%!TEX root = ../../blob1.tex

\section{Comparing \texorpdfstring{$n$}{n}-category definitions}

\label{sec:comparing-defs}

In \S\ref{sec:example:traditional-n-categories(fields)} we showed how to construct

a topological $n$-category from a traditional $n$-category; the morphisms of the 

topological $n$-category are string diagrams labeled by the traditional $n$-category.

In this appendix we sketch how to go the other direction, for $n=1$ and 2.

The basic recipe, given a disk-like $n$-category $\cC$, is to define the $k$-morphisms

of the corresponding traditional $n$-category to be $\cC(B^k)$, where

$B^k$ is the {\it standard} $k$-ball.

One must then show that the axioms of \S\ref{ss:n-cat-def} imply the traditional $n$-category axioms.

One should also show that composing the two arrows (between traditional and disk-like $n$-categories)

yields the appropriate sort of equivalence on each side.

We emphasize that we are just sketching some of the main ideas in this appendix ---

it falls well short of proving the definitions are equivalent.

%\nn{cases to cover: (a) ordinary $n$-cats for $n=1,2$; (b) $n$-cat modules for $n=1$, also 2?;

%(c) $A_\infty$ 1-cat; (b) $A_\infty$ 1-cat module?; (e) tensor products?}

\subsection{1-categories over \texorpdfstring{$\Set$ or $\Vect$}{Set or Vect}}

\label{ssec:1-cats}

Given a disk-like $1$-category $\cX$ we construct a $1$-category in the conventional sense, $c(\cX)$.

This construction is quite straightforward, but we include the details for the sake of completeness, 

because it illustrates the role of structures (e.g. orientations, spin structures, etc) 

on the underlying manifolds, and 

to shed some light on the $n=2$ case, which we describe in \S \ref{ssec:2-cats}.

Let $B^k$ denote the \emph{standard} $k$-ball.

Let the objects of $c(\cX)$ be $c(\cX)^0 = \cX(B^0)$ and the morphisms of $c(\cX)$ be $c(\cX)^1 = \cX(B^1)$.

The boundary and restriction maps of $\cX$ give domain and range maps from $c(\cX)^1$ to $c(\cX)^0$.

Choose a homeomorphism $B^1\cup_{pt}B^1 \to B^1$.

Define composition in $c(\cX)$ to be the induced map $c(\cX)^1\times c(\cX)^1 \to c(\cX)^1$ 

(defined only when range and domain agree).

By isotopy invariance in $\cX$, any other choice of homeomorphism gives the same composition rule.

Also by isotopy invariance, composition is strictly associative.

Given $a\in c(\cX)^0$, define $\id_a \deq a\times B^1$.

By extended isotopy invariance in $\cX$, this has the expected properties of an identity morphism.

We have now defined the basic ingredients for the 1-category $c(\cX)$.

As we explain below, $c(\cX)$ might have additional structure corresponding to the

unoriented, oriented, Spin, $\text{Pin}_+$ or $\text{Pin}_-$ structure on the 1-balls used to define $\cX$.

For 1-categories based on unoriented balls, 

there is a map $\dagger:c(\cX)^1\to c(\cX)^1$

coming from $\cX$ applied to an orientation-reversing homeomorphism (unique up to isotopy) 

from $B^1$ to itself.

(Of course our $B^1$ is unoriented, i.e.\ not equipped with an orientation.

We mean the homeomorphism which would reverse the orientation if there were one;

$B^1$ is not oriented, but it is orientable.)

Topological properties of this homeomorphism imply that 

$a^{\dagger\dagger} = a$ ($\dagger$ is order 2), $\dagger$ reverses domain and range, and $(ab)^\dagger = b^\dagger a^\dagger$

($\dagger$ is an anti-automorphism).

Recall that in this context 0-balls should be thought of as equipped with a germ of a 1-dimensional neighborhood.

There is a unique such 0-ball, up to homeomorphism, but it has a non-identity automorphism corresponding to reversing the

orientation of the germ.

