--- a/preamble.tex Thu May 27 17:35:56 2010 -0700
+++ b/preamble.tex Thu May 27 17:39:11 2010 -0700
@@ -13,6 +13,7 @@
\usepackage[section]{placeins}
\usepackage{leftidx}
+\usepackage{stmaryrd} % additional math symbols, e.g. \mapsfrom
\SelectTips{cm}{}
% This may speed up compilation of complex documents with many xymatrices.
--- a/text/appendixes/explicit.tex Thu May 27 17:35:56 2010 -0700
+++ /dev/null Thu Jan 01 00:00:00 1970 +0000
@@ -1,97 +0,0 @@
-%!TEX root = ../../blob1.tex
-
-Here's an alternative proof of the special case in which $P$, the parameter space for the family of diffeomorphisms, is a cube. It is much more explicit, for better or worse.
-
-\begin{proof}[Alternative, more explicit proof of Lemma \ref{extension_lemma}]
-
-
-Fix a finite open cover of $X$, say $(U_l)_{l=1}^L$, along with an
-associated partition of unity $(r_l)$.
-
-We'll define the homotopy $H:I \times P \times X \to X$ via a function
-$u:I \times P \times X \to P$, with
-\begin{equation*}
-H(t,p,x) = F(u(t,p,x),x).
-\end{equation*}
-
-To begin, we'll define a function $u'' : I \times P \times X \to P$, and
-a corresponding homotopy $H''$. This homotopy will just be a homotopy of
-$F$ through families of maps, not through families of diffeomorphisms. On
-the other hand, it will be quite simple to describe, and we'll later
-explain how to build the desired function $u$ out of it.
-
-For each $l = 1, \ldots, L$, pick some $C^\infty$ function $f_l : I \to
-I$ which is identically $0$ on a neighborhood of the closed interval $[0,\frac{l-1}{L}]$
-and identically $1$ on a neighborhood of the closed interval $[\frac{l}{L},1]$. (Monotonic?
-Fix a bound for the derivative?) We'll extend it to a function on
-$k$-tuples $f_l : I^k \to I^k$ pointwise.
-
-Define $$u''(t,p,x) = \sum_{l=1}^L r_l(x) u_l(t,p),$$ with
-$$u_l(t,p) = t f_l(p) + (1-t)p.$$ Notice that the $i$-th component of $u''(t,p,x)$ depends only on the $i$-th component of $p$.
-
-Let's now establish some properties of $u''$ and $H''$. First,
-\begin{align*}
-H''(0,p,x) & = F(u''(0,p,x),x) \\
- & = F(\sum_{l=1}^L r_l(x) p, x) \\
- & = F(p,x).
-\end{align*}
-Next, calculate the derivatives
-\begin{align*}
-\partial_{p_i} H''(1,p,x) & = \partial_{p_i}u''(1,p,x) \partial_1 F(u(1,p,x),x) \\
-\intertext{and}
-\partial_{p_i}u''(1,p,x) & = \sum_{l=1}^L r_l(x) \partial_{p_i} f_l(p).
-\end{align*}
-Now $\partial_{p_i} f_l(p) = 0$ unless $\frac{l-1}{L} < p_i < \frac{l}{L}$, and $r_l(x) = 0$ unless $x \in U_l$,
-so we conclude that for a fixed $p$, $\partial_p H''(1,p,x) = 0$ for all $x$ outside the union of $k$ open sets from the open cover, namely
-$\bigcup_{i=1}^k U_{l_i}$ where for each $i$, we choose $l_i$ so $\frac{l_i -1}{L} \leq p_i \leq \frac{l_i}{L}$. It may be helpful to refer to Figure \ref{fig:supports}.
