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Classification of 1-Manifolds

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Classification of 1-Manifolds Classification of 1-Manifolds The aim of this document is to clarify the proof of the following theorem from Appendix 2 of “Differential Topology" by Guillemin and Pollack: Theorem 1. Let X be a smooth, compact, non-empty, Hausdorff, connected 1- manifold, possibly with boundary. Then X is diffeomorphic to either a circle or a closed interval. The proof of this theorem given in ibid. is basically correct, but many details are not provided and the use of their \Smoothing Lemma" is not well-explained, and, indeed, is rather misleading. (They actually prove a slightly stronger statement than they give in the statement of the \Smoothing Lemma," and they need this stronger version of the \Smoothing Lemma" at the end of their proof because the function F they construct on Page 211 will not generally be smooth.) Another thing the reader will find frustrating in their proof is that the theorem is clearly not true without the “Hausdorff” assumption, though it is incredibly hard to see where this assumption is actually used in the proof. (The word “Hausdorff” does not appear in the proof.) We will basically follow their proof, using some lemmas which will fill in the details I find \missing" in their exposition. Most of these details are topological in nature. We will use some basic notions of \connectedness," which we now recall for the reader's convenience. If X is a topological space, two points x; y 2 X are said to be path connected (notation: x ∼ y) iff there is a continuous map γ : [0; 1] ! X with γ(0) = x, γ(1) = y. (Such a map γ is often called a path from x to y.) One easily checks that ∼ is an equivalence relation on X. A space X is called path connected (resp. connected) iff x ∼ y for every x; y 2 X (resp. iff ; and X are the only subsets of X which are both open and closed). Path-connected implies connected. Write [x] for the ∼ equivalence class of x. Clearly [x] is path connected. Suppose X satisfies the following assumption: (∗) For every x 2 X, there is a basis U for the neighborhoods of x in X such that every U 2 U is path-connected. Then it is easy to see that [x] is both open and closed in X for every x 2 X, hence X can be written as a coproduct (disjoint union) of path connected (hence connected) subspaces, called the path components of X. In particular, the notions \connected," and \path-connected" are the same for spaces satisfying (∗). n n−1 Since (∗) can be checked locally and (∗) clearly holds for R and R × R≥0, every manifold (even every manifold with boundary) satisfies (∗). For a sequence of points x1; x2;::: in a topological space X, and a point x 2 X, we write \xn ! x as n ! 1" to mean that any neighborhood of x in X contains all but finitely many of the xn. If X is Hausdorff, then \xn ! x as n ! 1" can be true for at most one x 2 X. We will refer to this simple fact as the \uniqueness of limits in Hausdorff spaces." 2 Lemma 2. Let X be a connected 1-manifold, possibly with boundary, but not neces- sarily Hausdorff or second countable. Then for any x 2 X, the 1-manifold X n fxg has at most two path components. Proof. The proof is slightly different depending on whether x is a boundary point or not. We will treat the case where x is an interior point, which is slightly harder, and leave the case of a boundary point to the reader. (When x is a boundary point, one in fact shows that X n fxg is path-connected.) Since x is an interior point of X, we can find a neighborhood U of x in X and a homeomorphism f : R ! U with f(0) = x. Set u± := f(±1) 2 X n fxg. To complete the proof, it suffices to show that X n fxg = [u+] [ [u−]; where [u±] denotes the ∼ equivalence class of u± in X n fxg (i.e. the set of points of X n fxg which can be be connected by a path in X n fxg to u±). Fix a point y 2 X n fxg. We need to show that y is in [u+] or [u−] (or both). If y 