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

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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:y→y,\partial\alpha: y\to y,

I want to consider two endomorphisms

s α:x→xs_\alpha: x\to x

and

t α:y→yt_\alpha: y\to y

satisfying

t α∘f=g∘s 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(a→b)=α as_\alpha(a\to b) = \alpha_a

and

t α(a→b)=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:x→yf,g:x\to y and a 2-morphism α:f⇒g\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:y→y\partial\alpha: y\to y

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

{}

∂α∘f=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″:x→yf,f',f'':x\to y and two 2-morphisms α:f→f′\alpha:f\to f' and α′:f′→f″\alpha':f'\to f''. The two 2-morphisms can be vertically composed giving a third 2-morphism α′α:f→f″\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

∂α∘f=f′,\partial\alpha\circ f = f',

{}

∂α′∘f′=f″,\partial\alpha'\circ f' = f'',

and

∂(α′α)∘f=f″.\partial(\alpha'\alpha)\circ f = f''.

Putting this together indicates the following rule for vertical composition:

{}

∂(α′α)∘f=∂α′∘∂α∘f.\partial(\alpha'\alpha)\circ f = \partial\alpha'\circ\partial\alpha\circ f.

{}

Horizontal Composition

Now, consider four morphisms f 1,f 1:y→zf_1,f_1:y\to z and f 2,f 2′:x→yf_2,f_2':x\to y together with two 2-morphisms α 1:f 1→f 1′\alpha_1:f_1\to f_1' and α 2:f 2→f 2′\alpha_2:f_2\to f_2'. The two 2-morphisms can be horizontally composed giving a third 2-morphism α 1∘α 2:(f 1∘f 2)→(f 1′∘f 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 1∘f 1=f 1′,\partial\alpha_1\circ f_1 = f_1',

{}

∂α 2∘f 2=f 2′,\partial\alpha_2\circ f_2 = f_2',

and

∂(α 1∘α 2)∘f 1∘f 2=f 1′∘f 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 1∘α 2)∘f 1∘f 2=∂α 1∘f 1∘∂α 2∘f 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″:y→zf_1,f_1',f_1'':y\to z and f 2,f 2′,f 2″:x→yf_2,f_2',f_2'':x\to y and 4 2-morphisms α 1:f 1→f 1′\alpha_1:f_1\to f_1', α 1′:f 1′→f 1″\alpha_1':f_1'\to f_1'', α 2:f 2→f 2′\alpha_2:f_2\to f_2', and α 2′:f 2′→f 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 1′α 1)∘(α 2′α 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 1′∘α 2′)(α 1∘α 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 1′α 1)∘(α 2′α 2)=(α 1′∘α 2′)(α 1∘α 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 1′α 1)∘(α 2′α 2)]∘f 1∘f 2 =∂(α 1′α 1)∘f 1∘∂(α 2′α 2)∘f 2 =∂α 1′∘∂α 1∘f 1∘∂α 2′∘∂α 2∘f 2 =∂α 1′∘f 1′∘∂α 2′∘f 2′ =∂(α 1′∘α 2′)∘f 1′∘f 2′ =∂(α 1′∘α 2′)∘∂α 1∘f 1∘∂α 2∘f 2 =∂(α 1′∘α 2′)∘∂(α 1∘α 2)∘f 1∘f 2 =∂[(α 1′∘α 2′)(α 1∘α 2)]∘f 1∘f 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 1′α 1)∘(α 2′α 2)(\alpha_1'\alpha_1)\circ(\alpha_2'\alpha_2) on f 1∘f 2f_1\circ f_2 is the same as the action of (α 1′∘α 2′)(α 1∘α 2)(\alpha_1'\circ\alpha_2')(\alpha_1\circ\alpha_2) on f 1∘f 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 1′α 1)∘(α 2′α 2)=(α 1′∘α 2′)(α 1∘α 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 1′α 1)∘(α 2′α 2)]=∂[(α 1′∘α 2′)(α 1∘α 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=g∘f −1\partial\alpha = g \circ f^{-1} for α:f⇒g\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 1″∘f 2″∘f 2 −1∘f 1 −1=(f 1″∘f 2″)∘(f 1∘f 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=g∘f −1.\partial\alpha = g\circ f^{-1}.

For vertical composition, we obtain

∂(α′α)=∂α′∘∂α\partial(\alpha'\alpha) = \partial\alpha'\circ\partial\alpha

and for horizontal composition

∂(α′∘α)=∂α′∘f∘∂α∘f −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:f→g\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

∂α∘s(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

∂(α′∘α)=∂α′∘∂α=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:C→DF,G:C\to D, and a natural transformation α:F⇒G\alpha:F\Rightarrow G, we’d like to understand the endofunctor (if it exists)

∂α:D→D\partial\alpha:D\to D

satisfying

∂α∘F=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

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