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Classification of 1- The aim of this document is to clarify the proof of the following theorem from Appendix 2 of “Differential ” by Guillemin and Pollack:

Theorem 1. Let X be a smooth, compact, non-empty, Hausdorff, connected 1- , possibly with . Then X is diffeomorphic to either a or a closed .

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 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 “,” which we now recall for the reader’s convenience. If X is a , two points x, y ∈ X are said to be 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 on X. A space X is called path connected (resp. connected) iff x ∼ y for every x, y ∈ X (resp. iff ∅ and X are the only of X which are both open and closed). Path-connected implies connected. Write [x] for the ∼ of x. Clearly [x] is path connected. Suppose X satisfies the following assumption: (∗) For every x ∈ X, there is a basis U for the neighborhoods of x in X such that every U ∈ U is path-connected. Then it is easy to see that [x] is both open and closed in X for every x ∈ X, hence X can be written as a coproduct (disjoint ) 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 of points x1, x2,... in a topological space X, and a point x ∈ X, we write “xn → x as n → ∞” 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 → ∞” can be true for at most one x ∈ 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 ∈ X, the 1-manifold X \{x} 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 \{x} is path-connected.) Since x is an interior point of X, we can find a neighborhood U of x in X and a f : R → U with f(0) = x. Set u± := f(±1) ∈ X \{x}. To complete the proof, it suffices to show that

X \{x} = [u+] ∪ [u−], where [u±] denotes the ∼ equivalence class of u± in X \{x} (i.e. the set of points of X \{x} which can be be connected by a path in X \{x} to u±).

Fix a point y ∈ X \{x}. We need to show that y is in [u+] or [u−] (or both). If y ∈ U, this is clear because f −1(y) ∈ R\{0} can clearly be connected to 1 or −1 by a path in R\{0}, hence y can be connected to u+ or u− by a path in U \{x} ⊆ X \{x}. Now suppose y∈ / 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 of [0, 1] containing 0, but not containing 1, so it has a maximum point a ∈ [0, 1). Replacing γ by γ|[a, 1] if necessary, we can (and do) assume the path γ satisfies γ(0) = x, γ(1) = y, γ(t) ∈ X \{x} for t ∈ (0, 1]. By continuity of γ, γ−1(U) is a neighborhood of 0 in [0, 1], so there must be some t ∈ (0, 1] with γ(t) =: z ∈ U. Note that t cannot be 1 because γ(1) = y∈ / U. Now γ|[t, 1] is a path in X \{x} from z to y, so we have z ∼ y. But we know z ∼ u+ or z ∼ u− by what we proved above since z ∈ 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 \ S has at most |S| + 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 ∈ X. Consider the set A of all open subsets of X satisfying the following three properties: (1) p ∈ U (2) U is path connected (3) f|U 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 ∈ B.

Step 1. If B ⊆ A is a non-empty chain, then U := ∪V ∈BV is in A, hence B has a supremum in the A. Clearly p ∈ U since p ∈ V for some (in fact every) V ∈ 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 f|U is one-to-one, suppose f(v1) = f(v2) for some v1, v2 ∈ U. We want to show v1 = v2. For i = 1, 2, we have vi ∈ Vi for some Vi ∈ 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 f|V2 is one-to-one, f(v1) = f(v2), and v1, v2 ∈ 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 ∈ U \ U. Let c := f(x). Claim 3(a): c∈ / f(U). Suppose, toward a contradiction, that c = f(u) for some u ∈ U. Since X is first countable and x ∈ U, there is a sequence u1, u2, · · · ∈ U with un → x as n → ∞. By continuity of f we have f(un) → f(x) = c = f(u) as n → ∞. But f|U : U → f(U) is a homeomorphism (since it is a bijective local homeomorphism) and f(un) → f(u) as n → ∞, so we must have un → u as n → ∞. We conclude that u = x by uniqueness of limits in the Hausdorff space X. This is a contradiction because u ∈ U, while x∈ / U. This proves Claim 3(a). From Claim 3(a) and the fact that U ∈ A, we know that f(U) is a path-connected (the 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) ∈ f(U) \ 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 < . Claim 3(b): f(U ∩ V ) = (c, c + δ). We know f(V ∩ U) is an open subset of f(V ) ∩ f(U) = (c, c + δ) containing c = f(x) in its , so f(V ∩ U) must contain (c, c + d) for some d ∈ (0, δ]. Let e be the supremum of all such d. Then we have (c, c + e) ⊆ f(U ∩ V ) ⊆ (c, c + δ) 4 but c + e∈ / f(U ∩ V ). It remains to show that e = δ, so, suppose, toward a contradiction, that e < δ. Then we can choose a sequence y1, y2, · · · ∈ U ∩ V with f(yn) → c + e as n → ∞. Since f|V : V → f(V )(1) f|U : U → f(U) are and c + e ∈ f(U) ∩ f(V ), we have yn → u as n → ∞ for some u ∈ U and yn → v as n → ∞ for some v ∈ V . Since f is continuous we must have f(u) = f(v) = c + e. Now, as usual, we use the fact that limits are unique in the Hausdorff space X to conclude that u = v ∈ U ∩ V and f(u = v) = c + e, contradicting c + e∈ / f(U ∩ V ). This proves Claim 3(b). Finally we claim that f|(U ∪ V ) is one-to-one. Since U ∪ V is path-connected (because U and V are path-connected and U ∩ V is non-empty) and contains p (since p ∈ U), this will contradict the assumption that U ∈ A is maximal—this is the contradiction we’ve been seeking since the beginning of Step 3 and will therefore complete Step 3. Since f|U and f|V are one-to-one, we just need to verify that f(U \ V ) and f(V \ U) are disjoint. For this, we make the following calculations using Claim 3(b) and the fact that the maps (1) are homeomorphisms: f(U \ V ) = f(U \ (U ∩ V )) = f(U) \ f(U ∩ V ) = (c, c + ) \ (c, c + δ) = (c + δ, c + ) f(V \ U) = f(V \ (U ∩ V )) = f(V ) \ f(U ∩ V ) = (c − δ, c + δ) \ (c, c + δ) = (c − δ, c).

