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\centerline{\bf Vector Fields and Classical Theorems of Topology}
\medskip
\centerline{\bf by Daniel Henry Gottlieb}
\bigskip

In this talk we prove a collection of classical theorems using the concept
of the index of a vector field.  They are:
\bigskip
\item{}
The Intermediate Value Theorem

\item{}
The Fundamental Theorem of Algebra

\item{}
Rouche's Theorem

\item{}
The Gauss-Lucas Theorem

\item{}
The Gauss-Bonnet Theorem

\item{}
The Brouwer Fixed Point Theorem

\item{}
The Borsuk-Ulam Theorem

\item{}
The Jordan Curve Theorem

\item{}
Gottlieb's Theorem

\item{}
The Poincare-Hopf Theorem.
\bigskip

The reason for this exercise is to argue that the following equation,
which we call the ``Law of Vector Fields,'' is fated to play a central role
in mathematics.  These theorems are all easy consequences of the law of
vector fields.  The proofs are so mechanical that one could say that the
Law of Vector Fields is a generalization of each of them.

The Law of Vector Fields is the following:  Let $M$ be a compact smooth
manifold and let $V$ be a vector field on $M$ so that $V(m)\not=\vec 0$ for
all $m$ on the boundary $\partial M$ of $M$.  Then $\partial M$ contains
an open set $\P_-M$ which consists of all $m\in\P M$ so that $V(m)$ points
inside.  We define a vector field, denoted $\P-V$ on $\P_-M$, so that
for every $m\in\P_-M$ we have $\P_-V(m)=$ Projection of $V(m)$ tangent
to $\P_-M$.  Under these conditions we have
$$
\TI\ V+ \TI\ \partial_-V=\chi(M)\tag1
$$
where $\TI\ V$ is the index of the vector field and $\chi(M)$ is the Euler
characteristic of $M$.  (\cite{M}, \cite{G${}_2$-G${}_5$}, \cite{P}).

The Law of Vector Fields can be used to define the index of vector fields,
so the whole of index theory follows from (1).  The definition of index is
not difficult, but proving it is well-defined is a little involved \cite{G-S}.
The definition proceeds as follows:  

\item{a)} 
The index of an empty vector field is
zero.  

\item{b)}  
If $M$ is a finite set of points and $V$ is defined on all of the
$M$ (the vectors are necessarily zero), then $\TI(V)=\text{ number of points 
in } M$.

\item{c)}
If $V$ is a {\it proper} vector field on a compact $M$, by which we mean $V$
has no zeros on $\partial M$, then we set
$$
\TI\ V=\chi(M)-\TI(\partial_-V).
$$

\item{d)}
If $V$ is defined on the closure of an open subset $U$ of a smooth manifold
$M$ so that the set of zeros $Z$ is compact and $Z\subset U$, then we say
$V$ is a {\it proper vector} field.  The index $\TI\ V$ is defined to be
$\TI(V|M)$ where $M$ is any compact manifold such that $Z\subset M\subset U$.

\item{e)}
If $C$ is a connected component of $Z$ and $C$ is compact and open in $Z$ define
$\TI_C(V)$ to be the index of $V$ restricted to an open set containing $C$
and no other zeros of $V$.

A key idea in proving this definition is well-defined is a generalization
of the concept of homotopy which we call otopy.  An {\it otopy} is what
$\partial_-V$ undergoes when $V$ undergoes a homotopy.  The formal definition
is as follows:  An {\it otopy} is a vector field $V$ defined on the closure
of an open set $T\subset M\times I$ so that $V(m,t)$ is tangent to the slice
$M\times t$.  The otopy is {\it proper} if the set of zeros $Z$ of $V$ is
compact and contained in $T$.  The restriction of $V$ to $M\times 0$ and
$M\times 1$ are said to be properly otopic vector fields.  Proper otopy is
an equivalence relation.