-
-\begin{figure}[!ht]
-\begin{equation*}
-\mathfig{0.5}{explicit/supports}
-\end{equation*}
-\caption{The supports of the derivatives {\color{green}$\partial_p f_1$}, {\color{blue}$\partial_p f_2$} and {\color{red}$\partial_p f_3$}, illustrating the case $k=2$, $L=3$. Notice that any
-point $p$ lies in the intersection of at most $k$ supports. The support of $\partial_p u''(1,p,x)$ is contained in the union of these supports.}
-\label{fig:supports}
-\end{figure}
-
-Unfortunately, $H''$ does not have the desired property that it's a homotopy through diffeomorphisms. To achieve this, we'll paste together several copies
-of the map $u''$. First, glue together $2^k$ copies, defining $u':I \times P \times X$ by
-\begin{align*}
-u'(t,p,x)_i & =
-\begin{cases}
-\frac{1}{2} u''(t, 2p_i, x)_i & \text{if $0 \leq p_i \leq \frac{1}{2}$} \\
-1-\frac{1}{2} u''(t, 2-2p_i, x)_i & \text{if $\frac{1}{2} \leq p_i \leq 1$}.
-\end{cases}
-\end{align*}
-(Note that we're abusing notation somewhat, using the fact that $u''(t,p,x)_i$ depends on $p$ only through $p_i$.)
-To see what's going on here, it may be helpful to look at Figure \ref{fig:supports_4}, which shows the support of $\partial_p u'(1,p,x)$.
-\begin{figure}[!ht]
-\begin{equation*}
-\mathfig{0.4}{explicit/supports_4} \qquad \qquad \mathfig{0.4}{explicit/supports_36}
-\end{equation*}
-\caption{The supports of $\partial_p u'(1,p,x)$ and of $\partial_p u(1,p,x)$ (with $K=3$) are subsets of the indicated region.}
-\label{fig:supports_4}
-\end{figure}
-
-Second, pick some $K$, and define
-\begin{align*}
-u(t,p,x) & = \frac{\floor{K p}}{K} + \frac{1}{K} u'\left(t, K \left(p - \frac{\floor{K p}}{K}\right), x\right).
-\end{align*}
-
-\todo{Explain that the localisation property survives for $u'$ and $u$.}
-
-We now check that by making $K$ large enough, $H$ becomes a homotopy through diffeomorphisms. We start with
-$$\partial_x H(t,p,x) = \partial_x u(t,p,x) \partial_1 F(u(t,p,x), x) + \partial_2 F(u(t,p,x), x)$$
-and observe that since $F(p, -)$ is a diffeomorphism, the second term $\partial_2 F(u(t,p,x), x)$ is bounded away from $0$. Thus if we can control the
-size of the first term $\partial_x u(t,p,x) \partial_1 F(u(t,p,x), x)$ we're done. The factor $\partial_1 F(u(t,p,x), x)$ is bounded, and we
-calculate \todo{err... this is a mess, and probably wrong.}
-\begin{align*}
-\partial_x u(t,p,x)_i & = \partial_x \frac{1}{K} u'\left(t, K\left(p - \frac{\floor{K p}}{K}\right), x\right)_i \\
- & = \pm \frac{1}{2 K} \partial_x u''\left(t, (1\mp1)\pm 2K\left(p_i-\frac{\floor{K p}}{K}\right), x\right)_i \\
- & = \pm \frac{1}{2 K} \sum_{l=1}^L (\partial_x r_l(x)) u_l\left(t, (1\mp1)\pm 2K\left(p_i-\frac{\floor{K p}}{K}\right)\right)_i. \\
-\intertext{Since the target of $u_l$ is just the unit cube $I^k$, we can make the estimate}
-\norm{\partial_x u(t,p,x)_i} & \leq \frac{1}{2 K} \sum_{l=1}^L \norm{\partial_x r_l(x)}.
-\end{align*}
-The sum here is bounded, so for large enough $K$ this is small enough that $\partial_x H(t,p,x)$ is never zero.
-
-\end{proof}
\ No newline at end of file
--- a/text/appendixes/famodiff.tex Thu May 27 17:35:56 2010 -0700
+++ b/text/appendixes/famodiff.tex Thu May 27 17:39:11 2010 -0700
@@ -9,14 +9,10 @@
(That is, $r_\alpha : X \to [0,1]$; $r_\alpha(x) = 0$ if $x\notin U_\alpha$;
for fixed $x$, $r_\alpha(x) \ne 0$ for only finitely many $\alpha$; and $\sum_\alpha r_\alpha = 1$.)
