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

Under Construction

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

$$\partial \alpha :y\to y,$$\partial\alpha: y\to y,

I want to consider two endomorphisms

$$s_{\alpha }:x\to x$$s_\alpha: x\to x

and

$$t_{\alpha }:y\to y$$t_\alpha: y\to y

satisfying

$$t_{\alpha }\circ f=g\circ s_{\alpha }.$$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

$$s_{\alpha }(a\to b)={\alpha }_a$$s_\alpha(a\to b) = \alpha_a

and

$$t_{\alpha }(a\to b)={\alpha }_b.$$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 $f,g:x\to y$ and a 2-morphism $\alpha :f\Rightarrow g$ as illustrated below

$$\begin{array}{rlr}y& \stackrel{f}{\leftarrow }& x\\ 1_y\downarrow & {}^{\alpha }\Downarrow & 1_x\downarrow \\ y& \underset{g}{\leftarrow }& x\end{array}$$\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 $\alpha$ as measuring the failure of this diagram to commute.

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

$$\partial \alpha :y\to y$$\partial\alpha: y\to y

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

$$

$$\partial \alpha \circ f=g.$$\partial\alpha\circ f = g.

$$

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

Vertical Composition

Next, consider three morphisms $f,f\prime ,f″:x\to y$ and two 2-morphisms $\alpha :f\to f\prime$ and $\alpha \prime :f\prime \to f″$. The two 2-morphisms can be vertically composed giving a third 2-morphism $\alpha \prime \alpha :f\to f″$.

$$\begin{array}{rlr}y& \stackrel{f}{\leftarrow }& x\\ 1_y\downarrow & {}^{\alpha }\Downarrow & 1_x\downarrow \\ y& \underset{f\prime }{\leftarrow }& x\\ 1_y\downarrow & {}^{\alpha \prime }\Downarrow & 1_x\downarrow \\ y& \underset{f″}{\leftarrow }& x\end{array}$$\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 $\alpha$ we have the following relations

$$\partial \alpha \circ f=f\prime ,$$\partial\alpha\circ f = f',

$$

$$\partial \alpha \prime \circ f\prime =f″,$$\partial\alpha'\circ f' = f'',

and

$$\partial (\alpha \prime \alpha )\circ f=f″.$$\partial(\alpha'\alpha)\circ f = f''.

Putting this together indicates the following rule for vertical composition:

$$

$$\partial (\alpha \prime \alpha )\circ f=\partial \alpha \prime \circ \partial \alpha \circ f.$$\partial(\alpha'\alpha)\circ f = \partial\alpha'\circ\partial\alpha\circ f.

$$

Horizontal Composition

Now, consider four morphisms $f_1,f_1:y\to z$ and $f_2,f_2\prime :x\to y$ together with two 2-morphisms ${\alpha }_1:f_1\to f_1\prime$ and ${\alpha }_2:f_2\to f_2\prime$. The two 2-morphisms can be horizontally composed giving a third 2-morphism ${\alpha }_1\circ {\alpha }_2:(f_1\circ f_2)\to (f_1\prime \circ f_2\prime )$.

$$\begin{array}{rlrlr}z& \stackrel{f_1}{\leftarrow }& y& \stackrel{f_2}{\leftarrow }& x\\ 1_z\downarrow & {}^{{\alpha }_1}\Downarrow & 1_y\downarrow & {}^{{\alpha }_2}\Downarrow & 1_x\downarrow \\ z& \underset{f_1\prime }{\leftarrow }& y& \underset{f_2\prime }{\leftarrow }& x\end{array}$$\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 $\alpha$ we have the following relations

$$\partial {\alpha }_1\circ f_1=f_1\prime ,$$\partial\alpha_1\circ f_1 = f_1',

$$

$$\partial {\alpha }_2\circ f_2=f_2\prime ,$$\partial\alpha_2\circ f_2 = f_2',

and

$$\partial ({\alpha }_1\circ {\alpha }_2)\circ f_1\circ f_2=f_1\prime \circ f_2\prime .$$\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:

$$

$$\partial ({\alpha }_1\circ {\alpha }_2)\circ f_1\circ f_2=\partial {\alpha }_1\circ f_1\circ \partial {\alpha }_2\circ f_2.$$\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 $f_1,f_1\prime ,f_1″:y\to z$ and $f_2,f_2\prime ,f_2″:x\to y$ and 4 2-morphisms ${\alpha }_1:f_1\to f_1\prime$, ${\alpha }_1\prime :f_1\prime \to f_1″$, ${\alpha }_2:f_2\to f_2\prime$, and ${\alpha }_2\prime :f_2\prime \to f_2″$ as illustrated below:

$$\begin{array}{rlrlr}z& \stackrel{f_1}{\leftarrow }& y& \stackrel{f_2}{\leftarrow }& x\\ 1_z\downarrow & {}^{{\alpha }_1}\Downarrow & 1_y\downarrow & {}^{{\alpha }_2}\Downarrow & 1_x\downarrow \\ z& \underset{f_1\prime }{\leftarrow }& y& \underset{f_2\prime }{\leftarrow }& x\\ 1_z\downarrow & {}^{{\alpha }_1\prime }\Downarrow & 1_y\downarrow & {}^{{\alpha }_2\prime }\Downarrow & 1_x\downarrow \\ z& \underset{f_1″}{\leftarrow }& y& \underset{f_2″}{\leftarrow }& x\end{array}$$\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

$$({\alpha }_1\prime {\alpha }_1)\circ ({\alpha }_2\prime {\alpha }_2).$$(\alpha_1'\alpha_1)\circ(\alpha_2'\alpha_2).