The following properties hold for the index:

\item{(2)}
Let $M$ be a connected manifold.  The proper otopy classes of proper vector
fields on $M$ are in one to one correspondence via the index to the
integers.  If $M$ is a compact manifold with a connected boundary, then a
vector field $V$ is properly homotopic to $W$ if and only if
$\TI\ V=\TI\ W$.
\medskip

\item{(3)}
$\TI(V|A\cup B)=\TI(V|A)+\TI(V|B)-\TI(V|A\cap B)$
\medskip

\item{(4)}
$\TI(V\times W)=\TI(V)\cdot\TI(W)$
\medskip

\item{(5)}
$\TI(-V)=(-1)^{\text{dim } M}(V)$
\medskip

\item{(6)}
If $V$ has no zeros, then $\TI(V)=0$
\medskip

\item{(7)}
$\TI(V)=\sum_C \TI_C(V)$ for all compact connected components $C$, assuming
$Z$ is the union of a finite number of compact connected components.
\bigskip
For certain vector fields the index is equal to classical invariants. Suppose
$f:{\Bbb R}^n\to{\Bbb R}^n$.  Let $M$ be a compact $n$ submanifold.  Define
$V^f$ by $V^f(m)=f(m)$.  If $f:\partial M\to{\Bbb R}^n-\vec 0$, then
\medskip

\item{(8)}
$\TI\ V^f=\text{deg }f$.
\medskip

Suppose $f:U\to{\Bbb R}^n$ where $U$ is an open set of ${\Bbb R}^n$.  Let
$V_f(m)=\vec m-\overset\longrightarrow\to{f(m)}$.  Then
\medskip
\item{(9)}
$\TI\ V_f=$ fixed point index of $f$ on $U$.
\bigskip

Suppose $f:M\to N$ is a smooth map between two Riemannian manifolds.  Let
$V$ be a vector field on $N$.  Let $f^*V$ be the pullback of $V$ on $M$.  We
define the pullback by
$$
\langle f^*V(m),\vec v_m\rangle=\langle V(f(m)), f_*(\vec v_m)\rangle.
$$
Note that for $f:M\to{\Bbb R}$ and $V=\frac d{dt}$, we have $f^*V=
\text{ gradient } f$.

Now suppose that $f:M^n\to {\Bbb R}^n$ where $M^n$ is compact and $V$ is a
vector field on ${\Bbb R}^n$ so that $f$ has no singular points near
$\partial M$ and $V$ has no zeros on $f(\partial M)$.  Then if $n>1$
\medskip
\item{(10)}
$\TI\ f^*V=\sum w_iv_i+(\chi(M)-\text{deg } \hat N)$
\medskip
\noindent
where $\hat N:\partial M\to S^{n-1}$ is the Gauss map defined by the
immersion of $\partial M$ if ${\Bbb R}^n$ under $f$, and 
$v_i=\text{\TI}_{c_i}(V)$ for the i${}^{\text{th}}$ zero of $V$ and $w_i$ 
is the winding number of the i${}^{\text{th}}$ zero
with respect to $f:\partial M\to{\Bbb R}^n$.  That is calculated by sending
a ray out from the i${}^{\text{th}}$ zero and noting where it hits the 
immersed $n-1$
manifold $\partial M$.  At each point of intersection the ray is either
passing inward or outward relative to the outward point normal $N$.  Add up
these point assigning $+1$ if the ray is going from inside to outside, and
$-1$ if the ray goes from outside to inside.
\bigskip
\noindent
{\bf \S1.  The Intermediate Value Theorem:}

We have a map $f:I\to R$ so that $f(0)>0$ and $f(1)<0$.  We must show
that $f(c)=0$ for some $c$.  Consider $V^f$ on $I$.  $\TI(\partial_-V^f)=2$.
Hence (1) says $\TI\ V^f=-1$.  Hence (6) implies that $V^f$ has a zero, hence
$f$ hits $0$.
\bigskip

