Classi(cid:12) ation of Surfa es via Morse Theory Abhijit Champanerkar Ajit Kumar S. Kumaresan Introdu tion In this arti le we lassify all ompa t surfa es up to homeomorphism using Morse theory. The single most important tool is the gradient like (cid:13)ow asso iated with a Morse fun tion. While it is well-known (and worked out in detail in [4℄ and [7℄) how the homotopy type hanges as when one passes a riti al point, the hange in homeomorphism type is perhaps not so well-known. Even in the books where it is done, te hni al details are inadequate and a beginning graduate student may have diÆ ulty in (cid:12)lling them. The highlights of this exposition are the two theorems (Theorem 13 and theorem 16) whi h tell us how the homeomorphism type hanges aswe passa riti al point of index 0,1 and 2. On e we prove these theorems the lassi(cid:12) ation follows easily. We have taken most of the ideas from [2℄. We have modi(cid:12)ed and supplied details for most of the proofs and hanged the style of presentation to make the exposition rigorous and lu id. Se tion 1deals with basi de(cid:12)nitions andMorseTheorem. In Se tion 2weintrodu e the gradient(cid:13)owanduseittoprovethatthedi(cid:11)eomorphism typeofamanifolddoesnot hange between the two levels whi h does not ontain a riti al point. This is the (cid:12)rst appli ation of the gradient (cid:13)ow and all the details have been painstakingly worked out. Se tion 3 in ludes modi(cid:12) ation of Morse fun tions and the proofs are an ex ellent illustration of the use of bump fun tions. Se tion 4 deals with one of the two important theorems regarding rossing of riti al levels of index 0 and 2. In Se tion 5 we prove theorems regarding passing of riti al level of index 1. For better understanding of the results proved, we have in luded lassi(cid:12) ation of losed 1-manifolds and on rete examples su h as sphere, torus and proje tive plane in Se tions 4 and 5. Se tion 6 in ludes onne ted sums. We have not dealt with thete hni alities ofunambiguity andasso iativity of onne ted sums. In thelast se tion we (cid:12)nish the lassi(cid:12) ation using the results from earlier se tions. The noteworthy point of the lassi(cid:12) ation is that the orientable as well as non orientable ases are treated simultaneously. Sin e we have aimed the arti le at fresh graduate students, we have supplied all details for most of the proofs. We also illustrate the theorems using examples and give simple appli ations of the theorem proved. We have in luded pi tures wherever ne essary to assist thegeometri understandingofideasandtheresults. Wehopethatthisarti lewillintrodu e the readers to some of the basi te hniques and ideas of di(cid:11)erential topology. 1 1 Criti al Points and Morse Lemma We assume all surfa es to be ompa t, onne ted, without boundary ( losed) unless stated otherwise. De(cid:12)nition 1. Let M be a smooth manifold and f: M ! R,a smooth fun tion on M. A point p 2 M is said to be a riti al point of f if Df(p) is singular on Tp(M). The image of p under f, that is, f(p) is alled a riti al value of f. Any real number whi h is not a riti al value is alled a regular value of f. Example 1. If f is a onstant fun tion on M then all points of M are riti al points. 