\classheader{2013-02-11}

I'll be following Warner's book from now on, so I won't have to type up notes and so you'll have something to refer to. We'll be talking about flows and vector fields soon.

\subsection*{Continuous Variants of $\Cinfty$ Theorems}

\begin{theorem-N}
Let $X$ and $Y$ be $\Cinfty$ manifolds, and assume that $X$ is compact. Let $f_0:X\to Y$ be continuous. Then $f_0$ can be approximated by $\Cinfty$ functions; you could either use a metric on $Y$, or use the compact-open topology. Either way, there are $\Cinfty$ functions $f_n:X\to Y$ that converge to $f_0$.
\end{theorem-N}
\begin{proof}
We will assume the Stone-Weierstrass theorem. Choose an embedding $X\hookrightarrow \R^n$. Because $X$ is compact, the Stone-Weierstrass theorem says that $\Cinfty(X)$ is dense in $C(X,\R)$ in the $\|\cdot\|_\infty$ norm. Thus, the statement is true if $Y=\R$. It will then follow for $Y=\R^n$, then for $Y=$ an open subset of $\R^n$, and then for any $Y=$ any $\Cinfty$ retract of an open subset of $\R^n$. But from the tubular neighborhood theorem, this last case includes any $Y$; there is an embedding $i:Y\hookrightarrow\R^n$ and a retract $r:U\to Y$ for some open $U\supset i(Y)$.
\end{proof}

\begin{theorem}[Partitions of unity]
Let $M$ be a $\Cinfty$ manifold, and let $\cal{U}$ be an open cover of $M$. Then for any $U\in\cal{U}$, there is a $\Cinfty$ function $\varphi_U:M\to \R$ such that
\begin{itemize}
\item[(a)] $\supp(\varphi_U)\subset U$
\item[(b)] $\varphi_U\geq 0$
\end{itemize}
and such that the collection $\{\varphi_U\mid U\in\cal{U}\}$ satisfies the condition
\[\text{for all }x\in M,\text{ there is some neighborhood }U(x)\text{ of }x\text{ such that }\{U\in\cal{U}\mid \varphi_U|_{U(x)}\neq 0\}\text{ is finite}\]
and $\sum_{U\in\cal{U}}\varphi_U(x)=1$ for all $x\in M$.
\end{theorem}

\begin{theorem-N}
Let $X$, $Y$, and $f_0$ be as in Theorem 1. The set
\[U:=\{x\in X\mid f_0|_{\text{some nbhd of }x}\text{ is }\Cinfty\}\]
is open in $X$. Let $K\subset U$ be a closed subset. Then the sequence of $f_n$'s from Theorem 1 can be chosen so that $f_n|_K=f_0|_K$.
\end{theorem-N}
\begin{proof}
We have a $\Cinfty$ map $\varphi$ with $\supp(\varphi)\subset U$, and a $\Cinfty$ map $\psi$ with $\supp(\psi)\subset X-K$, so that $\psi(x)=0$ for all $x\in K$, and hence $\varphi(x)=1$ for all $x\in V$, where $V$ is a neighborhood of $K$.

When $Y=\R^n$, we have the $f_n$'s as in Theorem 1: the $f_n$'s are $\Cinfty$ and converge to $f_0$.

Let $g_n=\varphi f_0+\psi f_n$ for all $n\in\N$. Then $\varphi f_0\to f_0$ on $U$, $f_0$ is $\Cinfty$ on $U$, and $\psi f_n$ is a $\Cinfty$ map, so $g_n$ is $\Cinfty$ on $X$ and $g_n(x)=f_0(x)$ for all $x\in K$. Then we deal with open subsets of $\R^n$, $\Cinfty$ retracts of open subsets, etc.