\noindent
{\bf \S2.  The Fundamental Theorem of Algebra:}

The proof will also work for a polynomial in the quaternions, provided the
homogeneous polynomial of top degree terms have an isolated zero.  Let
$f(z)=a_nz^n+a_{n-1}z^{n-1}+\dots+ a_0$.  Consider $V^f$ restricted to a
disk about the origin with radius $r$ so large that 
$|a_nz^n|>|a_{n-1}z^{n-1}+\dots+a_0|$.  Consider the homotopy $f(z)$ to
$g(z)=a_nz^n$, via $f_t(z)=a_nz^n+t(a_{n-1}z^{n-1}+\dots+a_0)$.  The
homotopy $V^{f_t}$ of the associated vector fields is proper.  No zero
can appear on the boundary.  Then homotopy $a_nz^n$ to $z^n$.  Let
$h(z)=z^n$, then contract $D$ through smaller and smaller radii until
$r=1$.  Then $\TI(V^h)=n\not=0$ using (1) for example.  Since the process
is a proper otopy, by (2), $\TI(V^f)\not=0$ and so by (6), $V^f$ has a zero,
so $f(z)$ has a root.
\bigskip
\noindent
{\bf \S3.  Rouche's Theorem}

If $f$ is analytic on a region $M$ so that $|f(z)|>|g(z)|$ for all
$z\in\partial M$, then $f(z)$ has as many zeros as $f(z)-g(z)$, if the
zeros are simple.  The homotopy $f_t(z)=f(t)-tg(z)$ is proper, so
$\TI\ V^f=\TI\ V^{f-g}$.  Now $\TI\ V^f=$ number of simple zeros.
\vfill\eject

\noindent
{\bf \S4. The Gauss-Lucas Theorem}

This states that if $f(z)$ is a polynomial, the zeros of $f'(z)$ are
contained in the convex hull of the zeros of $f(z)$.

First we study the case of a map $f:{\Bbb R}^m\to{\Bbb R}^n$ and a vector
field $V$ on ${\Bbb R}^n$.  Suppose $\vec x=(x_1,\dots,x_m)\in{\Bbb R}^m$
represents a point and $f_i$ and $V_j$ are the components of $f$ and $V$
respectively.

\proclaim{Theorem 1}
$f^*V(\vec x)=(V_1(f(\vec x)),\dots,V_n(f(\vec x))\left(\smallmatrix
\frac{\partial f_1}{\partial x_1}\dots \frac{\partial f_1}{\partial x_m}\\
\\ 
\frac{\partial f_n}{\partial x_1}\dots\frac{\partial f_n}{\partial x_m}
\endsmallmatrix\right)$  where $\left(\frac{\partial f_i}{\partial x_j}\right)$ is the Jacobian matrix.
\endproclaim

\demo{Proof}
Let $\vec\sigma_x=(\sigma_1,\dots,\sigma_m)$.  Then the defining relation
for $f^*V(m)$ given by $\langle f^*V(m),\vec\sigma\rangle=\langle V(f(m)),
f_*(\vec\sigma)\rangle$ yields
$$
\langle f^*V(m),\vec\sigma\rangle =
 (V_1(f(m)),\dots,V_n(f(m))\left(\matrix
\frac{\partial f_1}{\partial x_1}\dots \frac{\partial f_1}{\partial x_m}\\
\\ 
\frac{\partial f_n}{\partial x_1}\dots\frac{\partial f_n}{\partial x_m}
\endmatrix\right)\left(\matrix \sigma_1\\ \vdots\\ \sigma_m\endmatrix
\right).
$$
Hence the theorem is proved.
\enddemo

Now suppose $f:{\Bbb C}\to{\Bbb C}$.  Then $f(z)$ can be written as
$f(z)=u(z)+iv(z)$.

\proclaim{Theorem 2}
If $f$ is a complex analytic function and $V$ is a continuous vector field
on ${\Bbb C}$, then $f^*V(z)=\overline{f'(z)}\cdot V(f(z))$.
\endproclaim

