# A Rigorous Proof of the Index Theorem for Economists

Yuhki Hosoya
Faculty of Economics, Chuo University TEL: +81-90-5525-5142, E-mail:
ukki(at)gs.econ.keio.ac.jp742-1, Higashinakano, Hachioji-shi, Tokyo, 192-0393,
Japan.

###### Abstract

This paper provides a rigorous and gap-free proof of the index theorem used in
the theory of regular economy. In the index theorem that is the subject of
this paper, the assumptions for the excess demand function are only several
usual assumptions and continuous differentiability around any equilibrium
price, and thus it has a form that is applicable to many economies. However,
the textbooks on this theme contain only abbreviated proofs and there is no
known monograph that contains a rigorous proof of this theorem. Hence, the
purpose of this paper is to make this theorem available to more economists by
constructing a readable proof.

Keywords: index theorem, regular economy, transversality.

JEL codes: C65, D41, D51

MSC2020 codes: 91B24, 57R25

##  1 Introduction

How many equilibrium prices exist? Arrow and Debrew (1954) showed that at
least one equilibrium price exists. It seems that economists of this period
believed that the fewer the number of equilibria, the better. Therefore, Arrow
et al. (1959) proposed a sufficient condition for the existence of a unique
equilibrium price. However, their condition is very complicated and does not
seem to be valid in many economies.

Debreu (1970) proved that the number of equilibrium prices is at least finite
in “almost all” economies. In his view, it was difficult to prove that there
is exactly one equilibrium price, and thus he proved that at least there are
only a finite number of equilibria. This result had been refined and
reformulated by many economists. This research area is now known as the theory
of regular economy.

The most central result in the theory of regular economy is called the index
theorem. This theorem states that the sum of “indices” computed by a certain
method for equilibrium prices is equal to $1$. This result is powerful.
Because “indices” must be able to be summed, the number of equilibrium prices
must be a finite odd number. We can immediately reproduce the result of Debreu
(1970) using this theorem.

The problem is that no straightforward proof of this index theorem is known.
The most famous index theorem currently published in a form applicable to many
economies appears to be Proposition 5.6.1 of Mas-Colell (1985). However, the
proof of this theorem contains many gaps that ordinary readers will find
impossible to fill.

Therefore, in this paper, we focus on the proof of this result and provide a
rigorous and gap-free proof of this theorem in the shortest manner. The proof
presented in this paper requires basic knowledge of differential topology
supplied by Guillemin and Pollack (1974). Therefore, in this paper, we
construct a proof under the assumption that the contents of the above book is
known. However, since Guillemin-Pollack’s formulation of differential topology
is a bit peculiar, we have chosen to extract only the claims necessary for
this paper and offer them to readers.

The proof in this paper requires a classification theorem for one-dimensional
manifolds. However, the problem is that a well-written proof of the
classification theorem conteined in a book is missing. Therefore, in the
appendix of this paper, we present a proof of this classification theorem.

The structure of this paper is as follows. In Section 2, we explain a part of
the knowledge of differential topology contained in Guillemin-Pollack’s book
that is necessary to understand the proof. We do not prove any of the results
that appear in this section, but the proofs are either in Guillemin-Pollack or
can be easily reconstructed from what is written in Guillemin-Pollack. Section
3 contains the assertion of the main result and its proof. In Section 4, we
explain the relationships between Proposition 5.6.1 of Mas-Colell (1985) and
our main result, and the proof and its gaps in Mas-Colell’s book. In the
appendix, we prove the classification theorem for one-dimensional manifolds.

##  2 Preliminaries

In this paper, we assume that readers are sufficiently familiar with the
results in Guillemin and Pollack (1974). In particular, we assume that readers
know the definition of manifolds, tangent spaces and the tangent bundle, the
derivatives of smooth functions on manifolds, transversality, and oriented
manifolds. However, the definition of manifolds in Guillemin and Pollack
(1974) is not standard in differential topology, and with very few exceptions
such as Milnor (1965), another definition has been used. Thus, readers may not
be familiar with the manifolds treated in this paper, even though they have
appropriately deep knowledge on manifolds. Therefore, in this section, we
present the definitions of the terms in differential topology treated in this
paper and the assertion of some theorems. All of the results in this section
have been proved or can be easily verified from the results proved in
Guillemin and Pollack (1974).

Note that, although Guillemin and Pollack (1974) treated only $C^{\infty}$
manifolds and functions, in most cases, their proofs work well for $C^{k}$
manifolds and functions, even if $k\neq\infty$. However, several techniques
depend on the assumption $k=+\infty$. For example, Sard’s theorem does not
necessarily hold if the function is not $C^{\infty}$. The concept of Morse
function cannot be appropriately defined in $C^{1}$ manifolds. The tangent
bundle $T(X)$ is not necessarily a $C^{1}$ manifold when $X$ is a $C^{1}$
manifold. We avoid these technical problems in this paper. First, we use
Sard’s theorem only for a $C^{\infty}$ function defined on a $C^{\infty}$
manifold. Second, we consider the tangent bundle only for a $C^{\infty}$
manifold. Third, we avoid the use of Morse functions.

###  2.1 Manifolds

First, suppose that $X\subset\mathbb{R}^{N}$ and let $f:X\to\mathbb{R}^{M}$ be
a continuous function. Choose any $x\in X$ and let
$k\in\mathbb{N}\cup\\{\infty\\}$. We say that $f$ is $C^{k}$ around $x$ if and
only if there exists a $C^{k}$ function $F:U\to\mathbb{R}^{M}$ such that $U$
is an open neighborhood of $x$ in $\mathbb{R}^{N}$ and $F(y)=f(y)$ whenever
$y\in U\cap X$. As usual, a function that is $C^{k}$ everywhere is called a
$C^{k}$ function.

A function $f:X\to Y$ is called a homeomorphism if and only if $f$ is
bijective and both $f$ and $f^{-1}$ are continuous. Suppose that $f:X\to Y$ is
a homeomorphism. We say that $f$ is a $C^{k}$ diffeomorphism if and only if
both $f$ and $f^{-1}$ are $C^{k}$. A nonempty set $X\subset\mathbb{R}^{N}$ is
called an $n$-dimensional $C^{k}$ manifold if and only if for every $x\in X$,
there exists a $C^{k}$ diffeomorphism $\phi:U\to V$ such that $U$ is an open
set in $\mathbb{R}^{n}$ and $V$ is an open neighborhood of $x$ in $X$. Such a
$\phi$ is called a local parametrization around $x$. The number $n$ is called
the dimension of $X$ and denoted by $\dim X$. It is known that if $X$ is a
$C^{k}$ manifold, then $\dim X$ is uniquely determined.

Sometimes, $C^{0}$ manifolds are considered. A function is said to be $C^{0}$
if and only if it is continuous, and a set $X\subset\mathbb{R}^{N}$ is called
an $n$-dimensional $C^{0}$ manifold if and only if every $x\in X$ has an open
neighborhood that is homeomorphic to some open set in $\mathbb{R}^{n}$. A
$C^{0}$ manifold is sometimes called a topological manifold. However, in
$C^{0}$ manifolds, we cannot define the tangent space, which is explained
later. This is very restrictive for our argument, and thus we do not use a
$C^{0}$ manifold in this paper, except in the appendix.

We define $\mathbb{R}^{0}=\\{0\\}$. Hence, $0$-dimensional manifolds can be
defined. A set $X$ is a $0$-dimensional $C^{k}$ manifold if and only if it is
a discrete set.

Let $X$ be an $n$-dimensional $C^{k}$ manifold, and choose any $x\in X$.
Choose any local parametrization $\phi$ around $x$. Let $\phi(z)=x$ and
suppose that $D\phi(z)$ denotes the Fréchet derivative of $\phi$ at $z$. Let
$T_{x}(X)$ denote the range of $D\phi(z)$. Then, it is known that $T_{x}(X)$
is independent of the choice of $\phi$. We call this linear space $T_{x}(X)$
the tangent space of $X$ at $x$. It is known that the dimension of $T_{x}(X)$
is $n$, and for every $v\in\mathbb{R}^{N}$, $v\in T_{x}(X)$ if and only if
there exist $\varepsilon>0$ and a $C^{1}$ curve
$c:(-\varepsilon,\varepsilon)\to X$ such that $c(0)=x$ and $c^{\prime}(0)=v$.

Next, let $X,Y$ be $C^{k}$ manifolds, and let $f:X\to Y$ be a $C^{k}$
function. Choose $x\in X$, and a local parametrization $\phi$ around $x$. Let
$\phi(z)=x$. Then, $g=f\circ\phi$ is a $C^{k}$ function defined on an open set
$U\subset\mathbb{R}^{n}$, where $n$ is the dimension of $X$. Hence, the
Fréchet derivative $Dg(z)$ can be defined. Choose any $v\in T_{x}(X)$, and let
$D\phi(z)w=v$. Then, we can define $df_{x}(v)=Dg(z)w$, and it can be shown
that the left-hand side is independent of the choice of $\phi$. Moreover,
$df_{x}(v)\in T_{f(x)}(Y)$ for all $x\in X$ and $v\in T_{x}(X)$. This linear
function $df_{x}$ is called the derivative of $f$ at $x$. Suppose that $v\in
T_{x}(X)$ and $c:(-\varepsilon,\varepsilon)\to X$ is a $C^{1}$ function that
satisfies $c(0)=x$ and $c^{\prime}(0)=v$. For a $C^{k}$ function $f:X\to Y$,
if we define $d=f\circ c$, then we can check that $df_{x}(v)=d^{\prime}(0)$.

Suppose that $X,Y$ are $C^{k}$ manifolds whose dimensions are the same, and
$f:X\to Y$ is a $C^{k}$ function. Choose $x\in X$ and suppose also that
$df_{x}$ is a bijection from $T_{x}(X)$ onto $T_{f(x)}(Y)$. Then, there exists
an open neighborhood $U$ of $x$ such that $F(U)$ is an open neighborhood of
$f(x)$ and the restriction of $f$ is a $C^{k}$ diffeomorphism between $U$ and
$f(U)$. This result is known as the inverse function theorem.