Since $X$ is compact, we will further assume that $r_\alpha = 0$ (globally)
-for all but finitely many $\alpha$.
+for all but finitely many $\alpha$. \nn{Can't we just assume $\cU$ is finite? Is something subtler happening? -S}
-Let
-\[
- CM_*(X, T) \deq C_*(\Maps(X\to T)) ,
-\]
-the singular chains on the space of continuous maps from $X$ to $T$.
-$CM_k(X, T)$ is generated by continuous maps
+Consider $C_*(\Maps(X\to T))$, the singular chains on the space of continuous maps from $X$ to $T$.
+$C_k(\Maps(X \to T))$ is generated by continuous maps
\[
f: P\times X \to T ,
\]
@@ -24,7 +20,7 @@
Recall that $f$ is {\it supported} on $S\sub X$ if $f(p, x)$ does not depend on $p$ when
$x \notin S$, and that $f$ is {\it adapted} to $\cU$ if
$f$ is supported on the union of at most $k$ of the $U_\alpha$'s.
-A chain $c \in CM_*(X, T)$ is adapted to $\cU$ if it is a linear combination of
+A chain $c \in C_*(\Maps(X \to T))$ is adapted to $\cU$ if it is a linear combination of
generators which are adapted.
\begin{lemma} \label{basic_adaptation_lemma}
@@ -40,14 +36,12 @@
the restrictions $F(1, \cdot, \cdot) : D_i\times X\to T$ are all adapted to $\cU$.
\item If $f$ has support $S\sub X$, then
$F: (I\times P)\times X\to T$ (a $k{+}1$-parameter family of maps) also has support $S$.
-Furthermore, if $Q\sub\bd P$ is a convex linear subpolyhedron of $\bd P$ and $f$ restricted to $Q$
-has support $S'$, then
+Furthermore, if $Q$ is a convex linear subpolyhedron of $\bd P$ and $f$ restricted to $Q$
+has support $S' \subset X$, then
$F: (I\times Q)\times X\to T$ also has support $S'$.
-\item If for all $p\in P$ we have $f(p, \cdot):X\to T$ is a
-[immersion, diffeomorphism, PL homeomorphism, bi-Lipschitz homeomorphism]
-then the same is true of $F(t, p, \cdot)$ for all $t\in I$ and $p\in P$.
-(Of course we must assume that $X$ and $T$ are the appropriate
-sort of manifolds for this to make sense.)
+\item Suppose both $X$ and $T$ are smooth manifolds, metric spaces, or PL manifolds, and let $\cX$ denote the subspace of $\Maps(X \to T)$ consisting of immersions or of diffeomorphisms (in the smooth case), bi-Lipschitz homeomorphisms (in the metric case), or PL homeomorphisms (in the PL case).
+ If $f$ is smooth, Lipschitz or PL, as appropriate, and $f(p, \cdot):X\to T$ is in $\cX$ for all $p \in P$
+then $F(t, p, \cdot)$ is also in $\cX$ for all $t\in I$ and $p\in P$.
\end{enumerate}
\end{lemma}
@@ -80,7 +74,7 @@
For each (top-dimensional) $k$-cell $C$ of each $K_\alpha$, choose a point $p(C) \in C \sub P$.
If $C$ meets a subpolyhedron $Q$ of $\bd P$, we require that $p(C)\in Q$.
(It follows that if $C$ meets both $Q$ and $Q'$, then $p(C)\in Q\cap Q'$.
-This puts some mild constraints on the choice of $K_\alpha$.)
+Ensuring this is possible corresponds to some mild constraints on the choice of the $K_\alpha$.)
Let $D$ be a $k$-handle of $\jj$.
For each $\alpha$ let $C(D, \alpha)$ be the $k$-cell of $K_\alpha$ which contains $D$
@@ -134,7 +128,7 @@
\right) .