Consequently, the objects of $c(\cX)$ are equipped with an involution, also denoted $\dagger$.

If $a:x\to y$ is a morphism of $c(\cX)$ then $a^\dagger: y^\dagger\to x^\dagger$.

For 1-categories based on oriented balls, there are no non-trivial homeomorphisms of 0- or 1-balls, and thus no 

additional structure on $c(\cX)$.

For 1-categories based on Spin balls,

the nontrivial spin homeomorphism from $B^1$ to itself which covers the identity

gives an order 2 automorphism of $c(\cX)^1$.

There is a similar involution on the objects $c(\cX)^0$.

In the case where there is only one object and we are enriching over complex vector spaces, this

is just a super algebra.

The even elements are the $+1$ eigenspace of the involution on $c(\cX)^1$, 

and the odd elements are the $-1$ eigenspace of the involution.

For 1-categories based on $\text{Pin}_-$ balls,

we have an order 4 antiautomorphism of $c(\cX)^1$.

For 1-categories based on $\text{Pin}_+$ balls,

we have an order 2 antiautomorphism and also an order 2 automorphism of $c(\cX)^1$,

and these two maps commute with each other.

In both cases there is a similar map on objects.

\noop{

\medskip

In the other direction, given a $1$-category $C$

(with objects $C^0$ and morphisms $C^1$) we will construct a disk-like

$1$-category $t(C)$.

If $X$ is a 0-ball (point), let $t(C)(X) \deq C^0$.

If $S$ is a 0-sphere, let $t(C)(S) \deq C^0\times C^0$.

If $X$ is a 1-ball, let $t(C)(X) \deq C^1$.

Homeomorphisms isotopic to the identity act trivially.

If $C$ has extra structure (e.g.\ it's a *-1-category), we use this structure

to define the action of homeomorphisms not isotopic to the identity

(and get, e.g., an unoriented disk-like 1-category).

The domain and range maps of $C$ determine the boundary and restriction maps of $t(C)$.

Gluing maps for $t(C)$ are determined by composition of morphisms in $C$.

For $X$ a 0-ball, $D$ a 1-ball and $a\in t(C)(X)$, define the product morphism 

$a\times D \deq \id_a$.

It is not hard to verify that this has the desired properties.

\medskip

The compositions of the constructions above, $$\cX\to c(\cX)\to t(c(\cX))$$ 

and $$C\to t(C)\to c(t(C)),$$ give back 

more or less exactly the same thing we started with.  

As we will see below, for $n>1$ the compositions yield a weaker sort of equivalence.

} %end \noop

\medskip

Similar arguments show that modules for disk-like 1-categories are essentially

the same thing as traditional modules for traditional 1-categories.

\subsection{Pivotal 2-categories}

\label{ssec:2-cats}

Let $\cC$ be a disk-like 2-category.

We will construct from $\cC$ a traditional pivotal 2-category.

(The ``pivotal" corresponds to our assumption of strong duality for $\cC$.)

We will try to describe the construction in such a way the the generalization to $n>2$ is clear,

though this will make the $n=2$ case a little more complicated than necessary.

Before proceeding, we must decide whether the 2-morphisms of our

pivotal 2-category are shaped like rectangles or bigons.

Each approach has advantages and disadvantages.

For better or worse, we choose bigons here.

Define the $k$-morphisms $C^k$ of $C$ to be $\cC(B^k) \trans E$, where $B^k$ denotes the standard

$k$-ball, which we also think of as the standard bihedron (a.k.a.\ globe).

(For $k=1$ this is an interval, and for $k=2$ it is a bigon.)

Since we are thinking of $B^k$ as a bihedron, we have a standard decomposition of the $\bd B^k$

into two copies of $B^{k-1}$ which intersect along the ``equator" $E \cong S^{k-2}$.

Recall that the subscript in $\cC(B^k) \trans E$ means that we consider the subset of $\cC(B^k)$

whose boundary is splittable along $E$.