2 U, this is clear because f −1(y) 2 Rnf0g can clearly be connected to 1 or −1 by a path in Rnf0g, hence y can be connected to u+ or u− by a path in U nfxg ⊆ X nfxg. Now suppose y2 = U. Since X is connected (hence path connected because X satisfies (∗) above), we can find a path γ : [0; 1] ! X with γ(0) = x, γ(1) = y. Since γ is continuous, γ−1(x) is a closed subset of [0; 1] containing 0, but not containing 1, so it has a maximum point a 2 [0; 1). Replacing γ by γj[a; 1] if necessary, we can (and do) assume the path γ satisfies γ(0) = x, γ(1) = y, γ(t) 2 X n fxg for t 2 (0; 1]. By continuity of γ, γ−1(U) is a neighborhood of 0 in [0; 1], so there must be some t 2 (0; 1] with γ(t) =: z 2 U. Note that t cannot be 1 because γ(1) = y2 = U. Now γj[t; 1] is a path in X n fxg from z to y, so we have z ∼ y. But we know z ∼ u+ or z ∼ u− by what we proved above since z 2 U. We conclude that y ∼ u+ or y ∼ u− by transitivity of ∼. Corollary 3. Let X be a connected 1-manifold, possibly with boundary, but not necessarily Hausdorff or second countable. Then for any finite subset S ⊆ X, the 1-manifold X n S has at most jSj + 1 path components. Lemma 4. Suppose X is a connected, Hausdorff 1-manifold and f : X ! (a; b) ⊆ R is a surjective, local homeomorphism (resp. local diffeomorphism). Then f is one-to- one (i.e. bijective, since we assume it is surjective) and is hence a homeomorphism (resp. diffeomorphism).1 Proof. Fix an arbitrary point p 2 X. Consider the set A of all open subsets of X satisfying the following three properties: (1) p 2 U (2) U is path connected (3) fjU is one-to-one. 1Again, we don't assume from the outset that X is second countable, but the conclusion of the lemma of course implies that any such X must be second countable. 3 We partially order A by inclusion. Recall that a subset B ⊆ A is called a chain iff V1 ⊆ V2 or V2 ⊆ V1 for any V1;V2 2 B. Step 1. If B ⊆ A is a non-empty chain, then U := [V 2BV is in A, hence B has a supremum in the partially ordered set A. Clearly p 2 U since p 2 V for some (in fact every) V 2 B. Clearly U is path connected because it is a union of path connected sets, each of which contains p, so any point of U can be connected by a path to p. To see that fjU is one-to-one, suppose f(v1) = f(v2) for some v1; v2 2 U. We want to show v1 = v2. For i = 1; 2, we have vi 2 Vi for some Vi 2 B. Since B is a chain, we can assume, after possibly reversing the roles of v1 and v2, that V1 ⊆ V2. Therefore we must have v1 = v2 because fjV2 is one-to-one, f(v1) = f(v2), and v1; v2 2 V2. Step 2. There exists a maximal element U in A. This follows from Zorn's Lemma and Step 1. (Note that A 6= ; because f is a local homeomorphism near p.) Step 3. Any maximal element U of A is closed in X. Suppose, toward a contradiction, that there is a point x 2 U n U. Let c := f(x). Claim 3(a): c2 = f(U). Suppose, toward a contradiction, that c = f(u) for some u 2 U. Since X is first countable and x 2 U, there is a sequence u1; u2; · · · 2 U with un ! x as n ! 1. By continuity of f we have f(un) ! f(x) = c = f(u) as n ! 1. But fjU : U ! f(U) is a homeomorphism (since it is a bijective local homeomorphism) and f(un) ! f(u) as n ! 1, so we must have un ! u as n ! 1. We conclude that u = x by uniqueness of limits in the Hausdorff space X. This is a contradiction because u 2 U, while x2 = U. This proves Claim 3(a). From Claim 3(a) and the fact that U 2 A, we know that f(U) is a path-connected (the image of a path-connected space under a continuous map is path-connected), open (any local homeomorphism is an open map) subset of (a; b) with c = f(x) 2 f(U) n f(U). We conclude that f(U) must be of the form (c − , c) or of the form (c; c + ) for some > 0. The argument will be essentially the same either way, so let us assume f(U) = (c; c + ). Since f is a local homeomorphism, we can find a neighborhood V of x in X such that f takes V homeomorphically onto (c − δ; c + δ) for some 0 < δ < .
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