Step 4. X ∈ A and hence the proof is complete. Indeed, by Step 2, A contains some maximal element U. This U is open and non-empty by definition of A and closed by Step 3, so it must be equal to X by the assumption that X is path-connected.  Remark 5. The assumption that X is Hausdorff in Lemma 4 cannot be removed. It is helpful to keep in mind the following (standard) “doubling construction” to understand why one must make use of the “Hausdorff” assumption: Consider an open interval (a, b) ∈ R and a point x ∈ (a, b). Let a X := ((a, b) \{x}) {x1, x2}. Topologize X as follows: A basic open neighborhood of any point y ∈ ((a, b)\{x}) ⊆ X is a neighborhood of y in (a, b) \{x} in the usual sense. For i = 1 or i = 2, a ` basic open neighborhood of xi in X is a subset of X of the form (V \{x}) {xi}, 5

where V is a neighborhood of x in (a, b) in the usual sense. For i = 1 or i = 2, {xi} is closed in X and the map fi : X \{xi} → (a, b) which is the identity on (a, b) \{x} and which takes xj (j 6= i) to x is a homeomorphism. (In particular, this shows that X is a 1-manifold, which is easily seen to be path-connected by making use of f1 and f2.) The maps f1 and f2 clearly agree on

(X \{x1}) ∩ (X \{x2}) = (a, b) \{x}, so they glue to a map f : X → (a, b) which is, by construction, a local homeomor- −1 phism, but which is not one-to-one since f (x) = {x1, x2}. One can even give X a smooth structure making f a local diffeomorphism by simply declaring f1 and f2 to form a smooth atlas on X. This makes sense because the “transition func- tion” between the two charts f1, f2 is just the identity map of (a, b) \{x}, which is undeniably a diffeomorphism. Lemma 6. (Smoothing Lemma) Let a < b < c be real numbers. Suppose f : [a, c] → R is a such that f 0 exists, is continuous, and is strictly positive on [a, c]\{b}. Assume also that f is smooth on [a, c]\{b}. Then there exists a smooth function g :[a, c] → R which agrees with f on a neighborhood of {a, c} and has strictly positive derivative on the entire interval [a, c].

Proof. The idea is to “flatten out” f near b so that f becomes a strictly increasing linear function near b. It is a little tricky to do this without messing anything else up. By scaling by a large factor and rescaling at the end, if necessary, we can assume a < a + 2 < b − 2 < b < b + 2 < c − 2 < c.

