Jae Choon Cha
POSTECH
In this lecture we generalize the notion of a smooth manifold as follows. Let
be the $n$dimensional half space. This is a topological space, as a subspace of $\R^n$.
Definition. A subset $M\subset \R^k$ is a smooth manifolds of dimension $n$ with boundary if for all $x\in M$, there is a diffeomorphism of an open subset of $\R^n_+$ onto an open neighborhood of $x$ in $M$. We call such a diffeomorphism a parametrization.
What is the difference from the earlier definition of a smooth manifold? In the above new definition, the neighborhood of $x$ is required to be diffeomorphic to an open subset in $\R^n_+$, instead of $\R^n$. Recall, from the definition of the subspace topology, an open set in $\R^n_+$ is of the form $U\cap \R^n_+$ for some open set $U$ in $\R^n$. If $U$ contains a point $p=(x_1,\ldots,x_n)$ with $x_n=0$, then $U\cap \R^n_+$ is no longer open in $\R^n$ while it is open in $\R^n_+$, since for any $\epsilon>0$, the open ball $B(p,\epsilon)$ in $\R^n$ is not contained in $U\cap \R^n_+$.
The above discussion tells us that the following subset is essential. Let
Definition. Suppose $M$ is a smooth manifold of dimension $n$ with boundary. The boundary $\partial M$ is defined to be the subset of the points $x\in M$ for which there is a parametrization $g\colon U\cap \R^n_+ \to M$ with $U\subset\R^n$ open, such that $x\in g(U\cap \partial \R^n_+)$. The complement $M\sm \partial M$ is called the interior of $M$, and denoted by $\inte M$.
Examples.

As the simplest example, $\R^n_+$ is a smooth manifold of dimension $n$ whose boundary is $\partial \R^n_+$. So, our earlier definition of $\partial \R^n_+$ agrees with the general case definition of the boundary.

Let $D^n=\{x\in \R^n \mid x\le 1\}$. It is straightforward to verify directly that $D^n$ is a smooth manifold of dimension $n$ with boundary $\partial D^n = S^{n1}$.
Lemma. Suppose $M$ is a smooth manifold of dimension $n$ with boundary. Then the following hold.

The boundary $\partial M$ is a smooth manifold of dimension $n1$ (without boundary).

The interior $\inte M$ is a smooth manifold of dimension $n$ (without boundary).
Since they are straightforward consequences of the definition, we omit the proof.
The following provides a useful method to obtain a smooth manifold with boundary.
Theorem. Suppose $M$ is a smooth manifold without boundary, and $f\colon M \to \R$ is a smooth map which has $0\in \R$ as a regular value. Then the subspace $P=\{x\in M\mid f(x)\ge 0\}$ is a smooth manifold with $\partial P=f^{1}(0)$.
Proof. The subset $\{x\in M \mid f(x)>0\}$ is open in $M$ and thus it is a smooth manifold. So, it remains to show that every $x\in P$ such that $f(x)=0$ has an open neighborhood in $P$ which is diffeomorphic to an open subset of $\R^m_+$, where $m$ is the dimension of $M$. Apply our earlier theorem on the inverse image of a regular value, to obtain a parametrization $g\colon U\to M$ such that $0\in U\subset \R^m$, $g(0)=x$ such that $fg(x_1,\ldots,x_m)=x_m$. Now $g(U\cap \R^m_+) = P\cap g(U)$. So $g_ {U\cap\R^m_+} \colon U\cap\R^m_+ \to P$ is a parametrization for $P$ such that $g(U\cap \partial\R^m_+) = f^{1}(0)\cap g(U)$. $\quad\square$
Example. Here is a proof that $D^n$ is a smooth manifold with $\partial D^n=S^{n1}$: the function $f\colon \R^n \to \R$ given by $f(x_1,\ldots,x_m)=1(x_1^2+\cdots+x_m^2)$ has $0$ as a regular value, and thus $f^{1}[0,\infty)=D^n$ is a smooth manifold with boundary $f^{1}(0)=S^{n1}$.
The following result generalizes our previous theorem on the inverse image of a regular value, to the case of manifolds with nonempty boundary.
Theorem. Suppose $f\colon M\to N$ is a smooth map, where $M$ is smooth manifold of dimension $m$ with boundary and $N$ is smooth manifold of dimension $n$. Suppose $y\in N$ is a regular value for both $f\colon M\to N$ and $f_{\partial M}\colon \partial M\to N$. Then $f^{1}(y)$ is a smooth manifold of dimension $mn$ with boundary, and $\partial f^{1}(y)=f^{1}(y)\cap \partial M$.
Proof. By applying our previous theorem on the inverse image of a regular value, to the smooth map $f_{\inte M}\colon \inte M \to N$, it follows that $f^{1}(y)\cap \inte M$ is a smooth manifold of dimension $mn$. So, it suffices to show that for every $x\in f^{1}(y) \cap \partial M$, there is a parametrization $g\colon U\cap \R^{mn}_+ \to f^{1}(y)$, with $U$ an open neighborhood of $0$, such that $g(0)=x$ and $g(U\cap \partial \R^{mn}_+)=f^{1}(y)\cap \partial M \cap g(U)$.
Choose parametrizations $k\colon W\cap \R^m_+ \to M$ with $0\in W\subset \R^m$ and $h\colon V\to N$ with $0\in V\subset \R^n$, such that $k(0)=x$, $h(0)=y$ and $f(k(W\cap \R^m_+)) \subset h(V)$. Then the composition $h^{1}fk\colon W\cap \R^m_+ \to V$ is defined and smooth. By the definition of a smooth map, we may assume (by shrinking $W$) that there is an extension $F\colon W\to V$ of $h^{1}fk$, that is, $F_ {W\cap \R^m_+} = h^{1}fk$.
We will find a parametrization $\ell\colon U\to W$, with $0\in U\subset \R^m$ and $\ell(0)=0$, which has the following two properties:

$F\ell(x_1,\ldots,x_m) = (x_1,\ldots,x_n)$

The last coordinate of $\ell(x_1,\ldots,x_m)\in W\subset \R^m$ is equal to $x_m$.
Recall that the first property was already appeared and used in the proof this theorem for the case of manifolds without boundary. The second property is essential for the case of manifolds with boundary, and in its proof, the additional hypothesis that $y$ is a regular value of not only $f$ but also $f_{\partial M}$ plays a crucial role, as we will see below.
Since $y\in N$ is a regular value for $f_ {\partial M}$, $0=h^{1}(y) \in V$ is a regular value for the restriction $F_ {W\cap \partial\R^m_+} \colon W\cap \partial\R^m_+ = W\cap \R^{m1}_+ \to V$. So, the $n\times (m1)$ submatrix $[\frac{\partial F_i}{\partial x_j}]_{1\le i\le n,1\le j\le m1}$ has rank $n$ at $0$. Define $G\colon W\to V\times \R \subset \R^{n+1}$ by $G(x_1,\ldots,x_m) = (F(x_1,\ldots,x_m),x_m)$. Then $G(0)=0$, and
So, $dG_0$ has rank $n+1$. By the lemma in Lecture Note 14, it follows that there is a parametrization $\ell\colon U\to W$ with $0\in U\subset \R^m$ such that $\ell(0)=0$ and $G\ell(x_1,\ldots,x_m)=(x_1,\ldots,x_n,x_m)$. Then $F\ell(x_1,\ldots,x_m) = (x_1,\ldots,x_n)$. Moreover, since the last coordinate of $G(x_1,\ldots,x_m)$ is $x_m$, the last coordinate of $\ell(x_1,\ldots,x_m)$ is $x_m$ too. So our parametrization $\ell$ has the promised properties.
From the above properties of $\ell$, it follows that
Therefore, $g=k\ell_ {U\cap (0\times \R^{mn}_ +)} \colon U \cap (0\times \R^{mn}_+)\to f^{1}(y)$ is a parametrization. Moreover, $g(0)=k\ell(0)=x$, and $g(U\cap (0\times \partial\R^{mn}_+)) = f^{1}(y)\cap \partial M$. $\quad\square$
SardBrown theorem
Now we know that the inverse image of a regular value is a manifold. Then, how often can we invoke this? In other words, are regular values found frequently? For instance, recall what happened in our proof of the fundamental theorem of algebra: the smooth map $f\colon S^2 \to S^2$ we constructed from a complex polynomial had finitely many critical points, and thus all points in $S^2$ but finitely many were regular values. In general, we do not have such a nice situation, but it turns out that regular values are not rare. The following statement is (a special form of) a wellknown result due to Sard and Brown.
Theorem. (SardBrown) Suppose $M$ and $N$ are smooth manifolds, and $f\colon M\to N$ is a smooth map. Then, regular values of $f$ form a dense subset in $N$.
Application: Brouwer’s fixed point theorem
As an application of the machinery developed above together with the SardBrown theorem, we will present a proof of Brouwer’s fixed point theorem, which is stated below.
Brouwer’s fixed point theorem. For every continuous map $f\colon D^n\to D^n$ ($n\ge 1$), there is $x\in D^n$ such that $f(x)=x$.