\end{proof}
\begin{theorem-N}
Let $X$, $Y$, and $f_0$ be as in Theorem 1. Let $\Omega$ be a tubular neighborhood of $Y$, i.e.
\begin{center}
\begin{tikzcd}
Y \ar[bend left,hook]{r}{i} & \Omega\ar[bend left]{l}{r} \ar[hook]{r} & \R^n
\end{tikzcd}
\end{center}
Let $A\hookrightarrow Y$ be a closed $\Cinfty$ submanifold. Then the set
\[U=\{x\in X\mid \text{there is some neighborhood }U(x)\text{ such that }f_0|_{U(x)}\text{ is transverse to }A\}\]
is open. Let $K\subset U$ be closed. Then there is a sequence $f_n:X\to Y$ of functions such that
\begin{itemize}
\item[(i)] $f_n\to f_0$ in the compact-open topology
\item[(ii)] the $f_n$'s are $\Cinfty$ and transverse to $A$
\item[(iii)] $f_n|_K=f_0|_K$ for all $n\in\N$
\end{itemize}
\end{theorem-N}
\begin{remark}
By Theorem 2, we have $\Cinfty$ maps $f_n:X\to Y$ such that $f_n\to f_0$ uniformly and $f_n|_K=f_0|_K$. Thus, it suffices to prove that the $f_n$'s can be approximated by $f_n'$ which are $\Cinfty$, transverse to $A$, and $f_n'|_K=f_n|_K$. Thus, it suffices to prove Theorem 3 when $f_0$ is $\Cinfty$.
\end{remark}
\begin{proof}
Recall our proof that $\Cinfty$ functions can be approximated by functions transverse to a given submanifold; we will modify this proof. We had our tubular neighborhood
\begin{center}
\begin{tikzcd}
Y \ar[bend left,hook]{r}{i} & \Omega\ar[bend left]{l}{r} \ar[hook]{r} & \R^n
\end{tikzcd}
\end{center}
and considered the map $X\times S\to\R^n$ defined by $F(x,s)=f_0(x)+s$, where $S=\R^n$. But then we restrict $S$ to be a neighborhood of 0 so that the image of $F$ is entirely contained in $\Omega$, and then redefine $F(x,s)=r(f_0(x)+s)$.

We have a partition of unity $(\varphi,\psi)$ for the open cover $(U,X-K)$. Now we define $F(x,s)$ as
\[F(x,s)=r(\varphi(x)f_0(x)+\psi(x)s)\]
for $x\in X$, $s\in$ a neighborhood of 0 in $\R^n$. Let $f_s(x)=F(x,s)$. Then we have $V=\{x\in X\mid \psi(x)\neq 0\}$ and clearly $F'(x,0)$ is surjective for all $x\in V$, so it will be transverse to absolutely anything, and $f_s(x)=f_0(x)$ for all $x\in X-V$, and we assumed that $f_0$ was transverse to $A$. Thus, we can conclude that $F:X\times S\to Y$ is transverse to $A$ on $X\times\{0\}$. Because $X$ is compact, shrinking the neighborhood $S$ if necessary we can assume that $F:X\times S\to Y$ is transverse to $A$ everywhere. Thus $F^{-1}(A)$ is a manifold, we have $F^{-1}(A)\hookrightarrow X\times S\to S$, and now apply Sard's theorem.
\end{proof}


Given an element $\xi\in[X,Y]$ (this denotes the set of homotopy classes of continuous maps from $X$ to $Y$) and a closed $\Cinfty$ submanifold $A\subset Y$, pick an $f\in\xi$ that is $\Cinfty$ and transverse to $A$. Suppose that $f_0,f_1:X\to Y$ are both $\Cinfty$ and transverse to $A$, and that they are homotopic to each other via some $h$.

Define $F:X\times\S^1\to Y$ by
\begin{center}
\begin{tikzpicture}
\draw[line width=1mm,blue!80] (315:1) arc (315:405:1); 
\node[label=right:$f_1$] at (1,0) {} ;
\draw[line width=1mm,red!80] (135:1) arc (135:225:1);
\node[label=left:$f_0$] at (-1,0) {} ;
\draw[line width=1mm,purple!50] (45:1) arc (45:135:1);
\node[label=above:$h$] at (0,1) {} ;
\draw[line width=1mm,purple!50] (225:1) arc (225:315:1);
\node[label=below:$h$] at (0,-1) {} ;
\draw[line width=0.2mm] (0,0) circle (1);
\draw[thick] (45:0.95) to (45:1.05);
\draw[thick] (135:0.95) to (135:1.05);
\draw[thick] (225:0.95) to (225:1.05);
\draw[thick] (315:0.95) to (315:1.05);
\end{tikzpicture}
\end{center}
Assume that $\dim(X)>0$.