In this case, we can easily find a smooth “bump” function ρ :[a, c] → [0, 1] ⊆ R with support contained in [a + 1, c − 1], which is identically 1 on a neighborhood of R c b and which satisfies a ρ(x)dx = 1. (Just start with a very steep bump function supported on a very small neighborhood of b which is identically 1 near b and has integral less than one, then enlarge the “plateau” near c until the integral becomes 1.) Define g by Z x g(x) := f(a) + kρ(t) + f 0(t)(1 − ρ(t))dt, a where the constant k is defined by Z c k := f(c) − f(a) − f 0(t)(1 − ρ(t))dt a Z c = f 0(t)ρ(t)dt. a (Note that we can certainly apply the Fundamental Theorem of Calculus there because f 0 is continuous by assumption. We shall do this several times.) The last 6

formula for k makes it clear that k > 0. For x ∈ [a, a + 1) we have ρ(t) = 0 for all t ∈ [a, x], so we have Z x g(x) = f(a) + k · 0 + f 0(t)(1 − 0)dt a = f(a) + f(x) − f(a) = f(x). Similarly, for x ∈ (c − 1, c] we have ρ(t) = 0 for all t ∈ [x, c], so we have Z x Z c Z c g(x) = f(a) + k ρ(t)dt + f 0(t)(1 − ρ(t))dt − f 0(t)(1 − ρ(t))dt a a x Z c Z c = f(a) + k ρ(t)dt + f(c) − f(a) − k − f 0(t)dt a x = f(a) + k + f(c) − f(a) − k − f(c) + f(x) = f(x). We also have g0(x) = kρ(x) + f 0(x)(1 − ρ(x)). Since k > 0, ρ(x) ≥ 0, f 0(x) ≥ 0, and ρ(x) ≤ 1, this is clearly positive whenever ρ(x) > 0 and when ρ(x) = 0 it is positive whenever f 0(x) > 0, which is true for all x 6= b by assumption. This formula for g0(x) also makes it clear that g0 (and hence also g) is smooth away from b. Now, near b, we have ρ(x) = 1 so, near b, we simply have g0(x) = k, which makes it clear that g is smooth near b with positive derivative near b.  Remark 7. The hypotheses on f in the Smoothing Lemma ensure that f is strictly increasing, hence f yields a homeomorphism f :[a, c] → [f(a), f(c)] which restricts to a diffeomorphism f :[a, c] \{b} → [f(a), f(c)] \{f(b)}. The function g obtained from the Smoothing Lemma has the same image as f and is a diffeomorphism onto that image since it is bijective (like f, g is strictly increasing) and has nowhere zero derivative. Remark 8. When we apply the Smoothing Lemma in the proof of Theorem 1, we will have the slightly stronger hypothesis on f that f|[a, b] is smooth and f|[b, c] is smooth.

Proof of Theorem 1 Step 1. We can find a smooth function f : X → R with finitely many critical points. We fix one such function for the rest of the proof. One way to see this is to use the Whitney Theorem to embed X into some RN , then use Sard’s Theorem to show that, for almost every a ∈ Rn, the map RN → R given by x 7→ aT x restricts to a Morse function on X. See §1.7 in the Guillemin and Pollack book. The critical points of a Morse function are isolated and X is compact, so any Morse function is as desired. 7

Step 2. Let S be the finite subset of X consisting of all critical points for f and all boundary points of X. (The latter are isolated and X is compact, so there are finitely many.) Then the 1-manifold X \ S has finitely many path components and for each path component L of X \ S, f|L is a diffeomorphism from L onto some open interval (a, b) ⊆ R. Since S is finite and X is connected, X \ S has finitely many path components by Corollary 3. The map f|L : L → R is a local diffeomorphism since it is a smooth map of 1-manifolds (without boundary) with nowhere zero derivative (because f has no critical points in L). Since L is connected, f(L) is a connected, open (since f|L is open, being a local diffeomorphism) subset of R contained in the compact subset f(X), so f(L) = (a, b) for some a < b, and f|L : L → (a, b) is a diffeomorphism by Lemma 4. Step 3. For any t ∈ R, f −1(t) is finite. By Step 2, f −1(t) ∩ L has at most one point for each of the finitely many path components L of X \ S. Since S is finite, we’re done. Step 4. Let L be a path component of X \ S, so that we have a diffeomorphism f|L : L → (a, b) by Step 2. Since the closure L of L in X is compact (since X is compact) and contains L as a dense subspace, we must have f(L) = [a, b]. In fact, the smooth function f|L : L → [a, b] is a homeomorphism. Since L is compact and f|L : L → [a, b] is continuous and surjective, it is enough to prove that f is injective. (Any bijective continuous map from a compact space to a Hausdorff space is a homemorphism.) Since we know from Step 2 that f|L is injective, injectivity of f|L will follow from Claims 3(a) and 3(b) below. Claim 4(a): There is no x ∈ L \ L such that f(x) ∈ (a, b) = f(L). Suppose there is such an x, so we have f(x) = f(l) for some l ∈ L. Choose a sequence l1, l2, · · · ∈ L with ln → x as n → ∞. Then f(ln) → f(x) = f(l) as n → ∞ by continuity of f. But f|L : L → (a, b) is a homeomorphism by Step 2, so we must have ln → l as n → ∞. We conclude x = l by uniqueness of limits in the Hausdorff space X. But this is absurd because l ∈ L, while x∈ / L.

