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%!TEX root = ../blob1.tex
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\section{Comparing $n$-category definitions}
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\label{sec:comparing-defs}
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In this appendix we relate the ``topological" category definitions of Section \ref{sec:ncats}
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to more traditional definitions, for $n=1$ and 2.
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\subsection{Plain 1-categories}
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Given a topological 1-category $\cC$, we construct a traditional 1-category $C$.
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(This is quite straightforward, but we include the details for the sake of completeness and
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to shed some light on the $n=2$ case.)
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Let the objects of $C$ be $C^0 \deq \cC(B^0)$ and the morphisms of $C$ be $C^1 \deq \cC(B^1)$,
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where $B^k$ denotes the standard $k$-ball.
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The boundary and restriction maps of $\cC$ give domain and range maps from $C^1$ to $C^0$.
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Choose a homeomorphism $B^1\cup_{pt}B^1 \to B^1$.
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Define composition in $C$ to be the induced map $C^1\times C^1 \to C^1$ (defined only when range and domain agree).
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By isotopy invariance in $\cC$, any other choice of homeomorphism gives the same composition rule.
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Also by isotopy invariance, composition is associative.
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Given $a\in C^0$, define $\id_a \deq a\times B^1$.
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By extended isotopy invariance in $\cC$, this has the expected properties of an identity morphism.
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\nn{(slash)id seems to rendering a a boldface 1 --- is this what we want?}
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\medskip
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For 1-categories based on oriented manifolds, there is no additional structure.
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For 1-categories based on unoriented manifolds, there is a map $*:C^1\to C^1$
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coming from $\cC$ applied to an orientation-reversing homeomorphism (unique up to isotopy)
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from $B^1$ to itself.
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Topological properties of this homeomorphism imply that
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$a^{**} = a$ (* is order 2), * reverses domain and range, and $(ab)^* = b^*a^*$
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(* is an anti-automorphism).
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For 1-categories based on Spin manifolds,
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the the nontrivial spin homeomorphism from $B^1$ to itself which covers the identity
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gives an order 2 automorphism of $C^1$.
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For 1-categories based on $\text{Pin}_-$ manifolds,
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we have an order 4 antiautomorphism of $C^1$.
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For 1-categories based on $\text{Pin}_+$ manifolds,
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we have an order 2 antiautomorphism and also an order 2 automorphism of $C^1$,
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and these two maps commute with each other.
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\nn{need to also consider automorphisms of $B^0$ / objects}
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\medskip
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In the other direction, given a traditional 1-category $C$
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(with objects $C^0$ and morphisms $C^1$) we will construct a topological
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1-category $\cC$.
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If $X$ is a 0-ball (point), let $\cC(X) \deq C^0$.
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If $S$ is a 0-sphere, let $\cC(S) \deq C^0\times C^0$.
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If $X$ is a 1-ball, let $\cC(X) \deq C^1$.
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Homeomorphisms isotopic to the identity act trivially.
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If $C$ has extra structure (e.g.\ it's a *-1-category), we use this structure
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to define the action of homeomorphisms not isotopic to the identity
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(and get, e.g., an unoriented topological 1-category).
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The domain and range maps of $C$ determine the boundary and restriction maps of $\cC$.
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Gluing maps for $\cC$ are determined my composition of morphisms in $C$.
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For $X$ a 0-ball, $D$ a 1-ball and $a\in \cC(X)$, define the product morphism
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$a\times D \deq \id_a$.
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It is not hard to verify that this has the desired properties.
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\medskip
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The compositions of the above two ``arrows" ($\cC\to C\to \cC$ and $C\to \cC\to C$) give back
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more or less exactly the same thing we started with.
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\nn{need better notation here}
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As we will see below, for $n>1$ the compositions yield a weaker sort of equivalence.
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\medskip
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Similar arguments show that modules for topological 1-categories are essentially
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the same thing as traditional modules for traditional 1-categories.
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\subsection{Plain 2-categories}
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Let $\cC$ be a topological 2-category.
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We will construct a traditional pivotal 2-category.
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(The ``pivotal" corresponds to our assumption of strong duality for $\cC$.)
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We will try to describe the construction in such a way the the generalization to $n>2$ is clear,
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though this will make the $n=2$ case a little more complicated than necessary.
