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 $\ds \int_a^b r = 1$.

Define:

$\ds \map g x = \map f a + \int_a^x \paren {k \map r s + \map {f'} s \paren {1 - \map r s} } \rd s$

where the constant $\ds k = \map f b = - \map f a - \int_a^b \map {f'} s \paren {1 - \map r s} \d s$.

$\Box$

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.

$\blacksquare$