###  2.2 Submanifolds and Transversality

Let $X$ be an $n$-dimensional $C^{k}$ manifold, and $Y\subset X$. We call $Y$
a submanifold of $X$ if $Y$ itself is a $C^{k}$ manifold. It is known that if
$Y$ is a submanifold of $X$, then $T_{x}(Y)\subset T_{x}(X)$. If $\dim Y=m$,
then $n-m$ is called the codimension of $Y$, and denoted by
$\mbox{codim}~{}Y$.

Suppose that $X,Y$ are $C^{k}$ manifolds, $Z$ is a submanifold of $Y$, and
$f:X\to Y$ is $C^{k}$. We say that $f$ is transversal to $Z$ if and only if
for every $x\in f^{-1}(Z)$,

$\mbox{Im}(df_{x})+T_{f(x)}(Z)=T_{f(x)}(Y).$

Then, the following fact is known.

######  Fact 1.

Suppose that $X,Y$ are $C^{k}$ manifolds, $Z$ is a $C^{k}$ submanifold of $Y$,
and $f:X\to Y$ is a $C^{k}$ function that is transversal to $Z$. If
$S=f^{-1}(Z)$ is nonempty, then $S$ is also a submanifold of $X$, and
$\mbox{codim}~{}S=\mbox{codim}~{}Z$.

###  2.3 Tangent Bundle

Suppose that $X$ is an $n$-dimensional $C^{\infty}$ manifold. The following
set

$T(X)=\\{(x,v)|x\in X,\ v\in T_{x}(X)\\}$

is called the tangent bundle of $X$. It is known that $T(X)$ is a
$2n$-dimensional $C^{\infty}$ manifold, and if $(x,v)\in T(X)$ and $\phi$ is a
local parametrization around $x$ in $X$, then

$(z,w)\mapsto(\phi(z),d\phi_{z}(w))$

is a local parametrization around $(x,v)$ in $T(X)$.

By construction, it is easy to show that $T_{(x,v)}(T(X))=(T_{x}(X))^{2}$.

###  2.4 Manifolds with Boundary

We need the definition of manifolds with boundary. First, let
$H^{n}=\\{x\in\mathbb{R}^{n}|x_{n}\geq 0\\}$. A set $X\subset\mathbb{R}^{N}$
is called an $n$-dimensional $C^{k}$ manifold with boundary if and only if for
every $x\in X$, there exists a $C^{k}$ diffeomorphism $\phi:U\to V$ such that
$U$ is an open set in $H^{n}$ and $V$ is an open neighborhood of $x$ in $X$.
The mapping $\phi$ is called a local parametrization around $x$ as usual.
Tangent space $T_{x}(X)$ can be defined in the usual manner. Note that,
$T_{x}(X)$ is an $n$-dimensional linear space, even when $X$ is not a usual
manifold but a manifold with boundary.

Suppose that $X$ is an $n$-dimensional $C^{k}$ manifold with boundary, and
define the set $\partial X$ as the set of all $x\in X$ such that there is no
local parametrization around $x$ whose domain is an open set in
$\mathbb{R}^{n}$. It can be shown that if $\partial X$ is nonempty, then it is
an $(n-1)$-dimensional $C^{k}$ manifold, and $T_{x}(\partial X)$ is a subspace
of $T_{x}(X)$.

Note that, $X\times Y$ is not necessarily a manifold with boundary, even if
both $X,Y$ are manifolds with boundary. For example, it is known that
$[0,1]^{2}$ is not a manifold with boundary. However, if $X$ is an
$n$-dimensional $C^{k}$ manifold with boundary and $Y$ is an $m$-dimensional
$C^{k}$ manifold without boundary, then $X\times Y$ is an $(n+m)$-dimensional
$C^{k}$ manifold with boundary, and $\partial(X\times Y)=\partial X\times Y$
and $T_{(x,y)}(X\times Y)=T_{x}(X)\times T_{y}(Y)$. In particular, because the
unit interval $[0,1]$ is a $1$-dimensional $C^{\infty}$ manifold with
boundary, if $X$ is an $n$-dimensional $C^{k}$ manifold, then $Y=[0,1]\times
X$ is an $(n+1)$-dimensional $C^{k}$ manifold with boundary, and $\partial
Y=\\{0,1\\}\times X$. If we write $X_{t}=\\{t\\}\times X$, then $\partial
Y=X_{0}\cup X_{1}$ in this sense.

Suppose that $X,Y$ are $C^{k}$ manifolds with boundary, and $f:X\to Y$ is
$C^{k}$. Then, we can define the derivative $df_{x}:T_{x}(X)\to T_{f(x)}(Y)$
in the same manner as in Subsection 2.1. Usually, $\partial f$ denotes the
restriction of $f$ into $\partial X$. If $Y$ has no boundary and $Z$ is a
submanifold of $Y$, we can define transversality as usual: that is, $f$ is
transversal to $Z$ if and only if for any $x\in f^{-1}(Z)$,

$\mbox{Im}(df_{x})+T_{f(x)}(Z)=T_{f(x)}(Y).$

Then, the following fact holds.

######  Fact 2.

Suppose that $X$ is a $C^{k}$ manifold with boundary, $Y$ is a $C^{k}$
manifold without boundary, and $Z$ is a submanifold of $Y$ without boundary.
Let $f:X\to Y$ be a $C^{k}$ function such that both $f$ and $\partial f$ are
transversal to $Z$. If $S=f^{-1}(Z)$ is nonempty, then $S$ is a $C^{k}$
manifold with boundary whose codimension is the same as that of $Z$, and
$\partial S=\partial X\cap S$.

###  2.5 Orientation

Suppose that $V$ is an $n$-dimensional linear space. Choose any two bases
$v=(v_{1},...,v_{n})$ and $w=(w_{1},...,w_{n})$. Then, there uniquely exists
an $(n\times n)$ matrix $A$ such that

$w_{i}=a_{i1}v_{1}+...+a_{in}v_{n}$

for all $i\in\\{1,...,n\\}$. We say that $v$ is equivalent to $w$ if and only
if $\det A>0$.111We use both expressions $\det A$ and $|A|$ to denote the
determinant of $A$.

Suppose that $X\subset\mathbb{R}^{N}$ is an $n$-dimensional $C^{k}$ manifold
with/without boundary. Let

$T^{n}(X)=\\{(x,v_{1},...,v_{n})|x\in X,\ v_{i}\in T_{x}(X)\\}.$

Choose any function $\mbox{sign}_{X}:T^{n}(X)\to\\{+1,-1\\}$. We say that
$\mbox{sign}_{X}$ is sign-preserving if for any two equivalent bases
$v=(v_{1},...,v_{n})$ and $w=(w_{1},...,w_{n})$ of $T_{x}(X)$,
$\mbox{sign}_{X}(x,v)=\mbox{sign}_{X}(x,w)$. Suppose that for every $x\in X$,
there exists a local parametrization $\phi$ around $x$ such that for
$v_{i}(z)=D\phi(z)e_{i}$, $\mbox{sign}_{X}(\phi(z),v_{1}(z),...,v_{n}(z))$ is
constant, where $e_{i}\in\mathbb{R}^{n}$ is the $i$-th unit vector. Then, we
call this sign-preserving function $\mbox{sign}_{X}$ an orientation of $X$.

We must note an exception. Suppose that $X$ is a $0$-dimensional manifold. By
definition, $X$ is a discrete set, and $T_{x}(X)=\\{0\\}$. In this case, any
function $\mbox{sign}_{X}:X\to\\{-1,+1\\}$ is called an orientation of $X$.

Note that, there is a manifold with no orientation: the Möbius strip is a
typical example. If $X$ is a manifold and there is an orientation, then we say
that $X$ is orientable. If we think that an orientation of $X$ is given, then
$X$ is said to be oriented. It is known that if $X$ is a connected and
orientable manifold, then there are exactly two possible orientations.

If $X,Y$ are manifolds and either $\partial X=\emptyset$ or $\partial
Y=\emptyset$, then $X\times Y$ is also a manifold. If, in addition, both $X,Y$
are oriented, then $X\times Y$ is also oriented, and the orientation is made
in the following manner. Suppose that $\mbox{sign}_{X},\mbox{sign}_{Y}$ denote
orientations. Then, there uniquely exists a sign-preserving function
$\mbox{sign}_{X\times Y}$ such that

$\displaystyle~{}\mbox{sign}_{X\times
Y}((x,y),(v_{1},0),...,(v_{n},0),(0,w_{1}),...,(0,w_{m}))$ $\displaystyle=$
$\displaystyle~{}\mbox{sign}_{X}(x,v_{1},...,v_{n})\mbox{sign}_{Y}(y,w_{1},...,w_{m})$

for all $(x,y)\in X\times Y$ and bases $v=(v_{1},...,v_{n})$ of $T_{x}(X)$ and
$w=(w_{1},...,w_{m})$ of $T_{y}(Y)$. We can easily check that this function
$\mbox{sign}_{X\times Y}$ is an orientation of $X\times Y$, and thus, $X\times
Y$ is oriented by this orientation.

Suppose that $X\subset\mathbb{R}^{N}$ is an $n$-dimensional $C^{k}$ oriented
manifold, and $Y\subset X$ is an $(n-1)$-dimensional submanifold. Choose any
orientation $\mbox{sign}_{X}$ of $X$. Suppose also that there exists a
$C^{k-1}$ function $n:Y\to\mathbb{R}^{N}$ such that $n(x)\in T_{x}(X)$,
$\|n(x)\|=1$, and $n(x)\cdot v=0$ for all $v\in T_{x}(Y)$. Then, there exists
an orientation $\mbox{sign}_{Y}$ of $Y$ such that

$\mbox{sign}_{Y}(x,v_{1},...,v_{n-1})=\mbox{sign}_{X}(x,n(x),v_{1},...,v_{n-1}).$

Hence, $Y$ is an oriented manifold.