\demo{Proof}
Now $f'(z)=\frac{df(z)}{dz}=u_x+iv_x=v_y-iu_y$ since the Cauchy-Riemann
equations hold if $f$ is analytic.  Hence $f'(z)=u_x(z)-iu_y(z)$.  Now from
Theorem 1
$$
\aligned
f^*V(z) &= (V_1(f(z)),V_2(f(z))\left(\matrix u_x & u_y\\
v_x & v_y\endmatrix\right)\big|_z\\
&=(V_1(f(z)),V_2(f(z))\left(\matrix u_x & u_y\\  -u_y & u_x\endmatrix\right)
\big|_z\quad\text{ by Cauchy Riemann}\\
&=(u_xV_1(f(z))-u_yV_2(f(z)),\quad u_yV_1(f(z))+u_xV_2(f(z)))\big|_z.
\endaligned
$$
In complex notation we may rewrite the equality above as
$$
\aligned
f^*V(z)&=[u_xV_1(f(z))-u_yV_2(f(z))]+i[u_yV_1(f(z))+u_xV_2(f(z))]\\
&=(u_x+iu_y)(V_1(f(z))+iV_2(f(z)))\\
&=\overline{(u_x-iu_y)}(V_1(f(z))+iV_2(f(z)))\\
&=\overline{f'(z)}V(f(z)).
\endaligned
$$
Thus $f^*V(z)=\overline{f'(z)}V(f(z))$.\quad $\square$
\enddemo

Now let $V$ be the vector field given by $V(z)=z$.  Then $f^*V(z)=
\overline{f'(z)}f(z)$.  Thus $f^*V$ has zeros at the zeros of $f(z)$ and at the
zeros of $f'(z)$.  If $f(z)$ is a polynomial with zeros  at the set
$a_1,\dots,a_n$, then 
$$
f^*V(z) = \overline{f'(z)} f(z) =|f(z)|^2\left(\sum_{i=1}^n\ \frac{z-a_i}
{|z-a_i|^2}\right).
$$
Now this vector field cannot have a zero outside of the convex hull of the
zeros $a_1,\dots,a_n$.  Thus the Gauss-Lucas Theorem holds.

Note that $W(z)=\frac z{|z|}$ is the pullback of $\text{grad}|z|=\nabla(|z|)$.
Thus $f^*W=\nabla|f(z)|$.  Now since $W$ and $V$ are both pointing in the
same directions, we see that $f^*V$ is orthogonal to the level curves
$|f(z)|=k$.  So if $M_k=\{z|f(z)|\le k\}$, then
$$
\aligned
\chi(M_k) &= (\# \text{ of components of } M_k)-(\# \text{ of holes in }
M_k)\\
&=(\# \text{ of zeros of } f(z))- (\# \text{ of zeros of $f'(z)$ in }
M_k).
\endaligned
$$
This is a theorem of Hurwitz.
\bigskip
\noindent
{\bf \S5. The Gauss-Bonnet Theorem}

Equation (10) is a generalization of the extrinsic Gauss-Bonnet Theorem. We
obtain a proof as follows.  The Gauss-Bonnet Theorem as proved by Hopf goes
as follows \cite{HO${}_1$} [S, P.386].  The curvature integrala of a closed
submanifold $\partial M$ of dimension $n-1$ in ${\Bbb R}^n$ is the degree
of the Gauss map $\hat N:\partial M\to S^{n-1}$ where $S^{n-1}$ is the
unit sphere.  Then Hopf generalized the Gauss-Bonnet Theorem by showing
$\text{deg }\hat N=\frac12 \chi(\partial M)$ if $n$ is odd.  Now if we
consider a map $f:M^n\to {\Bbb R}^n$ so that (10) holds, and if we let
$x:{\Bbb R}^n\to {\Bbb R}$ be projection onto the $x$ axis, then (10) becomes
\bigskip
\item{(11)}
$\TI\ \nabla(f\circ x)=\chi(M)-\text{ deg }\hat N$
\bigskip
\noindent
since $\nabla(f\circ x)=f^*(\nabla x)$ and $\nabla x$ has no zeros.  Now for
odd $n$, equation (5) states that $\TI(\nabla(f\circ(-x))=
-\TI\ \nabla(f\circ x)$.  Hence $\chi(M)=\text{deg }\hat N$.  But
$\chi(M)=\frac12 \chi(\partial M)$.  So the Gauss-Bonnet theorem is proven.
Note that the argument for closed manifolds $M$ of odd dimension have
$\chi(M)=0$ uses exactly the same path:  For closed manifolds $\TI\ V=\chi(M)$
and by (5) $\TI(-V)=-\TI(V)$.