\end{equation}
-This completes the definition of $u: I \times P \times X \to P$.
+This completes the definition of $u: I \times P \times X \to P$. The formulas above are consistent: for $p$ at the boundary between a $k-j$-handle and a $k-(j+1)$-handle the corresponding expressions in Equation \eqref{eq:u} agree, since one of the normal coordinates becomes $0$ or $1$.
Note that if $Q\sub \bd P$ is a convex linear subpolyhedron, then $u(I\times Q\times X) \sub Q$.
\medskip
@@ -150,7 +144,7 @@
Next we show that for each handle $D$ of $J$, $F(1, \cdot, \cdot) : D\times X \to X$
is a singular cell adapted to $\cU$.
Let $k-j$ be the index of $D$.
-Referring to Equation \ref{eq:u}, we see that $F(1, p, x)$ depends on $p$ only if
+Referring to Equation \eqref{eq:u}, we see that $F(1, p, x)$ depends on $p$ only if
$r_\beta(x) \ne 0$ for some $\beta\in\cN$, i.e.\ only if
$x\in \bigcup_{\beta\in\cN} U_\beta$.
Since the cardinality of $\cN$ is $j$ which is less than or equal to $k$,
@@ -176,7 +170,7 @@
\medskip
Now for claim 4 of the lemma.
-Assume that $X$ and $T$ are smooth manifolds and that $f$ is a family of diffeomorphisms.
+Assume that $X$ and $T$ are smooth manifolds and that $f$ is a smooth family of diffeomorphisms.
We must show that we can choose the $K_\alpha$'s and $u$ so that $F(t, p, \cdot)$ is a
diffeomorphism for all $t$ and $p$.
It suffices to
@@ -188,8 +182,8 @@
}
Since $f$ is a family of diffeomorphisms and $X$ and $P$ are compact,
$\pd{f}{x}$ is non-singular and bounded away from zero.
-Also, $\pd{f}{p}$ is bounded.
-So if we can insure that $\pd{u}{x}$ is sufficiently small, we are done.
+Also, since $f$ is smooth $\pd{f}{p}$ is bounded.
+Thus if we can insure that $\pd{u}{x}$ is sufficiently small, we are done.
It follows from Equation \eqref{eq:u} above that $\pd{u}{x}$ depends on $\pd{r_\alpha}{x}$
(which is bounded)
and the differences amongst the various $p(D_0,\alpha)$'s and $q_{\beta i}$'s.
@@ -200,7 +194,7 @@
through essentially unchanged.
Next we consider the case where $f$ is a family of bi-Lipschitz homeomorphisms.
-We assume that $f$ is Lipschitz in $P$ direction as well.
+Recall that we assume that $f$ is Lipschitz in the $P$ direction as well.
The argument in this case is similar to the one above for diffeomorphisms, with
bounded partial derivatives replaced by Lipschitz constants.
Since $X$ and $P$ are compact, there is a universal bi-Lipschitz constant that works for
@@ -214,15 +208,14 @@
\end{proof}
\begin{lemma}
-Let $CM_*(X\to T)$ be the singular chains on the space of continuous maps
-[resp. immersions, diffeomorphisms, PL homeomorphisms, bi-Lipschitz homeomorphisms]
-from $X$ to $T$, and let $G_*\sub CM_*(X\to T)$ denote the chains adapted to an open cover $\cU$
+Let $\cX_*$ be any of $C_*(\Maps(X \to T))$ or singular chains on the subspace of $\Maps(X\to T)$ consisting of immersions, diffeomorphisms, bi-Lipschitz homeomorphisms or PL homeomorphisms.
+Let $G_* \subset \cX_*$ denote the chains adapted to an open cover $\cU$
of $X$.
-Then $G_*$ is a strong deformation retract of $CM_*(X\to T)$.
+Then $G_*$ is a strong deformation retract of $\cX_*$.
\end{lemma}
\begin{proof}
-If suffices to show that given a generator $f:P\times X\to T$ of $CM_k(X\to T)$ with
-$\bd f \in G_{k-1}$ there exists $h\in CM_{k+1}(X\to T)$ with $\bd h = f + g$ and $g \in G_k$.