This allows us to define the domain and range of morphisms of $C$ using

boundary and restriction maps of $\cC$.

Choosing a homeomorphism $B^1\cup B^1 \to B^1$ defines a composition map on $C^1$.

This is not associative, but we will see later that it is weakly associative.

Choosing a homeomorphism $B^2\cup B^2 \to B^2$ defines a ``vertical" composition map 

on $C^2$ (Figure \ref{fzo1}).

Isotopy invariance implies that this is associative.

We will define a ``horizontal" composition later.

\begin{figure}[t]

\centering

\begin{tikzpicture}

\newcommand{\vertex}{node[circle,fill=black,inner sep=1pt] {}}

\newcommand{\nsep}{1.8}

\node[outer sep=\nsep](A) at (0,0) {

\begin{tikzpicture}

	\draw (0,0) coordinate (p1);

	\draw (4,0) coordinate (p2);

	\draw (2,1.2) coordinate (pu);

	\draw (2,-1.2) coordinate (pd);

	\draw (p1) .. controls (pu) .. (p2) .. controls (pd) .. (p1);

	\draw (p1)--(p2);

	\draw (p1) \vertex;

	\draw (p2) \vertex;

	\node at (2.1, .44) {$B^2$};

	\node at (2.1, -.44) {$B^2$};

\end{tikzpicture}

};

\node[outer sep=\nsep](B) at (6,0) {

\begin{tikzpicture}

	\draw (0,0) coordinate (p1);

	\draw (4,0) coordinate (p2);

	\draw (2,.6) coordinate (pu);

	\draw (2,-.6) coordinate (pd);

	\draw (p1) .. controls (pu) .. (p2) .. controls (pd) .. (p1);

	\draw[help lines, dashed] (p1)--(p2);

	\draw (p1) \vertex;

	\draw (p2) \vertex;

	\node at (2.1,0) {$B^2$};

\end{tikzpicture}

};

\draw[->, thick, blue!50!green] (A) -- node[black, above] {$\cong$} (B);

\end{tikzpicture}

\caption{Vertical composition of 2-morphisms}

\label{fzo1}

\end{figure}

Given $a\in C^1$, define $\id_a = a\times I \in C^2$ (pinched boundary).

Extended isotopy invariance for $\cC$ shows that this morphism is an identity for 

vertical composition.

Given $x\in C^0$, define $\id_x = x\times B^1 \in C^1$.

We will show that this 1-morphism is a weak identity.

This would be easier if our 2-morphisms were shaped like rectangles rather than bigons.

In showing that identity 1-morphisms have the desired properties, we will

rely heavily on the extended isotopy invariance of 2-morphisms in $\cC$.

This means we are free to add or delete product regions from 2-morphisms.

Let $a: y\to x$ be a 1-morphism.

Define 2-morphsims $a \to a\bullet \id_x$ and $a\bullet \id_x \to a$

as shown in Figure \ref{fzo2}.

\begin{figure}[t]

\centering

\begin{tikzpicture}

\newcommand{\rr}{6}

\newcommand{\vertex}{node[circle,fill=black,inner sep=1pt] {}}

\newcommand{\namedvertex}[1]{node[circle,fill=black,inner sep=1pt] (#1) {}}

\node(A) at (0,0) {

\begin{tikzpicture}

\node[red,left] at (0,0)  {$y$};

\draw (0,0) \vertex arc (-120:-105:\rr) node[red,below] {$a$} arc(-105:-90:\rr) \vertex node[red,below](x2) {$x$};

\draw (0,0) \vertex arc (120:105:\rr) node[red,above] {$a$} arc (105:90:\rr) \vertex node[red,above](x1) {$x$} -- (x2);

\begin{scope}

	\path[clip] (0,0) arc (-120:-60:\rr) arc (60:120:\rr);

	\foreach \x in {0,0.24,...,3} {

		\draw[green!50!brown] (\x,1) -- (\x,-1);