Next, suppose that $X\subset\mathbb{R}^{N}$ is a $C^{k}$ manifold with
boundary. Choose any $x\in\partial X$. Then, there exists $p\in T_{x}(X)$ such
that $\|p\|=1$ and $p\cdot v=0$ for all $v\in T_{x}(\partial X)$. Choose any
local parametrization $\phi$ around $x$, and suppose that $p=D\phi_{x}(q)$.
Then, $q_{n}\neq 0$. Define $n(x)=p$ if $q_{n}<0$ and $n(x)=-p$ if $q_{n}>0$.
This function $n:\partial X\to\mathbb{R}^{N}$ is called the outward normal
vector field of $\partial X$. We can show that $n(x)$ is $C^{k-1}$, and thus,
we can conclude that $\partial X$ is oriented by the following sign-preserving
function:

$\mbox{sign}_{\partial
X}(x,v_{1},...,v_{n-1})=\mbox{sign}_{X}(x,n(x),v_{1},...,v_{n-1}).$

Suppose that $X$ is a $C^{k}$ oriented manifold with boundary, $Y$ is a
$C^{k}$ oriented manifold without boundary, $Z$ is a $C^{k}$ oriented
submanifold of $Y$ without boundary, $f:X\to Y$ is a $C^{k}$ function such
that both $f$ and $\partial f$ are transversal to $Z$, and $S=f^{-1}(Z)$ is
nonempty. Then, $S$ is a manifold with boundary. Let $\ell=\mbox{codim}~{}Z$.
Choose any $x\in f^{-1}(Z)$ and suppose that $H$ is an $\ell$-dimensional
linear subspace of $T_{x}(X)$ such that

$df_{x}(H)\oplus T_{f(x)}(Z)=T_{f(x)}(Y).$

Then, we can show that

$H\oplus T_{x}(S)=T_{x}(X).$

Choose any basis $(v_{1},...,v_{\ell})$ of $H$, and define
$w_{i}=df_{x}(v_{i})$. Choose any basis $(w_{\ell+1},...,w_{m})$ such that
$\mbox{sign}_{Z}(f(x),w_{\ell+1},...,w_{m})=+1$. Define

$h(v_{1},...,v_{\ell})=\mbox{sign}_{Y}(f(x),w_{1},...,w_{m}),$

and

$\mbox{sign}_{S}(x,v_{\ell+1},...,v_{n})=\mbox{sign}_{X}(x,v_{1},...,v_{n})h(v_{1},...,v_{\ell}).$

Then, we can show that $\mbox{sign}_{S}$ is an orientation. We call this
orientation the preimage orientation.

###  2.6 Classification of One-dimensional Manifolds

Let

$S^{1}=\\{x\in\mathbb{R}^{2}|\|x\|=1\\}.$

Suppose that $X$ is a compact and connected $1$-dimensional $C^{k}$ manifold
with/without boundary. Then, it is known that $X$ is $C^{k}$ diffeomorphic to
either $S^{1}$ or $[0,1]$. In Guillemin and Pollack (1974), this
classification theorem was proved using a Morse function. However, Morse
function can be defined only if $k\geq 2$, and thus their proof cannot be
applied if $k=1$. And actually, we need this result for $k=1$. Hence, we prove
this theorem in appendix.

It is known that every connected component of a manifold is open. Suppose that
$X$ is a compact $1$-dimensional $C^{k}$ manifold with/without boundary.
Because $X$ is compact, the number of connected components of $X$ is finite.
Because each connected component $X_{i}$ is diffeomorphic to either $S^{1}$ or
$[0,1]$, either $\partial X_{i}$ is empty or it contains $2$ points. In
particular, $\partial X$ is a finite set that includes even number of points.

Moreover, suppose that $X$ is oriented. Then, for each connected component
$X_{i}$ that is diffeomorphic to $[0,1]$, if $\partial X_{i}=\\{a,b\\}$, then
$\mbox{sign}_{\partial X}(a)+\mbox{sign}_{\partial X}(b)=0$. This implies that

$\sum_{x\in\partial X}\mbox{sign}_{\partial X}(x)=0.$

This result is crucial for our main result.

##  3 Result

In this section, we fix $n\geq 2$. We define
$\mathbb{R}^{n}_{+}=\\{x\in\mathbb{R}^{n}|x_{i}\geq 0\mbox{ for all }i\\}$ and
$\mathbb{R}^{n}_{++}=\\{x\in\mathbb{R}^{n}|x_{i}>0\mbox{ for all }i\\}$. Let
$S=\\{p\in\mathbb{R}^{n}_{++}|\|p\|=1\\}$. It is known that $S$ is an
$(n-1)$-dimensional $C^{\infty}$ manifold. Moreover, because $g(p)=p$ is a
natural normal vector field on $S$, we can assume that $S$ is oriented. Our
main result is as follows.

######  Theorem 1.

Let $f:\mathbb{R}^{n}_{++}\to\mathbb{R}^{n}$ be a continuous mapping such that
the following five conditions hold.

  1. (i)

$p\cdot f(p)=0$ for all $p\in S$, and $f(ap)=f(p)$ for all $a>0$ and $p\in S$.

  2. (ii)

There exists $s>0$ such that $f_{i}(p)>-s$ for all $p\in S$ and
$i\in\\{1,...,n\\}$.

  3. (iii)

If $(p^{k})$ is a sequence on $S$ such that $p^{k}\to p\notin S$ as
$k\to\infty$, then $\|f(p^{k})\|\to\infty$ as $k\to\infty$.

  4. (iv)

If $f(p^{*})=0$, then $f$ is continuously differentiable around $p^{*}$.

  5. (v)

If $f(p^{*})=0$, then

$g(p^{*})=\begin{vmatrix}Df(p^{*})&p^{*}\\\ (p^{*})^{T}&0\end{vmatrix}\neq 0.$

Let $E^{*}=\\{p\in S|f(p)=0\\}$, and for each $p^{*}\in E^{*}$, define

$\mathrm{index}(p)=\begin{cases}+1&\mbox{if }(-1)^{n}g(p^{*})>0,\\\
-1&\mbox{if }(-1)^{n}g(p^{*})<0.\end{cases}$

Then, $E^{*}$ is a finite set and

$\sum_{p^{*}\in E^{*}}\mathrm{index}(p^{*})=+1.$

This result is actually the same as Proposition 5.6.1 of Mas-Colell (1985).
However, the proof of this proposition includes big gaps, and thus many
readers cannot judge whether this proposition is correct. Hence, we provide a
rigorous and gap-free proof for this theorem. This is the purpose of this
paper.

  * Proof.We separate the proof into nine steps.

######  Step 1.

The set $E^{*}$ is finite.

  * Proof of Step 1.Choose any $p^{*}\in E^{*}$, and let $A=Df(p^{*})$ and $V=\\{v\in\mathbb{R}^{n}|v\cdot p^{*}=0\\}$. If $v\in V,\ v\neq 0$ and $Av=0$, then

$\begin{pmatrix}Df(p^{*})&p^{*}\\\
(p^{*})^{T}&0\end{pmatrix}\begin{pmatrix}v\\\ 0\end{pmatrix}=0,$

which contradicts (v). Therefore, we have that $v\in V$ and $v\neq 0$ imply
$Av\neq 0$. Hence,

$m=\min\\{\|Av\|/\|v\||v\in V,\ v\neq 0\\}>0.$

Because of the definition of the Fréchet derivative, there exists
$\varepsilon>0$ such that if $v\in V$ and $0<\|v\|<\varepsilon$, then

$\|f(p^{*}+v)-f(p^{*})-Av\|<\frac{\|v\|m}{2}.$

Because $f(p^{*})=0$,

$\displaystyle\|v\|m\leq\|Av\|\leq$
$\displaystyle~{}\|f(p^{*}+v)-f(p^{*})-Av\|+\|f(p^{*}+v)\|$ $\displaystyle<$
$\displaystyle~{}\frac{\|v\|m}{2}+\|f(p^{*}+v)\|,$

which implies that $f(p^{*}+v)\neq 0$. Now, for any
$q\in(p^{*}+V)\cap\mathbb{R}^{n}_{++}$, define $P(q)=q/\|q\|$. Then, $P$ is a
$C^{\infty}$ mapping from $(p^{*}+V)$ into $S$, and it is easy to show that
$dP_{p^{*}}$ is the identity mapping. By the inverse function theorem, there
exists $\delta>0$ such that $P^{-1}$ is defined, injective on $\\{p\in
S|\|p-p^{*}\|<\delta\\}$, and if $p\in S$ and $\|p-p^{*}\|<\delta$, then
$\|P^{-1}(p)-p^{*}\|<\varepsilon$. By (i), $p\in S,0<\|p-p^{*}\|<\delta$
implies that $f(p)\neq 0$, and thus $E^{*}$ is a discrete set.

By (iii), there exists $\varepsilon>0$ such that if $p\in S$ and
$p_{i}\geq\varepsilon$ for all $i\in\\{1,...,n\\}$, then $f(p)\neq 0$. This
implies that $E^{*}$ is compact, and thus finite. This completes the proof of
Step 1. ∎

######  Step 2.

Let $(p^{k})$ be a sequence on $S$ such that $p^{k}$ converges to $p\notin S$
and $\frac{1}{\|f(p^{k})\|}f(p^{k})$ converges to $z\in\mathbb{R}^{n}$ as
$k\to\infty$. Then, $z\geq 0$, $z\neq 0$ and $p\cdot z=0$.