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

$$({\alpha }_1\prime \circ {\alpha }_2\prime )({\alpha }_1\circ {\alpha }_2).$$(\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

$$({\alpha }_1\prime {\alpha }_1)\circ ({\alpha }_2\prime {\alpha }_2)=({\alpha }_1\prime \circ {\alpha }_2\prime )({\alpha }_1\circ {\alpha }_2).$$(\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

$$\begin{array}{rl}\partial [({\alpha }_1\prime {\alpha }_1)\circ ({\alpha }_2\prime {\alpha }_2)]\circ f_1\circ f_2& =\partial ({\alpha }_1\prime {\alpha }_1)\circ f_1\circ \partial ({\alpha }_2\prime {\alpha }_2)\circ f_2\\ & =\partial {\alpha }_1\prime \circ \partial {\alpha }_1\circ f_1\circ \partial {\alpha }_2\prime \circ \partial {\alpha }_2\circ f_2\\ & =\partial {\alpha }_1\prime \circ f_1\prime \circ \partial {\alpha }_2\prime \circ f_2\prime \\ & =\partial ({\alpha }_1\prime \circ {\alpha }_2\prime )\circ f_1\prime \circ f_2\prime \\ & =\partial ({\alpha }_1\prime \circ {\alpha }_2\prime )\circ \partial {\alpha }_1\circ f_1\circ \partial {\alpha }_2\circ f_2\\ & =\partial ({\alpha }_1\prime \circ {\alpha }_2\prime )\circ \partial ({\alpha }_1\circ {\alpha }_2)\circ f_1\circ f_2\\ & =\partial [({\alpha }_1\prime \circ {\alpha }_2\prime )({\alpha }_1\circ {\alpha }_2)]\circ f_1\circ f_2.\end{array}$$\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 $({\alpha }_1\prime {\alpha }_1)\circ ({\alpha }_2\prime {\alpha }_2)$ on $f_1\circ f_2$ is the same as the action of $({\alpha }_1\prime \circ {\alpha }_2\prime )({\alpha }_1\circ {\alpha }_2)$ on $f_1\circ f_2$ 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 $({\alpha }_1\prime {\alpha }_1)\circ ({\alpha }_2\prime {\alpha }_2)=({\alpha }_1\prime \circ {\alpha }_2\prime )({\alpha }_1\circ {\alpha }_2)$; you've only proved that $\partial [({\alpha }_1\prime {\alpha }_1)\circ ({\alpha }_2\prime {\alpha }_2)]=\partial [({\alpha }_1\prime \circ {\alpha }_2\prime )({\alpha }_1\circ {\alpha }_2)]$. (Actually, you haven't even proved that, since you've also got those $f$s 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 $2$-groupoid (or even some more general strict $2$-category in which morphisms are invertible but $2$-morphisms might not be). Then there is a well-defined boundary operation, with $\partial \alpha =g\circ f^{-1}$ for $\alpha :f\Rightarrow g$ (as you know). But this is far from injective, since there may be many $2$-morphisms from $f$ to $g$. So the interchange law that you've proved amounts to $(f_1″\circ f_2″\circ f_2^{-1}\circ f_1^{-1}=(f_1″\circ f_2″)\circ (f_1\circ f_2{)}^{-1}$, which is true enough, but this does nothing to prove the actual interchange law, which is an equation between $2$-morphisms.

—Toby Bartels

2-Groupoids

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

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

For vertical composition, we obtain

$$\partial (\alpha \prime \alpha )=\partial \alpha \prime \circ \partial \alpha$$\partial(\alpha'\alpha) = \partial\alpha'\circ\partial\alpha

and for horizontal composition

$$\partial (\alpha \prime \circ \alpha )=\partial \alpha \prime \circ f\circ \partial \alpha \circ f^{-1}.$$\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 $\alpha :f\to g$, in addition to the boundary morphism $\partial \alpha$, we can consider source and target morphisms

$$s(\alpha )=f$$s(\alpha) = f

and

$$t(\alpha )=g.$$t(\alpha) = g.

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

$$\partial \alpha \circ s(\alpha )=t(\alpha ).$$\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. $s(\alpha )=1_x$, then the boundary of the 2-morphism is equal to the target

$$\partial \alpha =t(\alpha ).$$\partial\alpha = t(\alpha).

Horizontally composing two such 2-morphisms results in

$$\partial (\alpha \prime \circ \alpha )=\partial \alpha \prime \circ \partial \alpha =t(\alpha \prime )\circ t(\alpha )$$\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 $C$ and $D$, functors $F,G:C\to D$, and a natural transformation $\alpha :F\Rightarrow G$, we’d like to understand the endofunctor (if it exists)

$$\partial \alpha :D\to D$$\partial\alpha:D\to D

satisfying

$$\partial \alpha \circ F=G.$$\partial\alpha\circ F = G.

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

$$a\to b.$$a\to b.