2 3 Example 2. Let M = S (cid:18) R and f(x;y;z) = z. Then N = (0;0;1) and S = (0;0;(cid:0)1) are the two riti al points of f. Example 3. If M is a ompa t manifold then there exist at least two riti al points for any non onstant fun tion f on M, namely the maximum and minimum of f. Example 4. Let M = T1- the 2-dimensional torus and f be the height fun tion on M. Then there are four riti al points X1;X2;X3;X4. See Fig. 1. N X4 X3 PSfrag repla ements X2 S X1 Figure 1: Criti al Points on the torus and the sphere Let M be a smooth manifold of dimension n. Let (cid:30) : U ! M be a parameterizations, n where U (cid:18) R is an open set ontaining 0. If (cid:30)(0)= p, then we say that (cid:30) is entered at p. De(cid:12)nition 2. Let M be a smooth manifold of dimension n and f: M ! R, a smooth fun tion on M. Let ((cid:30);U) be a parameterization entered at p. De(cid:12)ne g := f Æ(cid:30). The Hessian of f with respe t to (cid:30) is a matrix de(cid:12)ned as 2  g H(cid:30)(f)= H(f Æ(cid:30)):= ( );1(cid:20) i;j (cid:20) n: xixj 2 Let M be a surfa e. Let ((cid:30);U) and ( ;V) be two parameterization entered at p. Let (x;y) and (u;v) be oordinates w.r.t. (cid:30) and respe tively. Let (cid:18)(u;v) = (x(u;v);y(u;v)) be the orresponding hange of oordinates. De(cid:12)ne h(u;v) := gÆ(cid:18)(u;v). Then h = f Æ and the partial derivatives of h are as follows: h = gxxu +gyyu; u h = gxxv +gyyv; v 2  h 2 2 2 = gxxxu +2gxyxuyu +gyyyu +gxxuu +gyyuu; u 2  h 2 2 2 = gxxxv +2gxyxvyv +gyyyv +gxxvv +gyyvv; v 2  h = gxxxuxv +gxyxuyv +gxxuv +gyxxvyu +gyyyuyv +gyyuv: uv If p is a riti al point of f then gx = 0 = gy and hen e the Hessians of f with respe t to (cid:30) and are related as follows : t H (f)= J ((cid:18))ÆH(cid:30)(f)ÆJ((cid:18)) (1) where J((cid:18)) denotes the Ja obian of (cid:18). De(cid:12)nition 3. Let f be a smooth fun tion on a surfa e M. A riti al point p of f is said to be non degenerate if H(cid:30)(f)(p) is non singular for any parameterization (cid:30) entered at p. It follows from Eq. 1 that this de(cid:12)nition is independent of the parameterization. Example 5. All riti al points in Example 1aredegenerate. All riti al pointsin Example 2 and 4 are non degenerate. De(cid:12)nition 4. The index of a non degenerate riti al point p of smooth fun tion f on a surfa e M is the dimension of the maximal subspa e of TpM on whi h H(f) is negative de(cid:12)nite. Theindexofa riti alpointisindependentoftheparameterizationfollowsfromSylvester's Law. Remark 1. Con epts of non degenera y and index also hold for any smooth manifold. Example 6. In Example 4 X1 is a riti al point of index 0, X2 and X3 are riti al points of index 1 and X4 is of index 2. Theorem 1 (Morse, 1932). Let M be a surfa e and f : M ! R, a smooth fun tion on M. Let p 2 M be a non degenerate riti al point of f. Then there exists a parameterization ((cid:30);U) entered at p and oordinates (X;Y) su h that f Æ(cid:30)(X;Y) = f(p)+gi(X;Y); 0 (cid:20) i(cid:20) 2; where i is the index of p and gi's are de(cid:12)ned as follows: 2 2 2 2 2 2 g0(X;Y)= X +Y , g1(X;Y) = X (cid:0)Y and g2(X;Y) =(cid:0)X (cid:0)Y : 3 De(cid:12)ne a map g := f Æ(cid:30): U !Rwhere (cid:30) is a parameterization entered at p. Then g is smooth and (0;0)is a non degenerate riti al point of g. It is enough to prove the following form of the above theorem. 