  * Proof of Step 2.By (ii), $f_{i}(p^{k})\to-\infty$ is impossible, and thus by (iii), we have that $z_{i}\geq 0$. Because $\|z\|=1$, $z\neq 0$. Finally, suppose that $p_{i}>0$. By (i),

$f_{i}(p^{k})=-\frac{1}{p^{k}_{i}}\left(\sum_{j\neq
i}p^{k}_{j}f_{j}(p^{k})\right).$

Because the right-hand side is bounded from above, we have that the sequence
$(f_{i}(p^{k}))$ is bounded, and thus $z_{i}=0$, which implies that $p\cdot
z=0$. This completes the proof of Step 2. ∎

######  Step 3.

Choose any $q\neq 0$. Suppose that $A$ is an $n\times n$ matrix such that
$v^{T}Av<0$ for all $v\in\mathbb{R}^{n}$ such that $v\cdot q=0$ and $v\neq 0$.
Then,

$(-1)^{n}\begin{vmatrix}A&q\\\ q^{T}&0\end{vmatrix}>0.$

  * Proof of Step 3.Let $A_{t}=(1-t)A+tA^{T}$. By assumption, we have that $v^{T}A_{t}v<0$ for all $t\in[0,1]$ and $v\in\mathbb{R}^{n}$ such that $v\cdot q=0$ and $v\neq 0$. Suppose that there exists $t\in[0,1]$ such that

$\begin{vmatrix}A_{t}&q\\\ q^{T}&0\end{vmatrix}=0,$

then, there exist $v\in\mathbb{R}^{n}$ and $w\in\mathbb{R}$ such that
$(v,w)\neq 0$ and

$\begin{pmatrix}A_{t}&q\\\ q^{T}&0\end{pmatrix}\begin{pmatrix}v\\\
w\end{pmatrix}=0.$

This implies that $q^{T}v=0$, and thus $v\cdot q=0$. Therefore,

$0=v^{T}0=v^{T}A_{t}v+v^{T}qw=v^{T}A_{t}v,$

which implies that $v=0$. However, this implies that $wq=0$, and thus $w=0$,
which is a contradiction. By the intermediate value theorem, we have that

$c(t)\equiv\begin{vmatrix}A_{t}&q\\\ q^{T}&0\end{vmatrix}$

has a constant sign on $[0,1]$. If we set $t=\frac{1}{2}$, then $A_{t}$ is
symmetric, and thus by Theorem 5 of Debreu (1952), $(-1)^{n}c(1/2)>0$.
Therefore, $(-1)^{n}c(0)>0$, as desired. This completes the proof of Step 3. ∎

######  Step 4.

Suppose that $p\in\mathbb{R}^{n}$ and $\|p\|=1$, and $A$ is an $n\times n$
matrix such that $Ap=0$ and

$\begin{vmatrix}A&p\\\ p^{T}&0\end{vmatrix}\neq 0.$

Let $x_{1},...,x_{n-1}\in\mathbb{R}^{n}$ satisfy $p\cdot x_{i}=0$ and
$\det(p,x_{1},...,x_{n-1})>0$, and define $y_{i}=Ax_{i}$. Then,

$\mathrm{sign}\begin{vmatrix}A&p\\\
p^{T}&0\end{vmatrix}=-\mathrm{sign}(\det(p,y_{1},...,y_{n-1})).$

  * Proof of Step 4.Let

$B=\begin{pmatrix}A&p\\\ p^{T}&0\end{pmatrix}.$

Then,

$B\begin{pmatrix}x_{i}\\\ 0\end{pmatrix}=\begin{pmatrix}y_{i}\\\
0\end{pmatrix},\ B\begin{pmatrix}p\\\ 0\end{pmatrix}=\begin{pmatrix}0\\\
1\end{pmatrix},\ B\begin{pmatrix}0\\\ 1\end{pmatrix}=\begin{pmatrix}p\\\
0\end{pmatrix}.$

Define

$C=\begin{pmatrix}0&x_{11}&...&x_{n-1,1}&p_{1}\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\ 0&x_{1n}&...&x_{n-1,n}&p_{n}\\\
1&0&...&0&0\end{pmatrix}.$

Then, $\det(C)<0$ and

$BC=\begin{pmatrix}p_{1}&y_{11}&...&y_{n-1,1}&0\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\ p_{n}&y_{1n}&...&y_{n-1,n}&0\\\
0&0&...&0&1\end{pmatrix},$

and thus

$\det(B)=\det(p,y_{1},...,y_{n-1})(\det(C))^{-1},$

which implies that our claim is correct. This completes the proof of Step 4. ∎

Choose any $q\in S$, and define $f^{q}(p)=\frac{\|p\|}{p\cdot
q}q-\frac{1}{\|p\|}p$. Then, $f^{q}$ is defined on
$\mathbb{R}^{n}_{+}\setminus\\{0\\}$.

######  Step 5.

$f^{q}$ satisfies $(\mathrm{i})$, $(\mathrm{ii})$, $(\mathrm{iv})$,
$(\mathrm{v})$ of this theorem.

  * Proof of Step 5.Clearly, $p\cdot f^{q}(p)=0$ and $f^{q}(p)=f^{q}(ap)$ for all $a>0$, and thus (i) holds. Let $\bar{S}$ be the closure of $S$ in $\mathbb{R}^{n}$. Then, $f^{q}$ is defined on $\bar{S}$ and continuous, which implies that (ii) holds. Clearly, $f^{q}$ is $C^{\infty}$ and thus (iv) holds. It is obvious that $f^{q}(p)=0$ if and only if $p=aq$ for some $a>0$. Let $v\cdot q=0$ and $v\neq 0$, and define $p(t)=q+tv$. Then, $p(t)\cdot q=\|q\|^{2}=1$, and thus

$\frac{d}{dt}f^{q}(p(t))=\frac{t\|v\|^{2}}{\|p(t)\|}q+\frac{t\|v\|^{2}}{2\|p(t)\|^{3}}p(t)-\frac{1}{\|p(t)\|}v,$

which implies that

$v^{T}Df^{q}(q)v=v\cdot\left(\left.\frac{d}{dt}f^{q}(p(t))\right|_{t=0}\right)=-\|v\|^{2}<0.$

Therefore, $v\cdot q=0$ and $v\neq 0$ imply that $v^{T}Df^{q}(q)v<0$, and by
Step 3,

$(-1)^{n}\begin{vmatrix}Df^{q}(q)&q\\\ q^{T}&0\end{vmatrix}>0.$ (1)

Because $f^{q}(ap)=f^{q}(p)$ for all $a>0$, we have that
$Df^{q}(aq)=a^{-1}Df^{q}(q)$, and thus (v) holds. This completes the proof of
Step 5. ∎

Define

$S_{\varepsilon}=\\{p\in S|p_{i}\geq\varepsilon\mbox{ for all
}i\in\\{1,...,n\\}\\},$ $\tilde{S}_{\varepsilon}=\\{p\in
S|p_{i}>\varepsilon\mbox{ for all }i\in\\{1,...,n\\}\\},$ $\partial
S_{\varepsilon}=S_{\varepsilon}\setminus\tilde{S}_{\varepsilon}.$

Note that, $S_{\varepsilon}$ and $\partial S_{\varepsilon}$ are not manifolds.
Choose any $\varepsilon^{\prime}>0$ such that
$\tilde{S}_{\varepsilon^{\prime}}\neq\emptyset$.

######  Step 6.

If $\varepsilon>0$ is sufficiently small, then
$\tilde{S}_{\varepsilon}\neq\emptyset$ and

$(1-t)f(p)+tf^{q}(p)\neq 0$

for all $t\in[0,1]$, $q\in S_{\varepsilon^{\prime}}$, and $p\in\partial
S_{\varepsilon}$.

  * Proof of Step 6.Suppose not. Then, there exist sequences $(t^{k})$ on $[0,1]$, $(p^{k})$ on $S$, and $(q^{k})$ on $S_{\varepsilon^{\prime}}$ such that $p^{k}\to p\notin S$ as $k\to\infty$ and

$(1-t^{k})f(p^{k})+t^{k}f^{q^{k}}(p^{k})=0$

for all $k$. Let $z^{k}=\frac{1}{\|f(p^{k})\|}f(p^{k})$. By taking
subsequences, we can assume that $z^{k}\to z$ as $k\to\infty$. By Step 2,
$z\geq 0$, $z\neq 0$ and $p\cdot z=0$. Choose any $i$ such that $z_{i}>0$.
Then, $p_{i}=0$, and $\min_{q\in S_{\varepsilon^{\prime}}}f^{q}_{i}(p)>0$.
This implies that $\min_{q\in S_{\varepsilon^{\prime}}}f^{q}_{i}(p^{k})>0$ for
sufficiently large $k$. Moreover, because $z_{i}>0$, $f_{i}(p^{k})>0$ for
sufficiently large $k$. Thus, for such a $k$,

$(1-t^{k})f_{i}(p^{k})+t^{k}f_{i}^{q^{k}}(p^{k})>0,$

which is a contradiction. This completes the proof of Step 6. ∎

Hereafter, we fix $\varepsilon>0$ that satisfies the requirements in Step 6,
$E^{*}\subset S_{\varepsilon}$, and $\varepsilon<\varepsilon^{\prime}$.

Choose any $\eta>0$ such that if $p^{*}\in E^{*}$, then $f$ is $C^{1}$ on
$\\{p\in\mathbb{R}^{n}_{++}|\|p-p^{*}\|<4\eta\\}$, and
$\\{p\in\mathbb{R}^{n}_{++}|f(p)=0,\
\|p-p^{*}\|<4\eta\\}\subset\\{ap^{*}|a>0\\}$. By Step 1, $E^{*}$ is finite,
and thus such an $\eta$ exists. Define

$\hat{S}_{\varepsilon,a}=\\{p\in S_{\varepsilon}|\forall p^{*}\in E^{*},\
\|p-p^{*}\|\geq a\\}.$

Then, $\hat{S}_{\varepsilon,a}$ is compact. We assume without loss of
generality that $\hat{S}_{\varepsilon,2\eta}$ is nonempty. Define

$\varphi(p)=\begin{cases}Ce^{-\frac{1}{1-\|p\|^{2}}}&\mbox{if }\|p\|<1,\\\
0&\mbox{if }\|p\|\geq 1,\end{cases}$

where $C>0$ is chosen as

$\int_{\mathbb{R}^{n}}\varphi(p)dp=1.$

For $\delta>0$, define

$\varphi_{\delta}(p)=\delta^{n}\varphi(p/\delta),$

and

$f^{\delta}(p)=\int_{\mathbb{R}^{n}}f(p-q)\varphi_{\delta}(q)dq.$

It is known that $f^{\delta}$ uniformly converges to $f$ on the compact set
$S_{\varepsilon}$ as $\delta\downarrow 0$. Moreover, $f^{\delta}$ is
$C^{\infty}$.222See Appendix C.5 of Evans (2010). Choose any $C^{\infty}$
function $\xi:\mathbb{R}\to[0,1]$ such that $\xi^{-1}(1)=]-\infty,1]$ and
$\xi^{-1}(0)=[2,+\infty[$, and for any $p^{*}\in E^{*}$, let

$t_{p^{*}}(p)=\xi(\eta^{-1}\|p-p^{*}\|).$

For every $p\in S_{\varepsilon}$, define333If necessary, we can define
$\tilde{f}^{\delta}(p)=\tilde{f}^{\delta}(p/\|p\|)$, and extend the domain of
$\tilde{f}^{\delta}$ to a cone generated by $S_{\varepsilon}$, which includes
an open neighborhood of $E^{*}$.

