Jae Choon Cha

POSTECH

Suppose $M$ and $N$ are smooth manifolds of dimension $m$ and $n$ and $f\colon M\to N$ is a smooth map. Recall that $TM_x$ and $TN_y$ are vector spaces with $\dim TM_x = m$ and $\dim TN_y =n$ for $x\in M$ and $y\in N$, and the derivative $df_x\colon TM_x \to TN_{f(x)}$ is linear.

**Definition.**

- $x\in M$ is a
*critical point*if $\rank df_x < n$. - $x\in M$ is a
*regular point*if $x$ is not a critical point, that is, $\rank df_x = n$. - $y\in N$ is a
*critical value*if $y=f(x)$ for some critical point $x\in M$. - $y\in N$ is a
*regular value*if $y$ is not a critical value, that is, every $x\in f^{-1}(y)$ is a regular point.

The following simplest special case best illustrates what happens at a regular value and its pre-image. Suppose $M=\R^m$, $N=\R^n$, $m\ge n$, and $f\colon M\to N$ is equal to the projection $pr(x_1,\ldots,x_m) = (x_{m-n+1},\ldots,x_m)$. Then, $df_x=pr$ is obviously surjective, so $df_x$ has rank $n$ for every $x\in M$. So, every $y\in N$ is a regular value. In particular, for the regular value $y=0$, we have $f^{-1}(y) = \R^{m-n}\times 0 \subset \R^m=M$. This is an $(m-n)$-dimensional manifold.

Indeed, we will show that the general case is locally identical with the above special case.

**Lemma.** Suppose $f\colon M \to N$ is smooth, $m=\dim M$, $n=\dim N$. Suppose
$y\in N$ is a regular value of $f$ and $h\colon V\to N$ is a parametrization
such that $0\in V\subset \R^n$ and $h(0)=y$. Then for every $x\in f^{-1}(y)$,
there is a parametrizations $g\colon U\to M$ with $0\in U\subset \R^m$ such that
$g(0)=x$, $f(g(U))\subset h(V)$ and

In other words, $f$ looks like, locally, the standard projection $pr\colon \R^m \to \R^n$ which forgets the first $m-n$ coordinates.

*Proof.* Choose a parametrization $k\colon W\to M$ such that $W\subset \R^m$ is
open, $0\in W$, $k(0)=x$ and $fk(W) \subset h(V)$. Consider the composition
$h^{-1}gf \colon W\to V \subset \R^n$. Since $y$ is a regular value, $df_x \colon TM_x \to TN_y=\R^n$ is onto, and thus $d(h^{-1}fk)_0 = (dh_0)^{-1} df_x dk_0$ is onto. Let $R=\Ker d(h^{-1}fk)_0 \subset \R^m$. Then $\dim R = m-n$.
Choose a linear map $\Phi\colon \R^m \to \R^{m-n}$ such that $\Phi|_R$ is 1-1.
Define

by $F(z)=(\Phi(z),h^{-1}fk(z))$. Then $F(0)=0$, and $dF_0\colon \R^m \to \R^m$ is given by

By the choice of $\Phi$, $dF_0$ is 1-1, and thus $dF_0$ is an isomorphism. So, by the inverse function theorem, $F$ is a local diffeomorphism near $0$, that is, there are open neighborhoods $W'$, $U$ of $0$ in $\R^m$ such that $F|_ {W'}\colon W'\to U$ is a diffeomorphism. Let $g=k(F|_{W'})^{-1}\colon U \to M$.

Then $g(0)=x$. Since $(F|_{W'})^{-1}$ is a diffeomorphism and $k$ is a parametrization, $g$ is a parametrization. Moreover, since $pr\circ F = h^{-1}fk$ by the definition of $F$, it follows that $pr=h^{-1}fk (F|_{W'})^{-1} = h^{-1}fg$. (See also the above diagram.) $\quad\square$

**Theorem.** Suppose $f\colon M\to N$ is smooth with $m=\dim M$, $n=\dim N$. If
$y\in N$ is a regular value, then $f^{-1}(y)\subset M$ is a smooth manifold of
dimension $m-n$.