Also note the following 2 corollaries.  (a)  For $n$ odd, $\TI\ \nabla(f\circ x)
=0$.  (b)  If $f$ is an immersion, then $\text{deg }\hat N=\chi(M)$.  This
last is a theorem of Haefliger \cite{Ha}.  It follows since if $f$ is an 
immersion, then $\nabla(f\circ x)$ has no zeros.  Hence by (6)
$\TI\ \nabla(f\circ x)=0$ and so equation (11) yields $\text{deg }\hat N=
\chi(M)$.  \cite{G${}_5$}.
\vfill\eject
\noindent
{\bf \S6.  The Brouwer Fixed Point Theorem}

Let $f:B\to B$ be a continuous map where $B$ is a unit ball in ${\Bbb R}^n$.
The Brouwer fixed point theorem asserts that $f$ has a fixed point.  We
consider $f:B\to B\subset{\Bbb R}^n$.  Let $V_f(m)=m-f(m)$ be the vector
field.

The convexity of $B$ implies $V_f$ always points inside $B$ for any $m$ on
the boundary.  Hence $\partial_-B=\partial B$.  So applying (1), we get
$$
\TI\ V_f+\chi(S^{n-1})=\chi(B),
$$
hence
$$\
\TI\ V_f=1-(1+(-1)^{n-1})=(-1)^n\not= 0.
$$
Hence $V_f$ has a zero, hence $f$ has a fixed point.

The Brouwer fixed point theorem can be greatly generalized.  For example,
let us say, that $f:M\to{\Bbb R}^n$, where $M$ is a compact $n$-dimensional
submanifold, is transversal to $\partial M$ if the line segment from $m$
to $f(m)$ is not tangent to $\partial M$ at $m$ for $m\in\partial M$.  Then
$f$ has a fixed point if $\chi(M)-\sum\chi(\partial M_i)\not= 0$, when
$\partial M_i$ are the components of $\partial M$ so that the line segment
from $m$ to $f(m)$ begins by entering $M$. \cite{G${}_5$}.
\bigskip
\noindent
{\bf \S7.  The Borsuk-Ulam Theorem}

The key lemma, indeed many people call it the Borsuk-Ulam theorem itself,
is that an odd map $f:S^n\to S^n$ has odd degree.  Considering $S^n$ as the
unit sphere in ${\Bbb R}^{n+1}$, we say $f$ is odd if $f(-x)=-f(x)$ for all
$x\in S^n$.

Using the covering homotopy property for the covering space 
$S^n\overset p \to\longrightarrow {\Bbb R}P^n$, we can homotopy $f$ to an
odd map $f_1$ so that there are only a finite number of pairs of Antipodal
Points which are fixed by $f_1$.  Now we extend $f_1:S^n\to S^n$ to
$g:B\to B$ by $f(r\vec s)=rf_1(\vec s)$ where $\vec s\in\partial B$ and
$r\in[0,1]$.  Then $V^g$ is a vector field defined by
$V^g(r\vec s)=g(r\vec s)=rf_1(\vec s)$.  Now $\TI\ V^g=\text{deg }g$ by (8)
and $\text{deg }g=\text{deg }f_1=\text{deg }f$.  Hence (1) becomes
$\text{deg }f+\TI\ \partial_-V^g=1$.  Now $\partial_-V^g$ has zeros exactly
at those $m\in\partial B$ where $f_1(m)=-m$.  But then $-m$ is also a zero
of $\partial_-V^g$, and the symmetry of $f_1$ and the fact, from (1), that
index only depends on pointing inside implies that the index at $m$ is equal
to the index at $-m$.  Hence $\TI\ \partial_-V^g$ is even.  Thus
$\text{deg }f=1-\TI\ \partial_-V^g$ is odd.
\bigskip
\noindent
{\bf \S8.  The Jordan Curve Theorem}

If $S^{n-1}\subset {\Bbb R}^n$, then ${\Bbb R}^n-S^n$ is split into two
components, the inside and the outside.  We will show that if $M^{n-1}\subset
{\Bbb R}^n$, where is a smooth connected submanifold of ${\Bbb R}^n$, then
${\Bbb R}^n-M$ splits into the inside and the outside.