+If suffices to show that given a generator $f:P\times X\to T$ of $\cX_k$ with
+$\bd f \in G_{k-1}$ there exists $h\in \cX_{k+1}$ with $\bd h = f + g$ and $g \in G_k$.
This is exactly what Lemma \ref{basic_adaptation_lemma}
gives us.
More specifically, let $\bd P = \sum Q_i$, with each $Q_i\in G_{k-1}$.
@@ -276,11 +269,3 @@
}
% end \noop
-\medskip
-\hrule
-\medskip
-
-\nn{do we want to keep this alternative construction?}
-
-\input{text/appendixes/explicit.tex}
-
--- a/text/intro.tex Thu May 27 17:35:56 2010 -0700
+++ b/text/intro.tex Thu May 27 17:39:11 2010 -0700
@@ -8,7 +8,7 @@
\item When $n=1$ and $\cC$ is just a 1-category (e.g.\ an associative algebra), the blob complex $\bc_*(S^1; \cC)$ is quasi-isomorphic to the Hochschild complex $\HC_*(\cC)$. (See Property \ref{property:hochschild} and \S \ref{sec:hochschild}.)
\item When $\cC$ is the polynomial algebra $k[t]$, thought of as an n-category (see \S \ref{sec:comm_alg}), we have
that $\bc_*(M; k[t])$ is homotopy equivalent to $C_*(\Sigma^\infty(M), k)$, the singular chains
-on the configurations space of unlabeled points in $M$.
+on the configuration space of unlabeled points in $M$.
%$$H_*(\bc_*(M; k[t])) = H^{\text{sing}}_*(\Delta^\infty(M), k).$$
\end{itemize}
The blob complex definition is motivated by the desire for a derived analogue of the usual TQFT Hilbert space (replacing quotient of fields by local relations with some sort of resolution),
--- a/text/ncat.tex Thu May 27 17:35:56 2010 -0700
+++ b/text/ncat.tex Thu May 27 17:39:11 2010 -0700
@@ -581,7 +581,16 @@
See also Theorem \ref{thm:map-recon} below, recovering $C_*(\Maps(M \to T))$ up to homotopy the blob complex of $M$ with coefficients in $\pi^\infty_{\le n}(T)$.
-\begin{example}[Blob complexes of balls (with a fiber)]
\rm
\label{ex:blob-complexes-of-balls}
Fix an $m$-dimensional manifold $F$ and system of fields $\cE$.
We will define an $A_\infty$ $(n-m)$-category $\cC$.
When $X$ is a $k$-ball or $k$-sphere, with $k<n-m$, define $\cC(X) = \cE(X\times F)$.
When $X$ is an $(n-m)$-ball,
define $\cC(X; c) = \bc^\cE_*(X\times F; c)$
where $\bc^\cE_*$ denotes the blob complex based on $\cE$.
\end{example}
+\begin{example}[Blob complexes of balls (with a fiber)]
+\rm
+\label{ex:blob-complexes-of-balls}
+Fix an $m$-dimensional manifold $F$ and system of fields $\cE$.
+We will define an $A_\infty$ $(n-m)$-category $\cC$.
+When $X$ is a $k$-ball or $k$-sphere, with $k<n-m$, define $\cC(X) = \cE(X\times F)$.
+When $X$ is an $(n-m)$-ball,
+define $\cC(X; c) = \bc^\cE_*(X\times F; c)$
+where $\bc^\cE_*$ denotes the blob complex based on $\cE$.
+\end{example}
This example will be essential for Theorem \ref{product_thm} below, which allows us to compute the blob complex of a product. Notice that with $F$ a point, the above example is a construction turning a topological $n$-category into an $A_\infty$ $n$-category. We think of this as providing a `free resolution' of the topological $n$-category. \todo{Say more here!} In fact, there is also a trivial way to do this: we can think of each vector space associated to an $n$-ball as a chain complex concentrated in degree $0$, and take $\CD{B}$ to act trivially.