	}

\end{scope}

\draw[red, decorate,decoration={brace,amplitude=5pt}] ($(x1)+(0.2,-0.2)$) -- ($(x2)+(0.2,0.2)$) node[midway, xshift=0.7cm] {$x \times I$};

\end{tikzpicture}

};

\node(B) at (-4,-4) {

\begin{tikzpicture}

\node[red,left] at (0,0) {$y$};

\draw (0,0) \vertex 

	arc (120:105:\rr) node[red,above] {$a$}

	arc (105:90:\rr) node[red,above] {$x$} \namedvertex{x1};

%	arc (90:75:\rr) node[red,above] {$x \times I$};

%	arc (75:60:\rr) \vertex node[red,right] {$x$}

%	arc (-60:-90:\rr) node[red,below] {$a$}

%	arc (-90:-120:\rr);

\draw (0,0)

	arc (-120:-90:\rr) node[red,below] {$a$}

	arc (-90:-61:\rr) \namedvertex{x2} node[red,right] {$x$};

\draw (x1) -- node[red, above=3pt] {$x \times I$} (x2);

\begin{scope}

	\path[clip] (0,0) arc (-120:-60:\rr) arc (60:120:\rr);

	\foreach \x in {0,0.48,...,9} {

		\draw[green!50!brown] (\x/4,1) -- (\x,-1);

	}

\end{scope}

\end{tikzpicture}

};

\node(C) at (4,-4) {

\begin{tikzpicture}[y=-1cm]

\node[red,left] at (0,0) {$y$};

\draw (0,0) \vertex 

	arc (120:105:\rr) node[red,below] {$a$}

	arc (105:90:\rr) node[red,below] {$x$} \namedvertex{x1};

%	arc (90:75:\rr) node[red,below] {$x \times I$}

%	arc (75:60:\rr) \vertex node[red,right] {$x$}

%	arc (-60:-90:\rr) node[red,above] {$a$}

%	arc (-90:-120:\rr);

\draw (0,0)

	arc (-120:-90:\rr) node[red,above] {$a$}

	arc (-90:-61:\rr) \namedvertex{x2} node[red,right] {$x$};

\draw (x1) -- node[red, below=3pt] {$x \times I$} (x2);

\begin{scope}

	\path[clip] (0,0) arc (-120:-60:\rr) arc (60:120:\rr);

	\foreach \x in {0,0.48,...,9} {

		\draw[green!50!brown] (\x/4,1) -- (\x,-1);

	}

\end{scope}

\end{tikzpicture}

};

\draw[->] (A) -- (B);

\draw[->] (A) -- (C);

\end{tikzpicture}

\caption{Producing weak identities from half pinched products}

\label{fzo2}

\end{figure}

As suggested by the figure, these are two different reparameterizations

of a half-pinched version of $a\times I$.

We must show that the two compositions of these two maps give the identity 2-morphisms

on $a$ and $a\bullet \id_x$, as defined above.

Figure \ref{fzo3} shows one case.

\begin{figure}[t]

\centering

\begin{tikzpicture}

\newcommand{\vertex}{node[circle,fill=black,inner sep=1pt] {}}

\newcommand{\nsep}{1.8}

\node(A) at (0,0) {

\begin{tikzpicture}

	\draw (0,0) coordinate (p1);

	\draw (3.6,0) coordinate (p2);

	\draw (2.3,1) coordinate (p3);

	\draw (2.3,-1) coordinate (p4);

	\begin{scope}

		\clip (p1) 	.. controls +(.5,-.5) and +(-.8,0)  .. (p4) -- 

				(p2) -- (p3) .. controls +(-.8,0) and +(.5,.5) .. (p1);

		\foreach \x in {0,0.26,...,4} {

			\draw[green!50!brown] (\x,0) -- (intersection cs: first line={(p2)--(p3)}, second line={(0,0)--(0,1)});

			\draw[green!50!brown] (\x,0) -- (intersection cs: first line={(p2)--(p4)}, second line={(0,0)--(0,1)});