$\hat{f}^{\delta}(p)=f^{\delta}(p)-(f^{\delta}(p)\cdot p)p,$
$\tilde{f}^{\delta}(p)=\left(1-\sum_{p^{*}\in
E^{*}}t_{p^{*}}(p)\right)\hat{f}^{\delta}(p)+\sum_{p^{*}\in
E^{*}}t_{p^{*}}(p)f(p).$

Because $f^{\delta}$ converges to $f$ uniformly on $S_{\varepsilon}$ as
$\delta\downarrow 0$, there exists $\delta>0$ such that

  1. 1)

For any $p\in S_{\varepsilon}$, $\tilde{f}^{\delta}(p)=0$ if and only if $p\in
E^{*}$.

  2. 2)

If $\tilde{f}^{\delta}(p)=0$, then $\tilde{f}^{\delta}=f$ on some neighborhood
of $p$.

  3. 3)

$p\cdot\tilde{f}^{\delta}(p)=0$ for all $p\in S_{\varepsilon}$.

  4. 4)

For all $p\in\partial S_{\varepsilon}$, $q\in S_{\varepsilon^{\prime}}$, and
$t\in[0,1]$, $(1-t)\tilde{f}^{\delta}(p)+tf^{q}(p)\neq 0$.

  5. 5)

$\tilde{f}^{\delta}$ is $C^{1}$ on $S_{\varepsilon}$.

Clearly, if $\tilde{f}^{\delta}$ satisfies the claim of this theorem, then $f$
also satisfies the same claim. Hence, hereafter we implicitly use
$\tilde{f}^{\delta}$ instead of $f$, and thus assume that $f$ is $C^{1}$ on
$S_{\varepsilon}$.

Recall the definition of the tangent bundle:

$T(\tilde{S}_{\varepsilon})=\\{(p,v)|p\in\tilde{S}_{\varepsilon},\ v\in
T_{p}(\tilde{S}_{\varepsilon})\\}.$

Define a function $F:[0,1]\times
S_{\varepsilon}\times\tilde{S}_{\varepsilon^{\prime}}\to
T(\tilde{S}_{\varepsilon})$ as

$F(t,p,q)=(p,(1-t)f(p)+tf^{q}(p)),$

and $\tilde{F}$ as the restriction of $F$ into
$[0,1]\times\tilde{S}_{\varepsilon}\times\tilde{S}_{\varepsilon^{\prime}}$.
Moreover, we define

$G_{q}(t,p)=F(t,p,q),\ \tilde{G}_{q}(t,p)=\tilde{F}(t,p,q).$

Finally, for any
$(t,p,q)\in[0,1]\times\tilde{S}_{\varepsilon}\times\tilde{S}_{\varepsilon^{\prime}}$,
define $\pi(t,p,q)=q$. Because
$[0,1]\times\tilde{S}_{\varepsilon}\times\tilde{S}_{\varepsilon^{\prime}}$ is
a $C^{\infty}$ manifold and $\pi$ is $C^{\infty}$, we can apply Sard’s theorem
to $\pi$ and $\partial\pi$, and thus there exists
$q\in\tilde{S}_{\varepsilon^{\prime}}$ such that it is a regular value of
$\pi$ and $\partial\pi$.

######  Step 7.

If $q$ is a regular value of both $\pi$ and $\partial\pi$, then both
$\tilde{G}_{q}$ and $\partial\tilde{G}_{q}$ are transversal to
$\tilde{S}_{\varepsilon}\times\\{0\\}$.

  * Proof of Step 7.First, we verify that $\tilde{F}$ and $\partial\tilde{F}$ are transversal to $\tilde{S}_{\varepsilon}\times\\{0\\}$. Note that

$T_{p}(\tilde{S}_{\varepsilon})=\\{v\in\mathbb{R}^{n}|v\cdot p=0\\},$
$T_{(p,0)}(\tilde{S}_{\varepsilon}\times\\{0\\})=T_{p}(\tilde{S}_{\varepsilon})\times\\{0\\},$
$T_{(p,0)}(T(\tilde{S}_{\varepsilon}))=T_{p}(\tilde{S}_{\varepsilon})\times
T_{p}(\tilde{S}_{\varepsilon}).$

Suppose that $\tilde{F}(t,p,q)=(p,0)$. Let
$c:]-\varepsilon^{\prime\prime},\varepsilon^{\prime\prime}[\to\tilde{S}_{\varepsilon}\times\tilde{S}_{\varepsilon^{\prime}}$
be a $C^{1}$ function such that $c(0)=(p,q)$ and $c^{\prime}(0)=(v,w)$, and
define $d(s)=\tilde{F}(t,c(s))$. Then, $v\cdot p=w\cdot q=0$, and thus

$d^{\prime}(0)=(v,(1-t)df_{p}(v)+tdf_{p}^{q}(v)+(t/p\cdot q)w),$

which implies that both $d\tilde{F}_{(t,p,q)}$ and
$d(\partial\tilde{F})_{(t,p,q)}$ are surjective when $t\neq 0$. Suppose that
$t=0$. By the same argument as in the proof of Step 1, we have that
$df_{p}(v)=0$ if and only if $v=0$, and thus for every $w\in
T_{p}(\tilde{S}_{\varepsilon})$, there exists $v\in
T_{p}(\tilde{S}_{\varepsilon})$ such that $df_{p}(v)=w$. Therefore, in this
case,

$\mbox{Im}(d\tilde{F}_{(0,p,q)})+T_{(p,0)}(\tilde{S}_{\varepsilon}\times\\{0\\})=T_{(p,0)}(T(\tilde{S}_{\varepsilon})),$
$\mbox{Im}(d(\partial\tilde{F})_{(0,p,q)})+T_{(p,0)}(\tilde{S}_{\varepsilon}\times\\{0\\})=T_{(p,0)}(T(\tilde{S}_{\varepsilon})),$

which implies that $\tilde{F}$ and $\partial\tilde{F}$ are transversal to
$\tilde{S}_{\varepsilon}\times\\{0\\}$. As in the proof of the transversality
theorem in Section 2.3 of Guillemin and Pollack (1974), if both $\pi$ and
$\partial\pi$ have a common regular value $q$, then $\tilde{G}_{q}$ and
$\partial\tilde{G}_{q}$ are transversal to
$\tilde{S}_{\varepsilon}\times\\{0\\}$. This completes the proof of Step 7. ∎

######  Step 8.

Define $X\equiv\tilde{G}_{q}^{-1}(\tilde{S}_{\varepsilon}\times\\{0\\})$.
Then, $X$ is a compact $1$-dimensional $C^{1}$ manifold with boundary, where
$\partial X=\\{(0,p^{*})|p^{*}\in E^{*}\\}\cup\\{(1,q)\\}$.444Note that,
because $\varepsilon<\varepsilon^{\prime}$, we have that
$q\in\tilde{S}_{\varepsilon}$.

  * Proof of Step 8.Because of Step 6,555Actually, we use property 4) of $\tilde{f}^{\delta}$. we have that $X=G_{q}^{-1}(S_{\varepsilon}\times\\{0\\})$. Because $G_{q}$ is continuous, $X$ is closed. Moreover, because $X\subset[0,1]\times S_{\varepsilon}$, $X$ is compact. On the other hand, because $\tilde{G}_{q}$ and $\partial\tilde{G}_{q}$ are transversal to $\tilde{S}_{\varepsilon}\times\\{0\\}$, we have that $X$ is $1$-dimensional $C^{1}$ manifold with boundary, where $\partial X=X\cap(\\{0,1\\}\times\tilde{S}_{\varepsilon})$. Then,

$X\cap(\\{0,1\\}\times\tilde{S}_{\varepsilon})=\\{(0,p^{*})|p^{*}\in
E^{*}\\}\cup\\{(1,q)\\},$

and our claim is correct. This completes the proof of Step 8. ∎

######  Step 9.

$\mathrm{sign}_{\partial X}(1,q)=(-1)^{n-1}$, and for all $p^{*}\in E^{*}$,

$\mathrm{sign}_{\partial X}(0,p^{*})=(-1)^{n}\mathrm{index}(p^{*}).$

  * Proof of Step 9.Let $n(p)=p$. Then, $n(p)$ is a normal vector field on $\tilde{S}_{\varepsilon}$. Therefore, using this vector field, we can obtain an orientation $\mbox{sign}_{\tilde{S}_{\varepsilon}}$: that is, for a linearly independent family $v_{1},...,v_{n-1}\in T_{p}(\tilde{S}_{\varepsilon})$,

$\mbox{sign}_{\tilde{S}_{\varepsilon}}(p,v_{1},...,v_{n-1})=\mbox{sign}(\det(p,v_{1},...,v_{n-1})).$

Note that $[0,1]$ can be oriented by the usual function

$\mbox{sign}_{[0,1]}(t,a)=\mbox{sign}(a).$

Therefore, $[0,1]\times\tilde{S}_{\varepsilon}$ is oriented. The tangent
bundle $T(\tilde{S}_{\varepsilon})$ is oriented by the following function
$\mbox{sign}_{T(\tilde{S}_{\varepsilon})}$:

$\displaystyle~{}\mbox{sign}_{T(\tilde{S}_{\varepsilon})}((p,v),v_{1},...,v_{n-1},w_{1},...,w_{n-1})$
$\displaystyle=$
$\displaystyle~{}\mbox{sign}(\det((p,0),v_{1},...,v_{n-1},(0,p),w_{1},...,w_{n-1})).$