Let $U=X\times U'$ where $U'$ is a neighborhood of the regions where $F$ is either $f_0$ or $f_1$, as follows:
\begin{center}
\begin{tikzpicture}
\draw[line width=1mm,green!80!black,line cap=round] (305:1) arc (305:415:1); 
\draw[line width=1mm,green!80!black,line cap=round] (125:1) arc (125:235:1);
\draw[line width=0.2mm] (0,0) circle (1);
\draw[thick] (45:0.95) to (45:1.05);
\draw[thick] (135:0.95) to (135:1.05);
\draw[thick] (225:0.95) to (225:1.05);
\draw[thick] (315:0.95) to (315:1.05);
\end{tikzpicture}
\end{center}
and let $K=X\times\{\pm 1\}$. Apply Theorem 3 to $F:X\times\S^1\to Y$, get $\widetilde{F}:X\times\S^1\to Y$ which is $\Cinfty$, transverse to $A$, and for all $x\in X$, we have
\[\widetilde{F}(x,-1)=f_0(x),\quad \widetilde{F}(x,1)=f_1(x).\]
Then $\widetilde{F}^{-1}(A)=B$ is a $\Cinfty$ submanifold of $X\times\S^1$. We have $B\transverse X\times\{1\}=f_1^{-1}(A)\times\{1\}$ and $B\transverse X\times\{-1\}=f_0^{-1}(A)\times\{-1\}$.

The symbol $\transverse$ means ``is transverse to''.

However, this use of $\S^1$ was really an artificiality on my part. We really should have said everything for manifolds with boundary, which I'll define now.

Consider the closed lower half plane
\[H=\{(x_1,\ldots,x_n)\in\R^n\mid x_n\leq 0\},\]
with its sheaf $C_H^\infty$. An $n$-manifold with boundary is a topological space $X$ such that for all $x\in X$ there is a neighborhood $U(x)$ of $x$ in $X$ and a neighborhood $U(h)$ of some $h\in H$, and a homeomorphism $U(x)\to U(h)$, and these homeomorphisms are compatible.

A closed subset $A$ of a $\Cinfty$ manifold with boundary $(M,\boundary M)$ if, for all $a\in A$, either $a\notin \boundary M$ (this is just the old definition), or if $a\in A\cap \boundary M$, then we require that $(M,\boundary M)$ has a chart to $H$, $(x_1,\ldots,x_n)$ with $x_n\leq 0$, such that in the chart, $A$ is the set where $x_1=\cdots=x_k=0$ for some $k\neq n$, and such that $A\cap\boundary M=\boundary A$.

\begin{center}
\begin{tabular}{c|c}
Submanifold with boundary? & Examples\\\hline
\begin{tikzpicture}
\node at (0,0) {Yes};
\path (-1.1,-1.1) rectangle (1.1,1.1);
\end{tikzpicture} & \begin{tikzpicture}
\begin{scope}
\clip (0,0) circle (1);
\filldraw[draw=blue!70!black,thick,fill=blue!60!white] (1.5,0) circle (1);
\end{scope}
\draw[thick]  (0,0) circle (1);
\path (-1.1,-1.1) rectangle (1.1,1.1);
\end{tikzpicture}\begin{tikzpicture}
\begin{scope}
\clip (0,0) circle (1);
\filldraw[draw=blue!70!black,thick,fill=blue!60!white] (0.1,-0.3) circle (0.5);
\end{scope}
\draw[thick]  (0,0) circle (1);
\path (-1.1,-1.1) rectangle (1.1,1.1);
\end{tikzpicture}
\\
\begin{tikzpicture}
\node at (0,0) {No};
\path (-1.1,-1.1) rectangle (1.1,1.1);
\end{tikzpicture} & \begin{tikzpicture}
\begin{scope}
\clip (0,0) circle (1);
\draw[red,thick] (290:1) arc (220:140:0.5) node[mypoint,minimum size=3pt] {};
\end{scope}
\draw[thick]  (0,0) circle (1);
\path (-1.1,-1.1) rectangle (1.1,1.1);
\end{tikzpicture} \begin{tikzpicture}
\begin{scope}
\clip (0,0) circle (1);
\node[mypoint,minimum size=3pt,red] at (320:0.1) {};
\draw[red,thick] (320:0.1) arc (130:50:0.5) node[mypoint,minimum size=3pt] {};
\end{scope}
\draw[thick]  (0,0) circle (1);
\path (-1.1,-1.1) rectangle (1.1,1.1);
\end{tikzpicture}
\end{tabular}
\end{center}

Closed submanifolds $B_0$ and $B_1$ of a $\Cinfty$ manifold $X$ are concordant if there is a closed submanifold $B$ with boundary of $X\times I$ such that $B\cap X\times\{0\}=B_0\times\{0\}$ and $B\cap X\times\{1\}=B_1\times \{1\}$. This is an equivalence relation; let $\cal{C}$ denote the set of equivalence classes. The content of what we've proved is that there is a well-defined map
\[[X,Y]\times\cal{C}(Y)\to\cal{C}(X)\]
taking closed submanifolds in $Y$ to their preimage in $X$.