		}

	\end{scope}

	\draw (p1) .. controls ($(p1) + (.5,.5)$) and ($(p3) + (-.8,0)$) .. node[red, above=3pt] {$a$} (p3);

	\draw (p1) .. controls ($(p1) + (.5,-.5)$) and ($(p4) + (-.8,0)$) .. node[red, below=3pt] {$a$} (p4);

	\draw (p3) -- node[red, above=7pt, right=1pt] {$x \times I$} (p2);

	\draw (p4) -- node[red, below=7pt, right=1pt] {$x \times I$} (p2);

	\draw (p1) -- (p2);

	\draw (p1) \vertex;

	\draw (p2) \vertex;

	\draw (p3) \vertex;

	\draw (p4) \vertex;

\end{tikzpicture}

};

\node[outer sep=\nsep](B) at (5.5,0) {

\begin{tikzpicture}

	\draw (0,0) coordinate (p1);

	\draw (3.6,0) coordinate (p2);

	\draw (2.3,1) coordinate (p3);

	\draw (2.3,-1) coordinate (p4);

	\draw (4.6,0) coordinate (p2b);

	\begin{scope}

		\clip (p1) 	.. controls +(.5,-.5) and +(-.8,0)  .. (p4) -- 

				(p2) -- (p3) .. controls +(-.8,0) and +(.5,.5) .. (p1);

		\foreach \x in {0,0.26,...,4} {

			\draw[green!50!brown] (\x,0) -- (intersection cs: first line={(p2)--(p3)}, second line={(0,0)--(0,1)});

			\draw[green!50!brown] (\x,0) -- (intersection cs: first line={(p2)--(p4)}, second line={(0,0)--(0,1)});

		}

	\end{scope}

	\begin{scope}

		\clip (p3)--(p2)--(p4)--(p2b)--cycle;

		\draw[blue!50!brown, step=.23] ($(p4)+(0,-1)$) grid +(3,3);

	\end{scope}

	\draw (p1) .. controls ($(p1) + (.5,.5)$) and ($(p3) + (-.8,0)$) .. node[red, above=3pt] {$a$} (p3);

	\draw (p1) .. controls ($(p1) + (.5,-.5)$) and ($(p4) + (-.8,0)$) .. node[red, below=3pt] {$a$} (p4);

	\draw (p3) -- (p2);

	\draw (p4) -- (p2);

	\draw (p3) -- node[red, above=7pt, right=1pt] {$x \times I$} (p2b);

	\draw (p4) -- node[red, below=7pt, right=1pt] {$x \times I$} (p2b);

	\draw (p1) \vertex;

	\draw (p2) \vertex;

	\draw (p3) \vertex;

	\draw (p4) \vertex;

	\draw (p2b) \vertex;

\end{tikzpicture}

};

\node[outer sep=\nsep](C) at (11,0) {

\begin{tikzpicture}

	\draw (0,0) coordinate (p1);

	\draw (2.3,0) coordinate (p2);

	\draw (2.3,1) coordinate (p3);

	\draw (2.3,-1) coordinate (p4);

	\draw (3.6,0) coordinate (p2b);

	\begin{scope}

		\clip (p1) 	.. controls +(.5,-.5) and +(-.8,0)  .. (p4) -- 

				(p2) -- (p3) .. controls +(-.8,0) and +(.5,.5) .. (p1);

		\foreach \x in {0,0.26,...,4} {

			\draw[green!50!brown] (\x,-1) -- (\x,1);

		}

	\end{scope}

	\begin{scope}

		\clip (p3)--(p2)--(p4)--(p2b)--cycle;

		\draw[blue!50!brown, step=.23] ($(p4)+(0,-1)$) grid +(3,3);

	\end{scope}

	\draw (p1) .. controls ($(p1) + (.5,.5)$) and ($(p3) + (-.8,0)$) .. node[red, above=3pt] {$a$} (p3);

	\draw (p1) .. controls ($(p1) + (.5,-.5)$) and ($(p4) + (-.8,0)$) .. node[red, below=3pt] {$a$} (p4);

