarrow G, we’d like to understand the endofunctor (if it exists)

α:DD\partial\alpha:D\to D

satisfying

αF=G.\partial\alpha\circ F = G.

To make things concrete, we can consider a simple example where CC consists of two objects and one non-identity morphism between them

ab.a\to b.
Revised on August 19, 2010 at 07:50:21 by Eric Forgy