*Proof.* Suppose $x\in f^{-1}(y)$. Apply the lemma to obtain parametrizations
$g\colon U\to M$ and $h\colon V\to N$ such that $g(0)=x$, $h(0)=y$ and
$h^{-1}fg=pr$. We have $h^{-1}fg(z)=0$ if and only if $z\in U\cap (\R^{m-n}\times 0)$. Since $h^{-1}(y)=0$, it follows that $f^{-1}(y)\cap g(U)=g(U\cap (\R^{m-n}\times 0))$. So, the restriction $g|_{U\cap (\R^{m-n}\times 0)}\colon U\cap (\R^{m-n}\times 0) \to f^{-1}(y)$ is a
parametrization. $\quad\square$

**Theorem.** Suppose $f\colon M\to N$ is smooth and $\dim M=\dim N$. If $y\in N$ is a regular value, then for all $x\in f^{-1}(y)$, there exist open
neighborhoods $U$ of $x$ and $V$ of $y$ such that $f(U)=V$ and $f|_U\colon U\to V$ is a diffeomorphism.

*Proof.* Apply the above lemma to the given $f$, to obtain parametrizations
$g\colon U'\to M$ and $h\colon V'\to N$ such that $fg(U)\subset h(V)$ and
$h^{-1}fg$ is the identity. Let $U=g(U')$ and $V=fg(U') \subset h(V)$. Then
$f(U)=V$. Since $h^{-1}fg\colon U'\to h^{-1}(V)$, $h^{-1}\colon h^{-1}(V) \to V$
and $g\colon U'\to U$ are diffeomorphisms, $f|_U\colon U\to V$ is a
diffeomorphism. $\quad\square$

**Corollary.** If $y$ is a regular value of a smooth map $f\colon M\to N$ with
$\dim M=\dim N$, then $f^{-1}(y)$ is a discrete subset of $M$. In addition, if
$M$ is compact, then $f^{-1}(y)$ is a finite subset.

*Proof.* By the above theorem, $f^{-1}(y)$ is discrete. If $M$ is compact,
then $f^{-1}(y)$ is compact since it is closed. Since $f^{-1}(y)$ is discrete
and compact, $f^{-1}(y)$ is finite. $\quad\square$

The above corollary enables us to define the following.

**Definition.** Suppose $M$ is a compact smooth manifold, and $f\colon M\to N$
is a smooth map to a smooth manifold $N$ with $\dim N=\dim M$. For a regular
value $y\in N$ for $f$, define $\#f^{-1}(y)$ to be the number of points in the
set $f^{-1}(y)$.

**Theorem.** $\#f^{-1}(y)$ is a locally constant function on the subspace of
regular values of $f$.

*Proof.* Fix a regular value $y$. Write $f^{-1}(y)= \{ x_1,\ldots,x_r\}$ where
$r=\#f^{-1}(y)$. Choose disjoint open neighborhoods $U_i$ of $x_i$ in $M$, and
choose open neighborhoods $V_i$ of $y$ such that $f|_{U_i} \colon U_i\to V_i$ is
a diffeomorphism. Let

Observe that $y\in V$. The subset $M\sm (U_1\cup \cdots\cup U_m)$ is closed in the compact space $M$, and thus compact. It follows that $f(M\sm (U_1\cup \cdots\cup U_m))$ is compact, and consequently it is closed since $N$ is Hausdorff. Therefore $V$ is open. For each $z\in V$, $f^{-1}(z)$ lies in $U_1\cup\cdots\cup U_m$ since no points in $M\sm (U_1\cup\cdots\cup U_m)$ is sent to $V$, and moreover, for each $i$, there is a unique point in $U_i$ which is sent to $z$ under $f$, since $f\colon U_i\to V_i$ is a bijection. It follows that $\#f^{-1}(z)=r$ for each $z\in V$. $\quad\square$

**Corollary.** $\#f^{-1}(y)$ is a constant function if the set of regular values
of $f$ is connected.

### Application: Fundamental theorem of algebra

As an application of the last corollary, we will prove the following well-known theorem.

**Fundamental theorem of algebra.** Every non-constant polynomial $p(z)$ with
complex coefficients has a zero in the complex plane.

*Proof.* Write $p(z)=a_n z^n+\cdots+a_1z+a_0$ with $n\ge 1$, $a_i\in \C$,
$a_n\ne 0$. Suppose $p(z)\ne 0$ for all $z\in\C$. We will derive a
contradiction from this.

