\classheader{2013-01-18}

There was a mistake in the statement of exercise 5.10.

One way of producing deformations of $f_0:X\to Y$ is via the tubular neighborhood theorem. This is an example of where we'll use existence and uniqueness of solutions to ODE's.

There are many forms of the theorem, but we'll state it for $Y$ compact.

\begin{theorem}[Tubular neighborhood theorem]
Let $Y\subseteq\R^{m+n}$ be a compact $C^\infty$ submanifold of dimension $n$. Let \[N=\{(y,v)\in Y\times\R^{m+n}\mid \langle w,v\rangle =0\text{ for all }w\in T_yY\},\]
and let
\[N_\epsilon=\{(y,v)\in N\mid \|v\|<\epsilon\}.\]
Then
\begin{enumerate}
\item $N$ is a closed $C^\infty$ submanifold of $Y\times\R^{m+n}$.
\item There is some $\epsilon>0$ such that $(y,v)\mapsto y+v$ for $(y,v)\in N_\epsilon$ gives a diffeomorphism from $N_\epsilon$ to an open subset of $\R^{m+n}$. We say that this open subset is a ``tubular neighborhood''.
\end{enumerate}
\end{theorem}
\begin{proof}
Because we can check the condition of being a submanifold locally, to show that $N$ is a submanifold it suffices to cover $Y$ by open subsets $U$ such that $p_1^{-1}(U)$ is a $C^\infty$ submanifold of $U\times\R^{m+n}$, where $p_1:N\to Y$ is the projection map. We may assume that each $U$ is of the form $\{(x,f(x))\mid x\in\Omega\}$ for some open $\Omega\subseteq\R^n$ and for some $C^\infty$ map $f:\Omega\to\R^m$ (after some permutation in $S_{n+m}$).

Let $y=(x,f(x))$, with $x\in\Omega$. Then
\[T_yY=\{(y,f'(x)v)\mid v\in\R^n\}.\]
Write $w=(w_1,w_2)$ where $w_1\in\R^n$ and $w_2\in\R^m$. Then
\[\langle v,w_1\rangle +\langle f'(x)v,w_2\rangle = \langle v,w_1\rangle +\langle v,f'(x)^*w_2\rangle = 0\]
where $f'(x)^*:\R^m\to\R^n$ is the adjoint of $f'(x):\R^n\to\R^m$. 

For all $v\in\R^n$, $\langle v,w_1+f'(x)^*w_2\rangle=0$, so $w_1=-f(x)^*w_2$. Thus
\[p_1^{-1}(U)=\{\underbrace{(x,f(x))}_{\in\,Y},(-f'(x)^*w_2,w_2)\mid x\in\Omega,\,w_2\in\R^m\}.\]
If $\varphi:M\to\R^n$ is $C^\infty$ then its graph in $M\times\R^n$ is a $C^\infty$ submanifold (with $M=U\times\R^m$, we get the result).

Now to part 2. Let $A:N\to\R^{m+n}$ be defined by $A(y,v)=y+v$ for all $(y,v)\in N$. Note that we have $p^{-1}(U)\cong U\times\R^m$.

Let's discuss $T_{(y_0,0)}N$. We have two submanifolds of $N$, namely $N_{y_0}=p_1^{-1}(y_0)$ and $Y\hookrightarrow N$ (this is just $y\mapsto (y,0)$).

We see that $N_{y_0}$ and $Y$ are submanifolds of $N$ that meet transversely at $(y_0,0)$, and in fact
\[T_{y_0}Y\oplus T_{(y_0,0)}(N_{y_0})=T_{(y_0,0)}N.\]
By definition, $A'(y_0,0)(\alpha,0)=\alpha$ for all $\alpha\in T_{y_0}Y$, and $A'(y_0,0)(0,w)=(0,w)$. Therefore $A'(y_0,w)=y_0+w$. It follows that $A'(y_0,0):T_{(y_0,0)}N\xrightarrow{\cong} T_{y_0}\R^{n+m}$.

Then $Z=\{(y,v)\mid A'(y,v)\text{ is a diffeomorphism}\}$ is an open subset $\supseteq Y\times 0$. The compactness of $Y$ implies that there is some $\epsilon>0$ such that $N_\epsilon\subseteq Z$.

Now, we claim that there is some $\epsilon>0$ such that $A|_{N_\epsilon}$ is injective. If not, we'd get $(y_n',v_n')\neq (y_n'',v_n'')$, both in $N_{1/n}$, such that $y_n'+v_n'=y_n''+v_n''$ ($\ast$). Since $Y$ is compact, we may assume that $\lim_{n\to\infty}y_n'=:y'$ and $\lim_{n\to\infty} y_n''=:y''$ exist. We now get $y'=y''$, by taking the limit of $\ast$ as $n\to\infty$.