We choose a continuous normal vector field $\vec N$ on $M$.  Let $V$ be the
Electric vector field generate by an electron $e$.  Bring $e$ so close to
$M$ that $M$ looks like a hyperplane near $e$.  Move $e$ to the other side of 
the hyperplane at a speed near that of light along the normal direction.  The
vector field $\partial V$, which is $V$ projected onto $M$, originally and
finally has an isolated zero at the foot of the normal.  Outside of a small
ball about this zero the vector field $\partial V$ does not change.  Hence
the index of the zero does not change.  But the motion changes the zero from
$\partial_+V$ to $\partial_- V$ (or vice versa).  Hence $\TI(\partial_-V)$
changes by $\pm 1$.  Now suppose there were a path $\sigma$ from the
original position of $e$ to the final position of $e$ which does not cross
$M$.  Move $e$ along $\sigma$.  Then $\partial_-V$ undergoes an otopy which
changes its index.  Hence it is not a proper otopy by (2).  Hence a zero
must be on the frontier if $\partial_-V$ at some time.  This can only happen
when $e$ is on $M$, contradicting the statement that $\sigma$ avoids $M$.

Now $M$ is the boundary between the component of ${\Bbb R}^n-M$ which
``contains'' $\infty$, and the bounded components.  If we put $e$ in one of
the bounded components, then $\TI\ V=1$, since the electron is the only
defect inside the union of the bounded components of ${\Bbb R}^n-M$.  Call
this union $W$.  If we put $e$ inside another component of $W$, then the
index of this new $V_1$ is $+1$ also.  Hence there is a proper homotopy
between $V$ and $V_1$.  The set of defects of all the $V_t$ contains the
first and last position of the electron.  So there is a small connected
open set $U$ containing $c$ and $e_1$ which does not intersect $M$.  So $e$
and $e'$ are in the same path component.  So $w$ is connected.
\bigskip

\noindent
{\bf \S 9.  Gottlieb's theorem}

This theorem, named by Stallings in \cite{St}, has been considerably 
generalized, first by Rossett and then by Cheeger and Gromov.  I hope the reader
will forgive me for calling the theorem by my own name, but I wanted to
prove theorems specified by short, commonly known names, using the Law
of Vector Fields.

The key lemma in the original proof is:  If $X$ is a compact $CW$-complex
so that $\chi(X)\not=0$, then $G_1(X)$ is trivial.  Then if $X$ is a
$\kappa(\pi,1)$ we know that $G(x)=$ center of $\pi$.  Thus we must show
that if $F:X\times S^1\to X$ so that $F|X\times *$ is the identity and if
$\chi(X)\not=0$, then $F|x\times S^1$ is homotopically trivial.  The original
proof used Nielsen-Wecken fixed point classes \cite{G${}_1$}.  These were
transformed by Stallings \cite{St} into an algebraic setting which is of
importance algebraically.  The present proof is more elementary and does
not need Nielsen-Wecken fixed point theory.

Suppose $M$ is a regular neighborhood of an embedding of $X$ in some
${\Bbb R}^n$.  Then we have a map $F:M\times I\to M$ so that $F(m,0)=m$
and $F(m,1)=m$ for all $m\in M$.  We may adjust $F$ so that
$F(m,t)\not= m$ for all $m\in \partial M$ and all $t$, and so that $F_0$
and $F_1$ are the identity outside a small collar neighborhood of the
boundary.

Now we define the vector field $T$ on $M\times I$ by
$T(m,t)=(\vec m-\displaystyle{\overrightarrow{F(m,t)}},t)$.  Now $T$ is a
homotopy on $M\times I$.  Let $Z$ be the zeros of $T$ on $M\times I$.  Now
$Z$ is compact and $Z\cap(\partial M\times I)$ is empty.  Let $U$ be the
open set of $M\times I$ so that $\Vert T(m,t)\Vert<\epsilon$ where
$\epsilon$ is so small that two paths $\alpha(t)$ and $\beta(t)$ on $M$ are
homotopic if the distance between $\alpha(t)$ and $\beta(t)$ is always less
than $\epsilon$.  Let $W$ be a path component of $U$ containing $m\times 0$.
Then $T|U$ is a proper otopy from $T_0$ to $T_1$.  Now $\TI\ T_0=
\TI\ T_1$ and $\TI\ T_0=\chi(M)$ and $\TI\ T_1=\chi(M)$.  So the path
connected $W$ contains $M\times 0$ and $M\times 1$.