As the first step of the proof, we will show the following result, which is intresting on its own. For $A\subset X$, we say that a map $r\colon X\to A$ is a retraction if $r(x)=x$ for all $x\in A$.
Theorem. Suppose $M$ is a smooth manifold with nonempty boundary $\partial M$. Then there is no smooth retraction $r\colon M\to \partial M$.
Proof. Suppose $r\colon M\to \partial M$ is smooth and $r(x)=x$ for all $x\in \partial M$. By the SardBrown theorem, there is a regular value $y\in \partial M$ for $r$. Since $r_{\partial M}$ is the identity, $y$ is also a regular value for $r_{\partial M}$. It follows that $P=r^{1}(y)$ is a 1dimensional manifold with $\partial P=r^{1}(y)\cap \partial M$. Since $r_{\partial M}$ is the identity, $\partial P=\{y\}$. On the other hand, $P$ is compact, and a compact 1dimensional manifold with boundary is a union of finitely many copies of $I$ or $S^1$. This implies that $\partial P$ consists of even number of points. This is a contradiction. $\quad\square$
Theorem. (Smooth version of Brouwer’s theorem) If $f\colon D^n\to D^n$ is smooth, then $f(x)=x$ for some $x\in D^n$.
Proof. Suppose $f(x)\ne x$ for all $x\in D^n$. Define $r\colon D^n\to \partial D^n=S^{n1}$ as follows: for $x\in D^n$, the ray which starts from $f(x)$ and passes through $x$ has a unique intersection with $\partial D^n$. Let $r(x)\in \partial D^n$ be the intersection. It is not difficult to prove that $r\colon D^n\to \partial D^n$ is smooth. Indeed, the ray is parametrized as $\alpha(t)=(1x)f(x)+tx$, $t\ge 0$, and $r(x)=\alpha(t)$ when $\alpha(t)^2=1$. Letting $v=f(x)$ and $u=xf(x)$, this is equivalent to $u^2t^2 – 2(u\cdot v)t+v^21=0$. Solving this for $t$, one obtains
Since $u\ne 0$, this is smooth and thus so is $r(x)$. Since $r(x)=x$ for $x\in \partial D^n$, $r$ is a smooth retraction. But there is no smooth retraction of $D^n$ onto $\partial D^n$, by the above theorem. This shows that $f(x)=x$ for some $x\in D^n$. $\quad\square$
Now we are ready to show Brouwer’s fixed point theorem for a continuous map, without assuming that it is smooth.
Proof of Brouwer’s fixed point theorem. Suppose $f\colon D^n\to D^n$ be a continuous map and $f(x)\ne x$ for all $x\in D^n$. Since $D^n$ is compact,
is positive. By Weierstrass’ theorem, there is a map $P\colon \R^n\to \R^n$ with polynomial coordinates such that $P(x)f(x) < \epsilon/2$ for all $x\in D^n$. (Here we use that $D^n$ is bounded.) Define, for $x\in D^n$,
Since $P(x) \le f(x)+P(x)f(x) \le 1+\frac{\epsilon}{2}$, we have $g(x) \le 1$. That is, $g$ is a smooth map $D^n\to D^n$. By the smooth version of Brouwer’s theorem, $g(x)=x$ for some $x\in D^n$. Now,
It contradicts the choice of $\epsilon$. $\quad\square$