@@ -1104,9 +1113,8 @@
\begin{eqnarray*}
\hom_A(X_B\ot {_BY_A} \to Z_A) &\cong& \hom_B(X_B \to \hom_A( {_BY_A} \to Z_A)) \\
f &\mapsto& [x \mapsto f(x\ot -)] \\
- {}[x\ot y \mapsto g(x)(y)] & \leftarrowtail & g .
+ {}[x\ot y \mapsto g(x)(y)] & \mapsfrom & g .
\end{eqnarray*}
-\nn{how to do a left-pointing ``$\mapsto$"?}
If $A$ and $Z_A$ are both the ground field $\k$, this simplifies to
\[
(X_B\ot {_BY})^* \cong \hom_B(X_B \to (_BY)^*) .
--- /dev/null Thu Jan 01 00:00:00 1970 +0000
+++ b/text/obsolete/explicit.tex Thu May 27 17:39:11 2010 -0700
@@ -0,0 +1,97 @@
+%!TEX root = ../../blob1.tex
+
+Here's an alternative proof of the special case in which $P$, the parameter space for the family of diffeomorphisms, is a cube. It is much more explicit, for better or worse.
+
+\begin{proof}[Alternative, more explicit proof of Lemma \ref{extension_lemma}]
+
+
+Fix a finite open cover of $X$, say $(U_l)_{l=1}^L$, along with an
+associated partition of unity $(r_l)$.
+
+We'll define the homotopy $H:I \times P \times X \to X$ via a function
+$u:I \times P \times X \to P$, with
+\begin{equation*}
+H(t,p,x) = F(u(t,p,x),x).
+\end{equation*}
+
+To begin, we'll define a function $u'' : I \times P \times X \to P$, and
+a corresponding homotopy $H''$. This homotopy will just be a homotopy of
+$F$ through families of maps, not through families of diffeomorphisms. On
+the other hand, it will be quite simple to describe, and we'll later
+explain how to build the desired function $u$ out of it.
+
+For each $l = 1, \ldots, L$, pick some $C^\infty$ function $f_l : I \to
+I$ which is identically $0$ on a neighborhood of the closed interval $[0,\frac{l-1}{L}]$
+and identically $1$ on a neighborhood of the closed interval $[\frac{l}{L},1]$. (Monotonic?
+Fix a bound for the derivative?) We'll extend it to a function on
+$k$-tuples $f_l : I^k \to I^k$ pointwise.
+
+Define $$u''(t,p,x) = \sum_{l=1}^L r_l(x) u_l(t,p),$$ with
+$$u_l(t,p) = t f_l(p) + (1-t)p.$$ Notice that the $i$-th component of $u''(t,p,x)$ depends only on the $i$-th component of $p$.
+
+Let's now establish some properties of $u''$ and $H''$. First,
+\begin{align*}
+H''(0,p,x) & = F(u''(0,p,x),x) \\
+ & = F(\sum_{l=1}^L r_l(x) p, x) \\
+ & = F(p,x).
+\end{align*}
+Next, calculate the derivatives
+\begin{align*}
+\partial_{p_i} H''(1,p,x) & = \partial_{p_i}u''(1,p,x) \partial_1 F(u(1,p,x),x) \\
+\intertext{and}
+\partial_{p_i}u''(1,p,x) & = \sum_{l=1}^L r_l(x) \partial_{p_i} f_l(p).
+\end{align*}
+Now $\partial_{p_i} f_l(p) = 0$ unless $\frac{l-1}{L} < p_i < \frac{l}{L}$, and $r_l(x) = 0$ unless $x \in U_l$,
+so we conclude that for a fixed $p$, $\partial_p H''(1,p,x) = 0$ for all $x$ outside the union of $k$ open sets from the open cover, namely
+$\bigcup_{i=1}^k U_{l_i}$ where for each $i$, we choose $l_i$ so $\frac{l_i -1}{L} \leq p_i \leq \frac{l_i}{L}$. It may be helpful to refer to Figure \ref{fig:supports}.