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\nn{Note: We have to decide whether our 2-morphsism are shaped like rectangles or bigons.
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Each approach has advantages and disadvantages.
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For better or worse, we choose bigons here.}
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\nn{maybe we should do both rectangles and bigons?}
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Define the $k$-morphisms $C^k$ of $C$ to be $\cC(B^k)_E$, where $B^k$ denotes the standard
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$k$-ball, which we also think of as the standard bihedron.
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Since we are thinking of $B^k$ as a bihedron, we have a standard decomposition of the $\bd B^k$
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into two copies of $B^{k-1}$ which intersect along the ``equator" $E \cong S^{k-2}$.
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Recall that the subscript in $\cC(B^k)_E$ means that we consider the subset of $\cC(B^k)$
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whose boundary is splittable along $E$.
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This allows us to define the domain and range of morphisms of $C$ using
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boundary and restriction maps of $\cC$.
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Choosing a homeomorphism $B^1\cup B^1 \to B^1$ defines a composition map on $C^1$.
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This is not associative, but we will see later that it is weakly associative.
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Choosing a homeomorphism $B^2\cup B^2 \to B^2$ defines a ``vertical" composition map
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on $C^2$ (Figure \ref{fzo1}).
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Isotopy invariance implies that this is associative.
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We will define a ``horizontal" composition later.
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\nn{maybe no need to postpone?}
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\begin{figure}[t]
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\begin{equation*}
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\mathfig{.73}{tempkw/zo1}
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\end{equation*}
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\caption{Vertical composition of 2-morphisms}
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\label{fzo1}
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\end{figure}
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Given $a\in C^1$, define $\id_a = a\times I \in C^1$ (pinched boundary).
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Extended isotopy invariance for $\cC$ shows that this morphism is an identity for
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vertical composition.
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Given $x\in C^0$, define $\id_x = x\times B^1 \in C^1$.
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We will show that this 1-morphism is a weak identity.
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This would be easier if our 2-morphisms were shaped like rectangles rather than bigons.
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Define let $a: y\to x$ be a 1-morphism.
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Define maps $a \to a\bullet \id_x$ and $a\bullet \id_x \to a$
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as shown in Figure \ref{fzo2}.
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\begin{figure}[t]
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\begin{equation*}
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\mathfig{.73}{tempkw/zo2}
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\end{equation*}
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\caption{blah blah}
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\label{fzo2}
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\end{figure}
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In that figure, the red cross-hatched areas are the product of $x$ and a smaller bigon,
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while the remained is a half-pinched version of $a\times I$.
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\nn{the red region is unnecessary; remove it? or does it help?
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(because it's what you get if you bigonify the natural rectangular picture)}
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We must show that the two compositions of these two maps give the identity 2-morphisms
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on $a$ and $a\bullet \id_x$, as defined above.
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Figure \ref{fzo3} shows one case.
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\begin{figure}[t]
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\begin{equation*}
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\mathfig{.83}{tempkw/zo3}
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\end{equation*}
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\caption{blah blah}
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\label{fzo3}
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\end{figure}
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In the first step we have inserted a copy of $id(id(x))$ \nn{need better notation for this}.
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\nn{also need to talk about (somewhere above)
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how this sort of insertion is allowed by extended isotopy invariance and gluing.
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Also: maybe half-pinched and unpinched products can be derived from fully pinched
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products after all (?)}
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Figure \ref{fzo4} shows the other case.
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\begin{figure}[t]
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\begin{equation*}
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\mathfig{.83}{tempkw/zo4}
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\end{equation*}
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\caption{blah blah}
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\label{fzo4}
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\end{figure}
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We first collapse the red region, then remove a product morphism from the boundary,
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We define horizontal composition of 2-morphisms as shown in Figure \ref{fzo5}.
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It is not hard to show that this is independent of the arbitrary (left/right) choice made in the definition, and that it is associative.
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\begin{figure}[t]
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\begin{equation*}
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\mathfig{.83}{tempkw/zo5}
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\end{equation*}
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\caption{Horizontal composition of 2-morphisms}
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\label{fzo5}
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\end{figure}
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\nn{need to find a list of axioms for pivotal 2-cats to check}
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\nn{...}
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\medskip
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\hrule
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\medskip
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\nn{to be continued...}
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\medskip
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