Classification of Compact One-Manifolds
Theorem
Every compact one-dimensional manifold is diffeomorphic to either a circle or a closed interval.
Corollary
Any compact one-manifold has an even number of points in its boundary.
Proof
Lemma 1
Let $f$ be a function on $\closedint a b$ that is smooth and has a positive derivative everywhere except one interior point, $c$.
Then there exists a globally smooth function $g$ that agrees with $f$ near $a$ and $b$ and has a positive derivative everywhere.
Proof of Lemma 1
Let $r$ be a smooth nonnegative function that vanishes outside a compact subset of $\openint a b$, which equals $1$ near $c$, and which satisfies $\displaystyle \int_a^b r = 1$.
Define:
- $\displaystyle \map g x = \map f a + \int_a^x \paren {k \map r s + \map {f'} s \paren {1 - \map r s} } \rd s$
where the constant $\displaystyle k = \map f b= - \map f a - \int_a^b \map {f'} s \paren {1 - \map r s} \d s$.
$\Box$
Let $f$ be a Morse function on a one-manifold $X$.
Let $S$ be the union of the critical points of $f$ and $\partial X$.
As $S$ is finite, $X - S$ consists of a finite number of one-manifolds, $L_1, L_2, \cdots, L_n$.
Lemma 2
$f$ maps each $L_i$ diffeomorphically onto an open interval in $\R$
Proof of Lemma 2
Let $L$ be any of the $L_i$.
Because $f$ is a local diffeomorphism and $L$ is connected, $f \sqbrk L$ is open and connected in $\R$.
We also have $f \sqbrk L \in f \sqbrk X$, the latter of which is compact.
Hence there are numbers $c$ and $d$ such that $f \sqbrk L = \openint c d$.
It suffices to show $f$ is one to one on $L$, because then $f^{-1}: \openint c d \to L$ is defined and locally smooth.
Let $p$ be any point of $L$.
Set $q = \map f p$.
It suffices to show that every other point $z \in L$ can be joined to $p$ by a curve $\gamma: \closedint q y \to L$ such that $f \circ \gamma$ is the identity and $\map \gamma y = z$.
Since $\map f z = y \ne q = \map f p$, this result shows $f$ is one to one.
So let $Q$ be the set of points $x$ that can be so joined.
Since $f$ is a local diffeomorphism, $Q$ is both open and Definition:Closed Set (Topology).
Hence $Q = L$.
$\Box$
Lemma 3
Let $L$ be a subset of $X$ diffeomorphic to an open interval in $\R$, where $\dim X = 1$.
Then the closure $\map \cl L$ contains at most two points not in $L$.
Proof of Lemma 3
Let $g$ be a diffeomorphism $g: \openint a b \to L$ and let $p \in \map \cl L - L$.
Let $J$ be a closed subset of $X$ diffeomorphic to $\closedint 0 1$ such that:
- $1$ corresponds to $p$
- $0$ corresponds to some $\map g t$ in $L$.
Consider the set $\set {s \in \openint a t: \map g s \in J}$.
This set is both open and closed in $\openint a b$.
Hence $J$ contains either $g \sqbrk {\openint a t}$ or $g \sqbrk {\openint t b}$.
$\Box$
Proof of Corollary
Follows trivially.
$\blacksquare$