We may find a path $\gamma:I\to U$ so that $\gamma(0)=*\times 0$ and
$\gamma(1)=*\times 1$.  We can homotopy $F:M\times S^1\to M$ to a $G$ such that
$G(\gamma(t),t)=\gamma(t)$.  Then 
$\gamma\sim\alpha\cdot\gamma$ where $\alpha(t)=G(*,t)$.  Hence $\alpha\sim 0$,
which was to be shown.
\bigskip
\noindent
{\bf \S 10. The Poincare-Hopf Theorem}

If $M$ is a close manifold and $V$ is a continuous vector field defined
entirely on $M$, then $\TI\ V=\chi(M)$.  This is a special case of (1)
because since the boundary $\partial M$ is empty, so is $\partial_-V$ and so
$\TI(\partial_-V)=0$.
\vfill\eject

\Refs

\ref\no G${}_1$\manyby
Daniel H. Gottlieb \paper A certain subgroup of the fundamental group
\jour Amer. J. Math. \vol 87 \yr 1966 \pages 1233--1237
\endref

\ref\no G${}_2$\bysame
\paper A de Moivre formula for fixed point theory
\jour ATAS do $5^\circ$ Encontro Brasiliero de Topologia, Universidade
de S\~ao Pavlo, S\~ao Carlos, S.P. Brasil \vol 31 \yr 1988 \pages 59--67
\endref

\ref\no G${}_3$\bysame
\paper A de Moivre like formula for fixed point theory \jour Proceedings
of the Fixed Point Theory Seminar at the 1986 International Congress of
Mathematicians, R. F. Brown (editor), Contemporary Mathematics, AMS
Providence, Rhode Island \vol 72 \pages 99--106
\endref

\ref\no G${}_4$\bysame
\paper On the index of pullback vector fields\jour Proc. of the 2nd Siegen
Topology Symposium, August 1987, Ulrick Koschorke (editor), Lecture Notes
of Mathematics, Springer Verlag, New York
\endref

\ref\no G${}_5$\bysame
\paper Zeroes of pullback vector fields and fixed point theory for bodies
\jour Algebraic topology, Proc. of Intl. Conference March 21--24, 1988,
Contemporary Mathematics \vol 96 \pages 168--180
\endref

\ref\no G-S\by
Daniel H. Gottlieb and Geetha Samaranayake \paper The Index of
Discontinuous Vector Fields \jour (In Preparation)
\endref

\ref\no HO${}_1$\manyby
Heinz Hopf \paper \"Uber die Curvetura integra geschlossener Hyperfl\"achen
\jour Math. Ann. \vol 95 \yr 1925/26 \pages 340--367
\endref

\ref\no HO${}_2$\bysame
\paper Vectorfelder in $n$-dimensionalin Mannigfaltikeiten \jour Math. Ann.
\vol 96 \yr 1926/27 \pages 225--250
\endref

\ref\no Ha\by
Andre Haefliger \paper Quelques remarques sur les applications differentiable
d'une surace dans le plan \jour Ann. Inst. Fourier, Grenoble \vol 10
\yr 1960 \pages 47--60
\endref

\ref\no M \by
Marston Morse \paper Singular points of vector fields under general boundary
conditions \jour Amer. J. Math \vol 51 \yr 1929 \pages 165--178
\endref

\ref\no P\by
Charles C. Pugh \paper A generalized Poincare index formula \jour Topology
\vol 7 \yr 1968 \pages 217--226
\endref

\ref\no Sp\by
Michael Spivak \book A Comprehensive Introduction to Differential Geometry
\publ Publish or Perish, Inc \publaddr Wilmington, Delaware \vol 5 \yr 1979
\endref

\ref\no St\by
John Stallings \paper Centerless Groups - An Algebraic Formulation of
Gottlieb's Theorem \jour Topology \vol 4\yr 1965 \pages 129--134
\endref
\vskip.75truein
Purdue University


\endRefs
\enddocument