Claim 4(b): If x1, x2 ∈ L \ L and f(x1) = f(x2), then x1 = x2.

This is the hardest part of the proof. By Claim 4(a) we know f(x1) = f(x2) = a or f(x1) = f(x2) = b. The proof will be similar in either case, so let us assume f(x1) = f(x2) = b. Suppose, towards a contradiction that x1 6= x2. By Step 3, f −1(b) is a finite subset of X, so we can write −1 f (b) = {x1, x2, . . . , xk} for distinct points x1, . . . , xk ∈ X. Since X is Hausdorff we can choose pairwise disjoint open subsets U1,...,Uk of X with xi ∈ Ui for i = 1, . . . , k. Since X is a manifold, we can assume, after possibly shrinking the Ui, that each Ui is path connected (because X satisfies (*)). We can also assume, after possibly shrinking further, that a is not in any of the f(Ui). 8

Since x1 ∈ L, there must be a point l1 ∈ L ∩ U1. Since U1 is path-connected, we can find a continuous map γ : [0, 1] → U1 with γ(0) = l1, γ(1) = x1. Let t be the −1 minimum element of the closed subset γ (x1) ⊆ [0, 1]. Replacing γ by γ|[0, t] if necessary, we can assume

(**) γ(c) 6= x1 for c ∈ [0, 1).

Since γ is continuous and L ∩ U1 is a neighborhood of l1 = γ(0) in X, there is some d ∈ (0, 1] such that γ([0, d)) ⊆ L ∩ U1. Let δ ∈ (0, 1] be the supremum of all such d. Then γ([0, δ)) ⊆ L ∩ U1 but y := γ(δ) ∈ U1 \ L.

Choose a sequence d1, d2, · · · ∈ [0, δ) with dn → δ as n → ∞. Since γ is continuous, γ(dn) → γ(δ) = y as n → ∞, so we have y ∈ (L ∩ U1) \ L. We must have f(y) = a or f(y) = b by Claim 4(a) because y ∈ L \ L, so, in fact, we must have f(y) = b since y ∈ U1 and a∈ / f(U1) by our choice of U1. Therefore y is one of the xi—but it is in U1, which is disjoint from the other Ui, so actually we must have y = x1. Therefore we must have δ = 1 by (**) since x1 = y = γ(δ).

Now we have γ([0, 1)) ⊆ L ∩ U1, so f(γ([0, 1))) ⊆ f(L ∩ U1). But f(γ([0, 1))) is a connected subset of f(L) = (a, b) containing f(l1) whose closure in R contains f(γ(1)) = f(x1) = b, thus we conclude that f(L∩U1) contains [f(l1), b). Now repeat the same argument, replacing x1 with x2, to see that f(L ∩ U2) contains [f(l2), b) for any l2 ∈ L ∩ U2. But then f(L ∩ U1) and f(L ∩ U2) are not disjoint, so, since f|L : L → f(L) = (a, b) is one-to-one by Step 2, the sets L ∩ U1 and L ∩ U2 cannot be disjoint, contradicting our choice of the Ui. This completes Step 4.

Step 5. Call a sequence L1,L2,...,Ln of distinct path components of X \ S a chain iff Li ∩ Li+1 6= ∅ for i = 1, . . . , n − 1. Since X \ S has only finitely many path components (by Step 2), there must exist a maximal chain, so fix such a maximal chain L1,L2,...,Ln. By Step 4 and the definition of a chain, we can find p0, p1, . . . , pn ∈ X such that Li \ Li = {pi−1, pi} for. The points pi must be distinct, except that we may have p0 = pn. Then every path component of X \ S must be one of the Li and hence X = L1 ∪ · · · ∪ Ln.

To see that the pi are distinct (except for the possibility that p0 might equal pn), it is enough to note that no point of S can be in L for more than two path components L of X \ S, because such a point would have a neighborhood in X homeomorphic to a wedge of more than two copies of [0, 1) (use Step 4 to see this), but such a wedge is not a 1-manifold at the wedge point.