	\draw[green!50!brown] (p3) -- (p4);

	\draw (p3) -- node[red, above=7pt, right=1pt] {$x \times I$} (p2b);

	\draw (p4) -- node[red, below=7pt, right=1pt] {$x \times I$} (p2b);

	\draw (p1) \vertex;

	\draw (p3) \vertex;

	\draw (p4) \vertex;

	\draw (p2b) \vertex;

\end{tikzpicture}

};

\draw[->, thick, blue!50!green] (A) -- (B);

\draw[->, thick, blue!50!green] (B) -- node[black, above] {$=$} (C);

\end{tikzpicture}

\caption{Composition of weak identities, 1}

\label{fzo3}

\end{figure}

In the first step we have inserted a copy of $(x\times I)\times I$.

Figure \ref{fzo4} shows the other case.

\begin{figure}[t]

\centering

\begin{tikzpicture}

\newcommand{\vertex}{node[circle,fill=black,inner sep=1pt] {}}

\newcommand{\nsep}{1.8}

\clip (-4,-1.25)--(12,-1.25)--(16,1.25)--(-1,1.25)--cycle;

\node[outer sep=\nsep](A) at (0,0) {

\begin{tikzpicture}

	\draw (0,0) coordinate (p1);

	\draw (4,0) coordinate (p2);

	\draw (2.4,0) coordinate (p2a);

	\draw (2,1.2) coordinate (pu);

	\draw (2,-1.2) coordinate (pd);

	\begin{scope}

		\clip (p1) .. controls (pu) .. (p2) .. controls (pd) .. (p1);

		\foreach \t in {0,.065,...,1} {

			\draw[green!50!brown] ($(p1)!\t!(p2a)$) -- +(90 - \t*90 + \t*6 : 4);

			\draw[green!50!brown] ($(p1)!\t!(p2a)$) -- +(-90 + \t*90 - \t*6 : 4);

		}

		\draw[dashed] ($(p2a) + (-.6,3)$)--(p2a)--($(p2a) + (-.6,-3)$);

	\end{scope}

	\draw (p1) .. controls (pu) .. (p2) .. controls (pd) .. (p1);

	\draw (p1)--(p2);

	\draw (p1) \vertex;

	\draw (p2) \vertex;

	\draw (p2a) \vertex;

\end{tikzpicture}

};

\node[outer sep=\nsep](B) at (5,0) {

\begin{tikzpicture}

	\draw (0,0) coordinate (p1);

	\draw (4,0) coordinate (p2);

	\draw (2.4,0) coordinate (p2a);

	\draw (2,1.2) coordinate (pu);

	\draw (2,-1.2) coordinate (pd);

	\begin{scope}

	\clip (-.1,3)--($(p2a) + (-.6,3)$)--(p2a)--($(p2a) + (-.6,-3)$)--(-.1,-3)--cycle;

	\begin{scope}

		\clip (p1) .. controls (pu) .. (p2) .. controls (pd) .. (p1);

		\foreach \t in {0,.065,...,1} {

			\draw[green!50!brown] ($(p1)!\t!(p2a)$) -- +(90 - \t*90 + \t*6 : 4);

			\draw[green!50!brown] ($(p1)!\t!(p2a)$) -- +(-90 + \t*90 - \t*6 : 4);

		}

		\draw ($(p2a) + (-.6,3)$)--(p2a)--($(p2a) + (-.6,-3)$);

	\end{scope}

	\draw (p1) .. controls (pu) .. (p2) .. controls (pd) .. (p1);

	%\draw (p1)--(p2);

	\end{scope}

	\begin{scope}

		\clip (p1) .. controls (pu) .. (p2) .. controls (pd) .. (p1);

		\draw ($(p2a) + (-.6,3)$)--(p2a)--($(p2a) + (-.6,-3)$);

	\end{scope}

	\draw (p1) \vertex;

	\draw (p2a) \vertex;