Therefore, $X$ is also oriented by the preimage orientation. Suppose that
$t\in\\{0,1\\}$, $(t,p)\in X$, and $(1,v)\in T_{(t,p)}(X)$. Choose
$v_{1},...,v_{n-1}\in T_{p}(\tilde{S}_{\varepsilon})$ such that
$\det(p,v_{1},...,v_{n-1})>0$. Because $\partial\tilde{G}_{(t,p)}$ is
transversal to $\tilde{S}_{\varepsilon}\times\\{0\\}$, there exists
$v_{1}^{\prime},...,v_{n-1}^{\prime}\in T_{p}(\tilde{S}_{\varepsilon})$ such
that $\det(p,v_{1}^{\prime},...,v_{n-1}^{\prime})>0$ and for
$(v_{i}^{\prime},w_{i}^{\prime})=d\tilde{G}_{(1,q)}(0,v_{i}^{\prime})$,
$x=((v_{1}^{\prime},w_{1}^{\prime}),...,(v_{n-1}^{\prime},w_{n-1}^{\prime}),(v_{1},0),...,(v_{n-1},0))$
consists of a basis of $T_{G(t,p)}(T(\tilde{S}_{\varepsilon}))$.666Note that
$d\tilde{G}_{(t,p)}(0,v^{\prime})=(v^{\prime},w^{\prime})$ by the definition
of $\tilde{G}$. Then,

$\displaystyle~{}\mbox{sign}_{X}((t,p),(1,v))$ $\displaystyle=$
$\displaystyle~{}\mbox{sign}_{T(\tilde{S}_{\varepsilon})}((p,0),x)\times\mbox{sign}_{[0,1]\times\tilde{S}_{\varepsilon}}((t,p),(0,v_{1}^{\prime}),...,(0,v_{n-1}^{\prime}),(1,v)).$

Note that, for $x_{1},...,x_{n}\in
T_{(t,p)}([0,1]\times\tilde{S}_{\varepsilon})$,

$\mbox{sign}_{[0,1]\times\tilde{S}_{\varepsilon}}((t,p),x_{1},...,x_{n})=\mbox{sign}(\det(x_{1},(0,p),x_{2},...,x_{n})),$

and thus, we have that

$\displaystyle~{}\mbox{sign}_{[0,1]\times\tilde{S}_{\varepsilon}}((t,p),(0,v_{1}^{\prime}),...,(0,v_{n-1}^{\prime}),(1,v))$
$\displaystyle=$
$\displaystyle~{}\mbox{sign}((-1)^{n+1}\det(p,v_{1}^{\prime},...,v_{n-1}^{\prime}))=(-1)^{n+1}.$

On the other hand,

$\displaystyle\mbox{sign}_{T(\tilde{S}_{\varepsilon})}((p,0),x)=$
$\displaystyle~{}\mbox{sign}((-1)^{n^{2}+1}\det(p,w_{1}^{\prime},...,w_{n-1}^{\prime})\det(p,v_{1},...,v_{n-1}))$
$\displaystyle=$
$\displaystyle~{}(-1)^{n+1}\mbox{sign}(\det(p,w_{1}^{\prime},...,w_{n-1}^{\prime})).$

Hence, we obtain that

$\mbox{sign}_{X}((t,p),(1,v))=\mbox{sign}(\det(p,w_{1}^{\prime},...,w_{n-1}^{\prime})).$

Suppose that $t=1$. Then, $p=q$. Because
$w_{i}^{\prime}=Df^{q}(q)v_{i}^{\prime}$, by Step 4 and (1), we have that
$\mbox{sign}_{X}((1,q),(1,v))=(-1)^{n-1}$, and thus $\mbox{sign}_{\partial
X}(1,q)=(-1)^{n-1}$.

Next, suppose that $t=0$. Then, $p=p^{*}$ for some $p^{*}\in E^{*}$. Because
$w_{i}^{\prime}=Df(p)v_{i}^{\prime}$, by Step 4, we have that
$\mbox{sign}_{X}((0,p^{*}),(1,v))=(-1)^{n-1}\mbox{index}(p^{*})$. Because
$(1,v)$ is an inward vector at $T_{(0,p^{*})}(X)$, we have that

$\mbox{sign}_{\partial
X}(0,p^{*})=\mbox{sign}_{X}((0,p^{*}),(-1,-v))=(-1)^{n}\mbox{index}(p^{*}).$

This completes the proof of Step 9. ∎

Because any compact $1$-dimensional $C^{1}$ manifold is a finite union of
connected components, and each of them is diffeomorphic to either $S^{1}$ or
$[0,1]$, we have that

$\sum_{(t,p)\in\partial X}\mbox{sign}_{\partial X}(t,p)=0.$

By Step 9,

$\sum_{p^{*}\in E^{*}}\mbox{index}(p^{*})=1,$

as desired. This completes the proof of Theorem 1. ∎

##  4 Remarks

Our Theorem 1 is, at least formally, different from Proposition 5.6.1 of Mas-
Colell (1985). That is, Mas-Colell treated a function $f$ whose domain is $S$,
whereas in our Theorem 1, we treat a function $f$ defined on
$\mathbb{R}^{n}_{++}$ that is homogeneous of degree zero. Moreover, the
definitions of $\mbox{index}(p^{*})$ are a little different. However, readers
can check that these differences are superficial. First, the restriction of a
function treated in Theorem 1 into $S$ satisfies all the assumptions of Mas-
Colell’s proposition. Conversely, a function treated in Mas-Colell’s
proposition can be extended to $\mathbb{R}^{n}_{++}$ by the following formula
$f(p)=f(p/\|p\|)$, and this extended function satisfies all requirements of
our Theorem 1. Furthermore, Mas-Colell mentioned in Section 5.3 of his book
that his definition of $\mbox{index}(p^{*})$ is equivalent to our definition.
Therefore, actually our Theorem 1 is equivalent to Mas-Colell’s proposition.

This result is closely related to the Poincaré-Hopf index theorem. This
theorem claims for every smooth vector field in a compact smooth manifold, the
sum of the “indices” of zeroes are the same as the Euler characteristic of
this manifold. This theorem is often explained as the existence theorem of
“hair whorl.” Let us look at the human head as a manifold and consider the
direction of hair growth as a vector field. The human head can be seen as the
upper hemisphere in $\mathbb{R}^{3}$, and its Euler characteristic is $1$.
Therefore, there must be a zero with a positive index. This is the hair whorl.
From this explanation, this theorem is also called the “hairy ball theorem.”

However, upon closer examination, we find something strange. What would happen
to the head of an alopecia areata patient? In this case, if we remove the part
of the head that is bald from the upper hemisphere, a new manifold is created.
This manifold is diffeomorphic to the compact domain of a large disc from
which a small disc is removed. It is easy to see that the Euler characteristic
of this manifold is $0$, because the rotating vector field has no zero. Hence,
there must either be no hair whorl or an even number of hair whorls, although
we have never heard of a hair whorl disappearing caused by alopecia areata. In
other words, if a region is removed by alopecia areata, the Poincaré-Hopf
index theorem does not appear to hold.

The reason for this lies in a hidden assumption in the Poincaré-Hopf index
theorem. If the manifold has a boundary, then this theorem requires that the
vector field is outward on that boundary. Normal human head hair satisfies
this condition, and thus this theorem is applicable. However, this condition
could not be satisfied on the boundary of the hair lost area caused by
alopecia areata, and thus this theorem is not applicable.

With this in mind, let us look at our Theorem 1. Unfortunately, $S$ is not
compact, and thus the index theorem is not applicable. On the other hand,
$S_{\varepsilon}$ is compact. Although there is a problem that
$S_{\varepsilon}$ is not a manifold, we ignore this problem and check whether
the index theorem is applicable. Then, we encounter the problem that we
indicated above. That is, our vector field $f$ is not necessarily outward on
$\partial S_{\varepsilon}$. Therefore, the index theorem is still
inapplicable.

To solve this problem, Mas-Colell prepared a function $f^{q}$, to which the
conclusion of the index theorem holds. It is known that the sum of indices is
a homotopy invariant. Because $f^{q}$ is homotopic to $f$, the sum of indices
of $f$ is equal to that of $f^{q}$. However, $f^{q}$ has a unique zero $q$,
whose index is $+1$. Hence, the sum of indices of $f$ is also $+1$. This is
the ‘first’ proof of this theorem provided by Mas-Colell. Curiously, Mas-
Colell made the ‘second’ proof of this theorem, but this proof is profound and
unfamiliar.

There are a number of immediately visible problems with this proof. First,
$S_{\varepsilon}$ is not a manifold. Although $\tilde{S}_{\varepsilon}$ is a
manifold, it is not compact. If we change the definition of $S_{\varepsilon}$,
then it may possibly be a manifold with boundary. However, even in this case,
$[0,1]\times S_{\varepsilon}$ is not a manifold with boundary. As we have
already seen in the proof, to show that the sum of indices is homotopy
invariant, we need to use the transversality of $G_{q}(t,p)$ to prove that $X$
is a one-dimensional manifold and that the sum of orientations of its boundary
is $0$. However, for the reasons mentioned above, the domain of $G_{q}$ is not
a manifold, and thus it is unclear whether the result of transversality is
applicable. Step 6 is essential for solving this problem. In fact, Mas-Colell
himself proved Step 6, but he did not explain why it is needed at all.

Second, to apply the transversality result to $G_{q}$, the range must be a
manifold. However, because $f$ is a vector field, the range of this function
is not a manifold. Therefore, it is necessary to make the range of $G_{q}$ a
manifold artificially by introducing the notion of tangent bundle, and then
apply the transversality theorem to $F$ and show that there exists $q$ such
that $G_{q}$ is transversal. However, Mas-Colell did not mention the tangent
bundle in his proof.

Hence, Mas-Colell’s proof has a number of big gaps. We think that our proof
fulfills these gaps completely, and thus it is worthwhile to many economists
that want to use the index theorem.

##  Appendix A Proof of the Classification Theorem

The classification theorem for one-dimensional manifolds can be found in
Guillemin and Pollack (1974). However, as already mentioned, their proof uses
the Morse function. Since the definition of the Morse function involves
second-order derivatives, this argument cannot be applied to $C^{1}$
manifolds.

Milnor (1965) provided another proof using arc length. However, this proof is
too difficult to understand. On the other hand, Exercise 1.4.9 in Hirsch
(1976) asks us to prove this theorem. However, no answer is provided there.
Moreover, Hirsch stated in the preface to his book that the exercises contain
open problems. Therefore, the existence of this exercise does not solve this
problem in any way. Gale (1987) treated this classification theorem. However,
he only proved that there is a function that is not a diffeomorphism, but only
a homeomorphism. Thus, this cannot be used for our purpose.