2 Theorem 2. Let U (cid:18) R be a neighborhood of (0;0) and f: U ! Rbe a smooth fun tion. Assume that (0;0) is a non degenerate riti al point of f of index i for 0 (cid:20) i (cid:20) 2. Then there exists a di(cid:11)eomorphism (cid:30): V ! (cid:30)(V) (cid:18) U, where V is an open set ontaining (0;0) 2 in R , (cid:30)(0;0)= (0;0) and a system of oordinates (X;Y) on U su h that f Æ(cid:30)(X;Y)= f(0;0)+gi(X;Y): Proof. By Taylor series expansion f near origin is of the form: 2 2 f(x;y)(cid:0)f(0;0)= R(x;y)x +2S(x;y)xy+T(x;y)y (2) where R;S and T are smooth fun tions de(cid:12)ned as: Z 1 R(x;y) = (1(cid:0)t)fxx(tx;ty)dt; R(0;0)= r; 0 Z 1 S(x;y) = (1(cid:0)t)fxy(tx;ty)dt; S(0;0)= s; 0 Z 1 T(x;y) = (1(cid:0)t)fyy(tx;ty)dt; T(0;0)= t: 0 1 1 1 Note that R(0;0)= 2fxx(0;0);S(0;0)= 2fxy(0;0)and T(0;0)= 2fyy(0;0):In lassi al 2 notation we say that (0;0) is a non degenerate riti al point of f if rt(cid:0)s 6= 0. 2 Case 1. Let rt(cid:0)s > 0 and r > 0. Sin e R is ontinuous, there exists a neighbourhood 2 U1 (cid:18) U of (0;0) in whi h R and RT (cid:0)S remain positive. So we an write Eq. 2 as: 2 S(x;y) 2 2R(x;y)T(x;y)(cid:0)S (x;y) f(x;y)(cid:0)f(0;0)= R(x;y)[x+y ℄ +y : (3) R(x;y) R(x;y) De(cid:12)ne p S(x;y) X = X(x;y):= R(x;y)[x+y ℄; q R(x;y)) R(x;y)T(x;y)(cid:0)S2(x;y) Y = Y(x;y):= y : R(x;y) 2 De(cid:12)ne (cid:18): U1 p! R as (cid:18)(x;y) = (X(x;y);Y(x;y)). Then (cid:18)(0;0) = (0;0) and Ja obian 2 J((cid:18))(0;0) = rt(cid:0)s 6= 0. Hen e by inverse mapping theorem (cid:18) is invertible in some (cid:0)1 neighbourhood V of (0;0). De(cid:12)ne (cid:30)(X;Y) = (cid:18) (X;Y) = (x(X;Y);y(X;Y)). Then Eq. 3 be omes: 2 2 f Æ(cid:30)(X;Y) = f(0;0)+X +Y = f(0;0)+g0(X;Y): Sin e r is positive, t is also positive and hen e the index of (0;0) is 0. 2 Case 2. Let rt(cid:0)s > 0 and r < 0. Then we write Eq. 3 as: 2 S(x;y) 2 2R(x;y)T(x;y)(cid:0)S (x;y) f(x;y)(cid:0)f(0;0)= (cid:0)((cid:0)R(x;y))[x+y ℄ (cid:0)y : R(x;y) (cid:0)R(x;y) 4 Then de(cid:12)ne p S(x;y) X = X(x;y):= (cid:0)R(x;y)[x+y ℄; q R(x;y)) R(x;y)T(x;y)(cid:0)S2(x;y) Y = Y(x;y):= y : (cid:0)R(x;y) Again pro eeding as before we get 2 2 f Æ(cid:30)(X;Y) = f(0;0)(cid:0)X (cid:0)Y = f(0;0)+g2(X;Y): Sin e r is negative, t is negative and hen e index of (0;0) is 2. 2 Case 3. Let rt(cid:0)s < 0 and r > 0. Using similar arguments as before we get: 2 2 f Æ(cid:30)(X;Y) = f(0;0)+X (cid:0)Y = f(0;0)+g1(X;Y): If r is negative, then we an write Eq. 3 as: 2 S(x;y) 2 2S (x;y)(cid:0)R(x;y)T(x;y) f(x;y)(cid:0)f(0;0)= (cid:0)((cid:0)R(x;y))[x+y ℄ +y : R(x;y) (cid:0)R(x;y) Then de(cid:12)ne q S2(x;y)(cid:0)R(x;y)T(x;y) X = X(x;y):= y ; (cid:0)R(x;y) p S(x;y) Y = Y(x;y):= (cid:0)R(x;y)[x+y ℄ R(x;y) Pro eeding as above we get 2 2 f Æ(cid:30)(X;Y) = f(0;0)+X (cid:0)Y = f(0;0)+g1(X;Y): If t is non zero then the same arguments go through. Lastly suppose both of them are zero (for example the ase of