Classification of Compact One-Manifolds

= Theorem =

Every compact one-dimensional manifold is diffeomorphic to either a circle or a closed interval.

Corollary: It follows trivially from the theorem that any compact one-manifold has an even number of points in its boundary.

= Proof =

Lemma 1: Let f be a function on [a,b] that is smooth and have 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 have positive derivative everywhere. Proof: Let r be a smooth nonnegative function that vanishes outside a compact subset of (a,b), which equals 1 near c, and which satisfies $$\int_{a}^{b} r = 1$$. Define $$g(x) = f(a) + \int_{a}^{x} (k r(s)+f'(s)(1-r(s)))ds$$ where the constant $$k=f(b)-f(a)-\int_{a}^{b} f'(s)(1-r(s))ds$$. QED

Now, let f be a Morse function on a one-manifold X, and S the union of the critical points of f and $$\partial X$$. Since S is finite, X-S consists of a finite number of one-manifolds, L1, L2, ..., Ln.

Lemma 2: f maps each Li diffeomorphically onto an open interval in $$\mathbb{R}$$ Proof: Let L be any of the Li. Because f is a local diffeomorphism and L is connected, f(L) is open and connected in  $$\mathbb{R}$$. We also have $$f(L) \in f(X)$$, the latter of which is compact, so there are numbers c,d, such that $$f(L) = (c,d)$$. It suffices to show f is one to one on L, because then $$f^{-1}:(c,d) \to L$$ is defined and locally smooth. Let p be any point of L and set q=f(p). It suffices to show that every other point $$z \in L$$ can be joined to p by a curve $$\gamma : [q,y] \to L$$, such that $$f \circ \gamma$$ is the identity and $$\gamma (y) = z$$ Since $$f(z) = y \neq q = f(p)$$, 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 closed and hence Q=L. QED

Lemma 3: Let L be a subset of X diffeomorphic to an open interval in $$\mathbb{R}$$, where dim X = 1. Then the closure Cl(L) contains at most two points not in L. Proof: Let g be a diffeomorphism $$g:(a,b) \to L$$ and let $$p \in Cl(L)-L$$. Let J be a closed subset of X  diffeomorphic to [0,1] such that 1 corresponds to p and 0 corresponds to some g(t) in L.  Consider the set $$\left\{{ s \in (a,t) | g(s) \in J }\right\}$$. This set is both open and closed in (a,b), hence J contains either $$g((a,t))$$ or $$g((t,b))$$. QED