+
+\begin{figure}[!ht]
+\begin{equation*}
+\mathfig{0.5}{explicit/supports}
+\end{equation*}
+\caption{The supports of the derivatives {\color{green}$\partial_p f_1$}, {\color{blue}$\partial_p f_2$} and {\color{red}$\partial_p f_3$}, illustrating the case $k=2$, $L=3$. Notice that any
+point $p$ lies in the intersection of at most $k$ supports. The support of $\partial_p u''(1,p,x)$ is contained in the union of these supports.}
+\label{fig:supports}
+\end{figure}
+
+Unfortunately, $H''$ does not have the desired property that it's a homotopy through diffeomorphisms. To achieve this, we'll paste together several copies
+of the map $u''$. First, glue together $2^k$ copies, defining $u':I \times P \times X$ by
+\begin{align*}
+u'(t,p,x)_i & =
+\begin{cases}
+\frac{1}{2} u''(t, 2p_i, x)_i & \text{if $0 \leq p_i \leq \frac{1}{2}$} \\
+1-\frac{1}{2} u''(t, 2-2p_i, x)_i & \text{if $\frac{1}{2} \leq p_i \leq 1$}.
+\end{cases}
+\end{align*}
+(Note that we're abusing notation somewhat, using the fact that $u''(t,p,x)_i$ depends on $p$ only through $p_i$.)
+To see what's going on here, it may be helpful to look at Figure \ref{fig:supports_4}, which shows the support of $\partial_p u'(1,p,x)$.
+\begin{figure}[!ht]
+\begin{equation*}
+\mathfig{0.4}{explicit/supports_4} \qquad \qquad \mathfig{0.4}{explicit/supports_36}
+\end{equation*}
+\caption{The supports of $\partial_p u'(1,p,x)$ and of $\partial_p u(1,p,x)$ (with $K=3$) are subsets of the indicated region.}
+\label{fig:supports_4}
+\end{figure}
+
+Second, pick some $K$, and define
+\begin{align*}
+u(t,p,x) & = \frac{\floor{K p}}{K} + \frac{1}{K} u'\left(t, K \left(p - \frac{\floor{K p}}{K}\right), x\right).
+\end{align*}
+
+\todo{Explain that the localisation property survives for $u'$ and $u$.}
+
+We now check that by making $K$ large enough, $H$ becomes a homotopy through diffeomorphisms. We start with
+$$\partial_x H(t,p,x) = \partial_x u(t,p,x) \partial_1 F(u(t,p,x), x) + \partial_2 F(u(t,p,x), x)$$
+and observe that since $F(p, -)$ is a diffeomorphism, the second term $\partial_2 F(u(t,p,x), x)$ is bounded away from $0$. Thus if we can control the
+size of the first term $\partial_x u(t,p,x) \partial_1 F(u(t,p,x), x)$ we're done. The factor $\partial_1 F(u(t,p,x), x)$ is bounded, and we
+calculate \todo{err... this is a mess, and probably wrong.}
+\begin{align*}
+\partial_x u(t,p,x)_i & = \partial_x \frac{1}{K} u'\left(t, K\left(p - \frac{\floor{K p}}{K}\right), x\right)_i \\
+ & = \pm \frac{1}{2 K} \partial_x u''\left(t, (1\mp1)\pm 2K\left(p_i-\frac{\floor{K p}}{K}\right), x\right)_i \\
+ & = \pm \frac{1}{2 K} \sum_{l=1}^L (\partial_x r_l(x)) u_l\left(t, (1\mp1)\pm 2K\left(p_i-\frac{\floor{K p}}{K}\right)\right)_i. \\
+\intertext{Since the target of $u_l$ is just the unit cube $I^k$, we can make the estimate}
+\norm{\partial_x u(t,p,x)_i} & \leq \frac{1}{2 K} \sum_{l=1}^L \norm{\partial_x r_l(x)}.
+\end{align*}
+The sum here is bounded, so for large enough $K$ this is small enough that $\partial_x H(t,p,x)$ is never zero.
+
+\end{proof}
\ No newline at end of file