In fact, as of November 2022, using Google search, we found a document that
proved this theorem rigorously on the server of the University of Lisbon. Upon
investigation, we found that the document was written by Gustavo Granja, who
provided it for his student. However, there is no signature of the author
inside the document, and thus this monograph is not in a position to be added
to the reference list. In addition, since it is a lecture document, it could
disappear at any time because of Granja’s transfer or other reasons.
Therefore, in this section, we reconstruct Granja’s proof only as we need it.

Recall the definition of $S^{1}$:

$S^{1}=\\{x\in\mathbb{R}^{2}|\|x\|=1\\}.$

Suppose that $k\in\mathbb{N}\cup\\{0,\infty\\}$ is given, and suppose that $X$
is a $1$-dimensional compact and connected $C^{k}$ manifold. Our goal is to
show the following theorem.

######  Theorem 2.

$X$ is $C^{k}$ diffeomorphic to either $S^{1}$ or $[0,1]$.

  * Proof.In this proof, we call a function $f:W\to X$ a local parametrization of $X$ if and only if it is a $C^{k}$ diffeomorphism from $W$ onto $f(W)$, where either $W=(a,b)$ or $W=[a,b)$ for some $a,b\in\mathbb{R}$ with $a<b$ and $f(W)$ is open in $X$. By definition, we can easily check that $W=[a,b)$ means that $f(a)\in\partial X$.

Let $\varphi:U\to X$ and $\psi:V\to X$ be given $C^{k}$ functions such that
$\varphi^{-1}$ and $\psi^{-1}$ are local parametrizations whose domains are
$I$ and $J$, respectively, and $U,V\subset X$ satisfies

$U\cap V\neq\emptyset,\ U\setminus V\neq\emptyset,V\setminus U\neq\emptyset.$
(2)

If a pair $(\varphi,\psi)$ of functions satisfies the above requirements, then
we say that $(\varphi,\psi)$ satisfies condition A.

We separate the proof into nine steps. In Steps 1-6, we assume that
$(\varphi,\psi)$ satisfies condition A. By definition, $I,J\subset\mathbb{R}$
are bounded intervals.777We call a subset of $\mathbb{R}$ an interval if this
set is a convex set that contains at least two points. If $I$ is a bounded
interval whose closure is $[a,b]$, then $a,b$ are called endpoints of $I$.
Define $\alpha:\varphi(U\cap V)\to\psi(U\cap V)$ as
$\alpha=\psi\circ\varphi^{-1}$. It is easy to show that $\varphi(U\cap V)$ is
a (possibly infinitely many) disjoint union of relatively open separated
intervals of I.

######  Step 1.

Suppose that $K$ is a connected component of $\varphi(U\cap V)$ and $a$ is an
endpoint of $K$. Then, $a\notin K$ and if $a\in I$, then

$c\equiv\lim_{t\to a,\ t\in K}\alpha(t)\notin J,$

and $c$ is an endpoint of $J$.

  * Proof of Step 1.Because $J$ is bounded, $\alpha(t)$ is bounded on $K$. Moreover, because $\alpha$ is a homeomorphism defined on $K$, it is monotone, which implies that $c$ is defined and a real number.

Because $K$ is relatively open in $I$, if $a$ is not an endpoint of $I$, then
$a\notin K$. Suppose that $a$ is an endpoint of $I$ and $a\in K$. Then, $a\in
I$ and thus $I=[a,b)$ for some $b>a$, which implies that
$\varphi^{-1}(a)\in\partial X$ and $\alpha(a)$ is also an endpoint of $J$. Let
$\sup K=d$. Suppose that $d=b$. Then, $U\subset V$, which contradicts (2).
Hence, $d<b$ and $a<d<b$. Let $\delta=\lim_{t\to d,\ t\in K}\alpha(t)$. If
$\delta\notin J$, then $J=[\alpha(a),\delta)$, and thus $V\subset U$, which
contradicts (2). Therefore, $\delta\in J$. Because $\varphi^{-1}$ and
$\psi^{-1}$ are continuous, $\varphi^{-1}(d)=\psi^{-1}(\delta)$. Since $\psi$
is a homeomorphism, $U\cap V$ must include a neighborhood of
$\psi^{-1}(\delta)$, and thus $K$ includes $[a,d+\varepsilon)$ for some
$\varepsilon>0$, which is a contradiction. Thus, we conclude that $a\notin K$.

Next, suppose that $c\in J$. If $\varphi^{-1}(a)=\psi^{-1}(c)$, then $a\in K$,
which is a contradiction. Therefore, $\varphi^{-1}(a)\neq\psi^{-1}(c)$, which
contradicts the continuity of $\varphi^{-1}$ and $\psi^{-1}$. Hence, we have
that $c\notin J$. This completes the proof. ∎

######  Step 2.

$\varphi(U\cap V)$ consists of either one or two open intervals, and if $K$ is
a connected component of $\varphi(U\cap V)$, then $K$ has at least one
endpoint in common with $I$.

  * Proof of Step 2.Suppose that $K$ is a connected component of $\varphi(U\cap V)$. By Step 1, we have that $K$ is an open interval, and thus $K=(a,b)$ for some $a,b$ such that $a<b$. Let $c,d$ be endpoints of $I$ such that $c<d$. Then, $c\leq a<b\leq d$. If $c<a$ and $b<d$, then $V=\varphi^{-1}(K)\subset U$, which contradicts (2). Therefore, we have that either $c=a$ or $b=d$, which implies that $K$ has at least one endpoint in common with $I$. Therefore, there are at most two such $K$, which completes the proof of Step 2. ∎

######  Step 3.

If $\varphi(U\cap V)$ consists of two connected components, then $I$ and $J$
are open and $U\cup V$ is homeomorphic to $S^{1}$.

  * Proof of Step 3.Suppose that $a,b$ are endpoints of $I$ such that $a<b$. Then, there are two connected components $K_{1}=(a,c)$ and $K_{2}=(d,b)$ of $\varphi(U\cap V)$ such that $c\leq d$. Define

$e=\lim_{t\uparrow c}\alpha(t),\ f=\lim_{t\downarrow d}\alpha(t).$

By Step 1, we have that $e,f\notin J$, and $e,f$ are endpoints of $J$. If
$e=f$, then $\alpha$ is monotone in $K_{1}\cup K_{2}$, and because $J$ is
convex, there exists $r\in V\setminus U$ such that $\psi(r)=e=f$. However,
this implies that $e,f\in J$, which is a contradiction. Therefore, $e\neq f$,
which implies that $J$ is an open interval. By symmetrical arguments, we can
show that $I$ is an open interval, and thus $I=(a,b)$.

Thus, either $e<f$ or $e>f$. Because arguments can be done symmetrically, we
assume that $e>f$, and thus $J=(f,e)$. Then, $\alpha$ is increasing on both
$K_{i}$, and $\alpha(K_{1})=(g,e)$ and $\alpha(K_{2})=(f,h)$ for

$g=\lim_{t\downarrow a}\alpha(t),\ h=\lim_{t\uparrow b}\alpha(t).$

Because $e>f$, we have that $g\geq h$. If $g=h$, then $(U\cup V)\setminus U$
is a singleton, and thus $U\cup V$ can be seen as a one-point compactification
of $U$, which is homeomorphic to $S^{1}$. If $g>h$, then choose an increasing
homeomorphism $\lambda:(0,\pi)\to(a,b)$ and $\mu:[\pi,2\pi]\to[h,g]$, and
define

$f(e^{i\theta})=\begin{cases}\varphi^{-1}(\lambda(\theta))&\mbox{if
}0<\theta<\pi,\\\ \psi^{-1}(\mu(\theta))&\mbox{if }\pi\leq\theta\leq
2\pi.\end{cases}$

We can easily check that $f$ is a homeomorphism between $S^{1}$ and $U\cup V$.
This completes the proof of Step 3. ∎

######  Step 4.

Suppose that $\varphi(U\cap V)=K$ for an open interval $K$ whose endpoints are
in common with $I$. Then, $I=[a,b)$ and $K=(a,b)$ for some $a,b$ with $a<b$,
and $U\cap\partial X\neq\emptyset$. If $V\cap\partial X=\emptyset$, then
$U\cup V$ is homeomorphic to $[0,1)$. If $V\cap\partial X\neq\emptyset$, then
$U\cup V$ is homeomorphic to $[0,1]$.

  * Proof of Step 4.Suppose that $K=(a,b)$. Because of (2), we have that $I\neq K$, and thus $I=[a,b)$. Therefore, $\varphi^{-1}(a)\in\partial X$, and thus $U\cap\partial X=\\{\varphi^{-1}(a)\\}$. Let $c,d$ be endpoints of $\psi(U\cap V)$, and

$c=\lim_{t\downarrow a}\alpha(t).$

By Step 1, $c\notin J$ and $c$ is an endpoint of $J$. Because of (2), we have
that $d\in J$, and thus $\psi^{-1}(d)\in V\setminus U$. Suppose that $e$ is
another endpoint of $J$. Then, $J$ is $(e,c)$, $(c,e)$, or $[e,c)$. If
$J=(e,c)$, then $J\cap\partial X=\emptyset$ and $e<d$. In this case, choose
homeomorphisms $\lambda:[0,1/2)\to[a,b)$ and $\mu:[1/2,1)\to(e,d]$, and define

$f(x)=\begin{cases}\varphi^{-1}(\lambda(x))&\mbox{if }0\leq x<1/2,\\\
\psi^{-1}(\mu(x))&\mbox{if }1/2\leq x<1.\end{cases}$

Then, we can easily check that $f$ is a homeomorphism between $[0,1)$ and
$U\cup V$. The cases $J=(c,e)$ and $J=[e,c)$ can be similarly treated. This
completes the proof. ∎

######  Step 5.

Suppose that $\varphi(U\cap V)=K$ for an open interval $K$ and there exists an
endpoint $b$ of $K$ such that it is not an endpoint of $I$. Then, $U\cup V$ is
homeomorphic to $Y$, where $Y$ is $(0,1),[0,1)$ or $[0,1]$. Suppose that
$f:Y\to(U\cup V)$ is the diffeomorphism. If $Y=(0,1)$, then $(U\cup
V)\cap\partial X=\emptyset$. If $Y=[0,1)$, then $(U\cup V)\cap\partial
X=\\{f(0)\\}$. If $Y=[0,1]$, then $(U\cup V)\cap\partial X=\\{f(0),f(1)\\}$.