Suppose L is a path component of X\S not among the Li. By Step 4, L\L = {p, q} for distinct p, q ∈ X. Neither p nor q can be p0 or pn by maximality of our chain L1,...,Ln. Neither p nor q can be one of the “intermediate points” p1, . . . , pn−1 by 0 the fact noted above. This means L is disjoint from X := L1 ∪ · · · ∪ Ln. Since this is true for any such L, the union of the closures of all of the finitely many such L would then form a closed subset of X complementary to the closed subset X0—this can’t happen because X is connected. 9

Step 6. Let (ai, bi) = f(Li) (cf. Steps 2,5). Choose any diffeomorphism τi :[ai, bi] → [i − 1, i]. Then the maps fi := τi(f|Li): Li → [i − 1, i] are smooth maps, as well as homeomorphisms, and restrict to diffeomorphisms fi : Li → (i − 1, i). Assume p0 6= pn. Then the fi agree on the overlaps of the Li, so they glue to yield a homeomorphism F : X → [0, n] ⊆ R such that F |Li = fi : Li → (i − 1, i) is a diffeomorphism. The map F is smooth when restricted to any Li (in particular near p0 and pn) and is continuously differentiable everywhere (with derivative zero at the intermediate points p1, . . . , pn−1, which were critical points for the original f, and has non-zero derivative everywhere else), so we can find a diffeomorphism G : X → [0, n] which agrees with F away from an arbitrarily small neighborhood of p1, . . . , pn−1. This follow from the Smoothing Lemma by working on small neighborhoods of p1, . . . , pn−1.

Step 7. Assume p0 = pn. Then the definition of F in Step 6 makes sense, except that F (p0) = 0, while F (pn) = n, so (cos(2πF/n), sin(2πF/n)) defines a function H : X → S1 with properties similar to those of F as in Step 6. One can modify H near the troublesome points p0 = pn, p1, . . . , pn−1 to find a diffeomorphism K : X → S1. This completes the proof of Theorem 1. Remark 9. In the Guillemin and Pollack textbook, the authors make the point of choosing τi in Step 6 above to be the unique affine linear diffeomorphism τi : [ai, bi] → [i − 1, i]. (Here “affine linear” means the map is of the form t 7→ ax + b for some a, b ∈ R.) Aside from the fact that this is a fairly natural way to choose the τi, it offers no actual tangible advantages that I am aware of, and hence it might actually be somewhat misleading because the reader might (wrongly) assume that the insistence on choosing τi in this manner offers some advantage over a completely random choice. Remark 10. To understand the necessity for the use of the Smoothing Lemma, it might be helpful to run through the proof of Theorem 1 in a simple example. Suppose X = [−1, 2] and the function f chosen in Step 1 is the (Morse) function f(x) = x2. The only critical point of f is 0 ∈ X and the S of Step 2 is S = {−1, 0, 2}. The assertions proved in Steps 3 and 4 are obvious in this example and one can find a maximal chain as in Step 5 by taking L1 = (−1, 0), L2 = (0, 2), so that p0 = −1, p1 = 0, p2 = 2. The assertion that L1 = [−1, 0] and L2 = [0, 2] cover X in Step 5 is again obvious. If we take τ1 and τ2 to be the affine linear diffeomorphisms as 2 suggested by Guillemin and Pollack, then f1 : L1 → [0, 1] is given by f1(x) = 1 − x 2 and f2 : L2 → [1, 2] is given by f2(x) = (1/4)x + 1. Our f1 and f2 of course glue to yield a homeomorphism F : X → [0, 2], which is indeed a diffeomorphism away from p1 = 0 ∈ X, but notice that it is not even twice differentiable at p1 = 0. The Smoothing Lemma is needed to fix this. Remark 11. One major difference between topology and differential topology is the following: If {Zi} is a (locally finite) covering of a topological space X by closed 10

subspaces Zi and fi : Zi → Y are continuous functions which agree on the “overlaps” Zi ∩ Zj, then the fi glue to a continuous function f : X → Y . If “continuous” is replaced by “smooth” everywhere (and “topological space” is replaced by “subset of RN ,” so that “smooth” has a meaning) then this is utterly false. Notice, however, that in the above proof we still basically use this idea, then circumvent the difficulties that arise by use of the Smoothing Lemma. Remark 12. The basic strategy of the proof of Theorem 1 is to cut X up into “standard pieces” (the Li) and then think about how these pieces can possibly be glued together to form a manifold. This is a fairly typical strategy for this sort of “classification” problem. For example, one can classify (compact, connected) 2- manifolds (without boundary, up to homeomorphism, say) in a very similar manner by thinking about how one can possibly glue triangles (“2-simplices”) together to form a 2-manifold. (The requirement that the glued up object be a manifold places significant restrictions on the possible gluings, just as this requirement placed re- ∼ strictions on the way the Li = [0, 1] (“1-simplices”) could be glued in Step 5 of the above proof.) I recommend the little Dover paperback book “Combinatorial Topology” by P. S. Alexandrov for a nice discussion of these points—I learned this material from Mark Hovey when I was an undergraduate at Wesleyan.