--- a/text/appendixes/comparing_defs.tex	Sun Jul 18 18:26:05 2010 -0600
+++ b/text/appendixes/comparing_defs.tex	Mon Jul 19 07:45:26 2010 -0600
@@ -3,11 +3,25 @@
 \section{Comparing $n$-category definitions}
 \label{sec:comparing-defs}
 
-In this appendix we relate the ``topological" category definitions of \S\ref{sec:ncats}
-to more traditional definitions, for $n=1$ and 2.
+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 topological $n$-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 topological $n$-categories)
+yields the appropriate sort of equivalence on each side.
+Since we haven't given a definition for functors between topological $n$-categories
+(the paper is already too long!), we do not pursue this here.
+\nn{say something about modules and tensor products?}
 
-\nn{cases to cover: (a) plain $n$-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?}
+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) plain $n$-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 $\Set$ or $\Vect$}
 \label{ssec:1-cats}
@@ -34,8 +48,7 @@
 If the underlying manifolds for $\cX$ have further geometric structure, then we obtain certain functors.
 The base case is for oriented manifolds, where we obtain no extra algebraic data.
 
-For 1-categories based on unoriented manifolds (somewhat confusingly, we're thinking of being 
-unoriented as requiring extra data beyond being oriented, namely the identification between the orientations), 
+For 1-categories based on unoriented manifolds, 
 there is a map $*:c(\cX)^1\to c(\cX)^1$
 coming from $\cX$ applied to an orientation-reversing homeomorphism (unique up to isotopy) 
 from $B^1$ to itself.
@@ -52,8 +65,9 @@
 For 1-categories based on $\text{Pin}_+$ manifolds,
 we have an order 2 antiautomorphism and also an order 2 automorphism of $c(\cX)^1$,
 and these two maps commute with each other.
-\nn{need to also consider automorphisms of $B^0$ / objects}
+%\nn{need to also consider automorphisms of $B^0$ / objects}
 
+\noop{
 \medskip
 
 In the other direction, given a $1$-category $C$
@@ -83,12 +97,14 @@
 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 topological 1-categories are essentially
 the same thing as traditional modules for traditional 1-categories.
 
+
 \subsection{Plain 2-categories}
 \label{ssec:2-cats}
 Let $\cC$ be a topological 2-category.

--- a/text/appendixes/smallblobs.tex	Sun Jul 18 18:26:05 2010 -0600
+++ b/text/appendixes/smallblobs.tex	Mon Jul 19 07:45:26 2010 -0600
@@ -175,7 +175,7 @@
 \begin{align*}
 & \sum_{m=0}^k  \sum_{\most(i) \in \{1, \ldots, k\}^{m-1} \setminus \Delta} \sum_{\substack{i_m = 1 \\ i_1 \not\in \most(i)}}^{k}  (-1)^{\sigma(\most(i) i_m) + m} \ev\left(\restrict{\phi_{\most(i)\!\downarrow_{i_m}(b_{i_m})}}{x_0 = 0}\tensor b_{\most(i) i_m}\right) \\
 & \quad = \sum_{m=0}^{k-1} \sum_{q=1}^k \sum_{i \in \{1, \ldots, k-1\}^m\setminus \Delta} (-1)^{\sigma(i)+q+1} \ev\left(\restrict{\phi_{i(b_{q})}}{x_0 = 0}\tensor (b_q)_i\right) \\
-\intertext{(here we've used Equation \eqref{eq:sigma(ab)} and renamed $i_m$ to $q$ and $most(i)$ to $i$, as well as shifted $m$ by one), which is just}
+\intertext{(here we've used Equation \eqref{eq:sigma(ab)} and renamed $i_m$ to $q$ and $\most(i)$ to $i$, as well as shifted $m$ by one), which is just}
 & \quad = \sum_{q=1}^k (-1)^{q+1} s(b_q) \\
 & \quad = s(\bdy b).
 \end{align*}