  * Proof of Step 5.This step can be proved in almost the same manner as the proof of Step 4. Hence, we omit the proof. ∎

######  Step 6.

If $U\cup V$ is homeomorphic to $S^{1}$ or $[0,1]$, then $U\cup V=X$.

  * Proof of Step 6.In this case, $U\cup V$ becomes a clopen set. Therefore, by connectedness of $X$, we have that $U\cup V=X$. This completes the proof of Step 6. ∎

Choose any pair $(\varphi,\psi)$ that satisfies condition A. Suppose that
there exist an open set $U^{\prime}$ of $X$ and an extension
$\varphi^{\prime}:U^{\prime}\to\mathbb{R}$ of $\varphi$ such that the closure
$\bar{U}$ of $U$ is included in $U^{\prime}$ and $(\varphi^{\prime},\psi)$
satisfies condition A. Then, we say that $(\varphi,\psi)$ satisfies condition
B.

In Steps 7-8, we suppose that $(\varphi,\psi)$ satisfies condition B and
$\varphi^{\prime}:U^{\prime}\to I^{\prime}$ is fixed.

######  Step 7.

Suppose that $k\geq 1$, and $\varphi(U\cap V)$ has two connected components.
Then, $U\cup V$ is $C^{k}$ diffeomorphic to $S^{1}$.

  * Proof of Step 7.As in the proof of Step 3, we define $a,b,c,d,e,f,g,h$. Then, $I=(a,b)$, $\varphi(U\cap V)=(a,c)\cup(d,b)$,

$\lim_{t\uparrow c}\alpha(t)=e,\ \lim_{t\downarrow d}\alpha(t)=f,$
$\lim_{t\downarrow a}\alpha(t)=g,\ \lim_{t\uparrow b}\alpha(t)=h,$

and $e\neq f$. Because arguments can be done symmetrically, we assume that
$e>f$, and thus $J=(f,e)$. Then, $\alpha$ is increasing on both $K_{i}$, and
$\alpha(K_{1})=(g,e)$ and $\alpha(K_{2})=(f,h)$. If $g=h$, then $U\cup
V=U\cup\\{\psi(g)\\}$. By Step 3, in this case $U\cup V$ is homeomorphic to
$S^{1}$, and thus by Step 6, $U\cup V=X$, which implies that $V\subset
X=U^{\prime}$. This contradicts that $U^{\prime},V$ satisfies (2). Therefore,
we must have that $h<g$. Choose an increasing linear function
$\lambda:[-\delta,\pi+\delta]\to[a-\varepsilon,b+\varepsilon]$ such that
$\delta<\pi/2$, $\lambda(0)=a,\ \lambda(\pi)=b$, and
$[a-\varepsilon,c)\cup(d,b+\varepsilon]\subset\varphi^{\prime}(U^{\prime}\cap
V)$, and thus the domain of $\alpha$ can be extended to
$[a-\varepsilon,c)\cup(d,b+\varepsilon]$. Define $\beta=\alpha\circ\lambda$.
Choose $p,q\in(h,g)$ such that

$\beta(\pi+\delta/2)<p<q<\beta(-\delta/2).$

Because $\beta(\pi+\delta)\leq\beta(-\delta)$, we have that such $p,q$ exist.
Let $\mu(\theta)=((2\pi-\theta)p+(\theta-\pi)q)/\pi$ on $[\pi,2\pi]$, and
choose any $C^{\infty}$ function $\nu:[\pi,2\pi]\to[0,1]$ such that
$\nu^{-1}(0)=[\pi,\pi+\delta/2]\cup[2\pi-\delta/2,2\pi]$,
$\nu^{-1}(1)=[\pi+\delta,2\pi-\delta]$ and $\nu^{\prime}(\theta)\neq 0$ if
$\theta\in(\pi+\delta/2,\pi+\delta)\cup(2\pi-\delta/2,2\pi)$, and define

$g(\theta)=(1-\nu(\theta))(\beta(\theta)+\beta(\theta-2\pi))+\nu(\theta)\mu(\theta),$

where we define $\beta(\theta)=0$ if $\theta\notin[-\delta,\pi+\delta]$. Then,
we can check that $g$ is $C^{k}$ on $[\pi,2\pi]$, $g^{\prime}(\theta)>0$ for
all $\theta\in[\pi,2\pi]$, and $g\equiv\beta$ on some neighborhood of
$\\{\pi,2\pi\\}$. Define

$f(e^{i\theta})=\begin{cases}\varphi^{-1}(\lambda(\theta))&\mbox{if
}0<\theta<\pi,\\\ \psi^{-1}(g(\theta))&\mbox{if }\pi\leq\theta\leq
2\pi.\end{cases}$

Then, $f$ is a $C^{k}$ diffeomorphism between $S^{1}$ and $U\cup V$. This
completes the proof of Step 7. ∎

######  Step 8.

Suppose that $k\geq 1$, and $\varphi(U\cap V)$ is an open interval. Then,
“homeomorphic” that appears in Steps 4-5 can be replaced with “$C^{k}$
diffeomorphic.”

  * Proof of Step 8.We treat only the case in which $I,J$ are open intervals and $\varphi(U\cap V)=K$ for an open interval $K$, because the remaining cases can be treated in almost the same manner. It is easy to show that there exists a $C^{\infty}$ diffeomorphism $g_{1}:I\to(0,1)$ and $g_{2}:J\to(1,2)$ such that $g_{1}(\varphi(U\cap V))=(1/2,1)$ and $g_{2}(\psi(U\cap V))=(1,3/2)$. Therefore, without loss of generality, we can assume that $I=(0,1),J=(1,2)$, and $\varphi(U\cap V)=(1/2,1),\psi(U\cap V)=(1,3/2)$. This implies that $\alpha$ is increasing. Because $(\varphi,\psi)$ satisfies condition B, $\alpha$ can be extended to $(1/2,1+\varepsilon]$ for some $\varepsilon\in(0,1/2)$ such that $\alpha(1+\varepsilon)<2$. Choose any $C^{\infty}$ function $\lambda:(0,3/2)\to[0,1]$ such that $\lambda(x)=1$ if $x\leq 1$, $\lambda(x)=0$ if $x\geq 1+\varepsilon$, and $\lambda^{\prime}(x)<0$ for all $x\in(1,1+\varepsilon)$. Define

$g(x)=\alpha(1+\varepsilon)+2(x-1)(2-\alpha(1+\varepsilon)),$

and

$f(x)=\begin{cases}\varphi^{-1}(x)&\mbox{if }x<1,\\\
\psi^{-1}(\lambda(x)\alpha(x)+(1-\lambda(x))g(x))&\mbox{if }x\geq
1.\end{cases}$

Then, $f$ is a $C^{k}$ diffeomorphism between $(0,3/2)\to U\cup V$. This
completes the proof of Step 8. ∎

######  Step 9.

$X$ is $C^{k}$ diffeomorphic to $S^{1}$ or $[0,1]$.

  * Proof of Step 9.In this proof, let $\bar{W}$ denote the closure of $W$. Choose any open covering $(U_{i})$ of $X$ such that there exists a $C^{k}$ diffeomorphism $\varphi_{i}:U_{i}\to A_{i}$ such that $A_{i}$ is either $(a,b)$ or $[a,b)$. Shrinking $U_{i}$ if needed, we can assume that $\varphi_{i}$ can be extended to some open set $U_{i}^{\prime}$ of $X$ that includes $\bar{U}_{i}$, and $\varphi_{i}$ is a $C^{k}$ diffeomorphism defined on $U_{i}^{\prime}$. Because $X$ is compact, there exists a finite subcovering $(U_{i_{j}})_{j=1}^{n}$. Define

$V_{j}=U_{i_{1}}\cup...\cup U_{i_{j}}.$

Renumbering and removing some $U_{i_{j}}$ if needed, we can assume that

$U_{i_{j}}\cap V_{j-1}\neq\emptyset,\ U_{i_{j}}\setminus
V_{j-1}\neq\emptyset,\ V_{j-1}\setminus U_{i_{j}}\neq\emptyset.$

Shrinking $U_{i_{1}}^{\prime},...,U_{i_{n}}^{\prime}$ if needed, we can assume
without loss of generality that

$V_{j-1}\setminus U_{i_{j}}^{\prime}\neq\emptyset.$

If $n=1$, then $X=U_{i_{1}}$ is $C^{k}$ diffeomorphic to either $[0,1)$ or
$(0,1)$. However, because $X$ is compact, it is impossible. Hence, we have
that $n\geq 2$. By Steps 7-8, there exists a $C^{k}$ diffeomorphism
$\psi_{2}:V_{2}\to I_{2}$ such that $I_{2}$ is $S^{1}$, $[0,1]$, $(0,1)$, or
$[0,1)$. Because $X$ is compact, $V_{2}=X$ implies that $I_{2}$ is either
$S^{1}$ or $[0,1]$, and by Step 6, the converse is also true. Therefore,
$I_{2}$ is $S^{1}$ or $[0,1]$ if and only if $n=2$, and if $n>2$, then $I_{2}$
is either $(0,1)$ or $[0,1)$. Moreover, if $I_{2}=[0,1)$, then
$\psi_{2}^{-1}(0)\in\partial X$. Hence, $(\varphi_{3},\psi_{2})$ satisfies
condition B. Inductively, we can assume that for each $j\in\\{2,...,n\\}$,
there is a $C^{k}$ diffeomorphism $\psi_{j}:V_{j}\to I_{j}$ that satisfies the
same condition as $\psi_{2}$, and because $V_{n}=X$, $I_{n}$ is either $S^{1}$
or $[0,1]$. This completes the proof of Step 9. ∎

Step 9 says that our initial claim is correct. This completes the proof. ∎

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