Both sequences eventually lie in a fixed neighborhood of $(y',0)\in N$, but this contradicts the inverse function theorem because $A'(y,0)$ is a diffeomorphism.

Thus $A|_{N_\epsilon}:N_\epsilon\to\R^{n+m}$ is injective, open, and $A'$ is injective for all points of $N_\epsilon$. Thus, $A(N_\epsilon)=U_\epsilon$ is open, and $A:N_\epsilon\to U_\epsilon$ is a diffeomorphism to an open subset $U_\epsilon\subset\R^{m+n}$.
\end{proof}

Now, let's prove the existence of rich deformations of $f_0:X\to Y$ where both are compact $C^\infty$ manifolds.
\begin{proof}
Let $i:Y\hookrightarrow \R^{m+n}$. We have $Y\subset U_\epsilon\subset\R^{m+n}$.

We know that $i\circ f_0$ has a rich deformation $G:X\times S\to\R^{m+n}$, i.e. $G'(x,0)$ is surjective for all $x\in X$, and $G(x,0)=f_0(x)$ for all $x\in X$. Then $G^{-1}(U_\epsilon)$ is open in $X\times S$, which contains $X\times\{0\}$.

Because $X$ is compact, there is an open neighborhood $S'\subseteq S$ of 0 such that $X\times S'\subseteq G^{-1}(U_\epsilon)$. Thus, $G(X\times S')\subseteq U_\epsilon$. 

There is a $C^\infty$ retraction $r:U_\epsilon\to Y$ (remember that $U_\epsilon$ is a ``tube'' around $Y$).
\begin{center}
\begin{tikzcd}
X\times S' \ar{r}{G} \ar[bend right]{rr}[swap]{F} & U_\epsilon \ar{r}{r} & Y
\end{tikzcd}
\end{center}
\[\underbrace{p_1(A|_{N_\epsilon})^{-1}}_{=r}y=y\text{ for all }y\in Y\]

\begin{center}
\begin{tikzcd}
N_\epsilon \ar{r}{A|_{N_\epsilon}}[swap]{\cong} \ar{d}[swap]{p_1} & U_\epsilon & Y\ar[hook]{l}\\
Y & Z&
\end{tikzcd}
\end{center}

For all $x\in X$, $f_0(x)\in Y$, and
\[F(x,0)=rG(x,0)=rf_0(x)=f_0(x).\]
$G'(x,0)$ is onto by assumption for all $x\in X$.
\begin{center}
\begin{tikzcd}
Y \ar[hook]{r} \ar[bend right]{rr}[swap]{\id} & U_\epsilon \ar{r}{r} & Y
\end{tikzcd}
\end{center}
so $r'(x)$ is onto for all $x\in Y$.

$G(x,0)\in Y$ so $r'(f_0(x))\cdot G'(x,0)=F'(x,0)$ is onto.
\end{proof}


Let $X$ be a compact $C^\infty$ manifold. We want a ``rich'' (our definition of this can vary) deformation of $f_0:X\to Y$ such that $f_s:X\to Y$ is an embedding.

Let $F:X\times S\to Y$ be a deformation of $f_0$.
\[B_0=\{(x_1,x_2)\in X\times X\mid x_1\neq x_2\text{ and }f_0(x_1)=f_0(x_2)\}\]
\[B=\{(x_1,x_2,s)\in X\times X\times S\mid x_1\neq x_2\text{ and }F(x_1,s)=F(x_2,s)\}\xrightarrow{\;\;q\;\;}S\]
$q^{-1}s$ is the ``bad set of $f_s:X\to Y$''. 
\begin{center}
\begin{tikzcd}[row sep=0in]
X\times X\times S \ar{r}{G} & Y\times Y\\
(x_1,x_2,s) \ar[mapsto]{r} & (F(x_1,s),F(x_2,s))
\end{tikzcd}
\end{center}
Then $G^{-1}(\Delta Y)=B$, where $\Delta Y$ is the image of the diagonal embedding $Y\to Y\times Y$. We may \textbf{hope} that $G$ is transverse to $\Delta Y$. Note that $\Delta Y$ is $C^\infty$ submanifold whose codimension in $Y\times Y$ is $\dim(Y)$. $B\subset X\times X\times S$ is a submanifold of codimension $=\dim(Y)$, i.e. $\dim(B)=(2\dim(X)-\dim(Y))+\dim(S)$.

Consider $q:B\to S$. Sard's theorem say that there are plenty of regular values of $q$. If $s$ is a regular value, then $B_s$ is a $C^\infty$ submanifold of $B$ of codimension $=\dim(S)$.
\[\dim(B_s)=\dim(B)-\dim(S)=2\dim(X)-\dim(Y)+\dim(S)-\dim(S).\]
Thus, under the assumption that $2\dim(X)<\dim(Y)$, we may hope that $f_s:X\to Y$ is injective. To get an embedding, we have to work a little more.