Identify $\R^3$ with $\C\times\R$, and write $S^2=\{(z,x_3)\in \C \times\R \mid |z|^2 + x_3^2 = 1\}$. Let $h_+\colon S^2 \sm \{ (0,1)\} \to \C$ be the streographic projection from the north pole $(0,1) \in S^2$, that is, for $x\in S^2\sm\{(0,1)\}$, $h(x)$ is the intersection of the ray from $(0,1)$ through $x$ with the $xy$-plane $\C$.

An elementary computation gives us $h_+(z,x_3) = \frac{z}{1-x_3}$. Similarly, the streographic projection $h_- \colon S^2\sm\{(0,-1)\} \to \C$ from the south pole $(0,-1)\in S^2$ is given by $h_-(z,x_3)=\frac{z}{1+x_3}$, since $-x_3$ plays the role of $x_3$.

It is also straightforward to show that

So, both $h_\pm$ and $h_\pm^{-1}$ are smooth, and thus $h_\pm$ is a diffeomorphism. In addition, $h_+h_-^{-1}\colon \C\sm \{0\} \to \C\sm\{0\}$ is given by $h_+h_-^{-1}(w)=\overline w{}^{-1}$. We will use the diffeomorphisms $h_\pm$ as coordinate systems.

Now, define $f\colon S^2\to S^2$ by

We claim that $f$ is smooth. It is obvious that $f$ is smooth away from $(0,1)\in S^2$. So it suffices to show that $f$ is smooth on a neighborhood of $(0,1)\in S^2$. Note that $f(x)=(0,-1)$ if and only if $x\ne(0,1)$ and $h_+^{-1}ph_+(x)=(0,-1)$, that is, $ph_+(x)=h_+(0,-1)=0$. Since $p$ has no zero, $f(x)\ne (0,-1)$ for all $x\in S^2$. This enables us to define $g\colon \C \to \C$ by $g(w)=h_- f h_-^{-1}(w)$. Since $h_-$ is a diffeomorphism and $h_-(0,1)=0$, it suffices to show that $g$ is smooth near $0\in \C$, to conclude that $f$ is smooth near $(0,1)\in S^2$. By the definition of $f$, we have

For $w\ne 0$, we have

Since the last expression is equal to $g(0)=0$ for $w=0$, $g(w)$ is given by the above equation for all $w\in \C$, including $w=0$. Note that the denominator of the last expression is nonzero for $w=0$ since $a_n\ne 0$, and is nonzero for $w\ne 0$ since if it were zero for some $w\ne 0$, then $\overline{w}{}^{-1}$ would be a zero for $p$. So, $g(w)$ is smooth. This shows that $f$ is smooth on $S^2$.

A key property of our $f$ is that its critical points are rare. To see this, recall that, when we view the polynomial $p$ as $p\colon \R^2\to \R^2$, the $2\times 2$ matrix $dp$ is nonsingular if and only if the complex derivative $p'(w)$ is nonzero. So, if $p'(w)\ne 0$, then $p$ is a local diffeomorphism near $w$, and consequently $f$ is a local diffeomorphism near $x=h_+^{-1}(w) \in S^2$, and thus $x$ is not a critical point. Since there are at most finitely many zeros for $p'$, it follows that there are at most finitely many critical points for $f$. This implies that $f$ has finitely many critical values. Therefore, the set $R$ of regular points for $f$ in $S^2$ is $S^2$ with finitely many points removed, so $R$ is connected.

By the above corollary, the function $\#f^{-1}(y)$ is constant on $R$. If $\#f^{-1}(y)$ is constantly zero, then $f(S^2)$ is contained in the set of critical values, which is a discrete set. So, in this case, $f(S^2)$ is a point, and thus $f$ is a constant. It contradicts that $p$ is a non-constant polynomial. Therefore, $\#f^{-1}(y)$ is a nonzero constant on $R$. Since a critical value of $f$ is contained in $f(S^2)$, it follows that $f$ is onto. Recall that when $p$ has no zero, $f(x)\ne (0,-1)$ for all $x\in S^2$. This is a contradiction. $\quad\square$