hyperbola f(x;y)= xy). Consider a map (cid:30)(x;y) = (u(x;y);v(x;y)), where u = x+y and v = x(cid:0)y. Now de(cid:12)ne a u+v u(cid:0)v map g(u;v)= f( ; ). Then 2 2 1 1 1 1 1 gu = fx+ fy and guu = fxx+ fxy + fyy: 2 2 4 2 4 1 2 Hen e guu(0;0) = 2fxy(0;0) = s 6= 0. Otherwise the rt(cid:0)s = 0, a ontradi tion. Again pro eeding as above for the fun tion g we get the required result. Che k that in any ase the index of (0;0) is 1. Remark 2. The onverse of the above theorem is also true. 2 Example 7. Let M = P be the proje tive plane obtained by identifying antipodal points 2 2 2 2 of S . Consider the map f: M ! Rde(cid:12)ned by f[(x;y;z)℄ = x +2y +3z . Che k that 2 this is a smooth fun tion on P. [(1;0;0)℄;[(0;1;0)℄and [(0;0;1)℄are the only riti al points with index 0;1 and 2 respe tively. 5 Let us illustrate this example. Let U1 = f[(x;y;z)℄: x 6= 0g, U2 = f[(x;y;z)℄: y 6= 0g and U3 = f[(x;y;z)℄: z 6= 0g. De(cid:12)ne maps 2 [(1;u;v)℄ (cid:30)1: R !U1 as (cid:30)1(u;v) = p ; 2 2 1+u +v 2 [(u;1;v)℄ (cid:30)2: R !U2 as (cid:30)2(u;v) = p and 2 2 1+u +v 2 [(u;v;1)℄ (cid:30)3: R !U3 as (cid:30)3(u;v) = p : 2 2 1+u +v 2 (cid:0)1 It is easy to he k that ((cid:30)i;R ) for i = 1;2;3 are parameterizations and that (Ui;(cid:30)i ) 2 for i = 1;2;3 form a hart on P. We will (cid:12)nd riti al points and their indi es using the 2 remark 2. De(cid:12)ne a map g = f Æ(cid:30)1: R !Ras 2 2 1 2u 3v g(u;v)= f Æ(cid:30)1(u;v)= 2 2 + 2 2 + 2 2 1+u +v 1+u +v 1+u +v Its is easy to he k that gu;gv vanishes only at (u;v) = (0;0). Hen e [(1;0;0)℄ is the only riti al point of f in U1. De(cid:12)ne p u v 2 U(u;v):= p and V(u;v):= p 2 2 2 2 1+u +v 1+u +v It is easy to he k that these are oordinates in some neighbourhood of (0;0). Let (cid:18) be the 2 2 inverse of (U;V). Then g Æ(cid:18)(U;V) = 1+U +V . Hen e by remark 2 [(1;0;0)℄ is a non degenerate riti al point of index 0. Similarly one an he k for the other riti al points. Remark 3. Morse Theorem is true for any smooth manifold. Corollary 3. Non degenerate riti al points of a smooth fun tion are isolated. The fun tion gi;i= 0;1;2 de(cid:12)ned in the theorem 1 are alled the model fun tions. 2 Model neighbourhood of gi's are neighbourhoods U(s);s > 0 of (0;0) in R de(cid:12)ned as follows: p For i = 0 and i= 2, U(s) is a dis of radius s. 2 2 2 U(s)= f(X;Y)2 R :X +Y (cid:20) sg For i= 1, 2 2 2 U(s)= f(X;Y)2 R :jX (cid:0)Y j(cid:20) s;jXYj(cid:20) sg: This is an o tagon whi h is homeomorphi to a re tangle. See Fig. 2. De(cid:12)nition 5. Let f: M !Rbe a smooth fun tion on a losed surfa e M. Let p be a non degenerate riti al point of f. Let ((cid:30);U) be a parameterization entered at p su h that f in U is of the form f((cid:30)(X;Y))= f(p)+gi(X;Y) where i is the index of p. Let V (cid:18) U be a model neighbourhood for gi. Then the neigh- bourhood (cid:30)(V)ofp is alled a anoni al neighbourhood ofp and((cid:30);V)is alled a anoni al parameterization. See Fig. 3. We will assume that boundary is in luded in anoni al neighbourhoods. 6 PSfrag repla ements i= 0 i= 1 i= 2 Figure 2: Model Neighbourhoods PSfrag repla ements i= 2 i= 1 i= 0 Figure 3: Canoni al Neighbourhoods 2 Morse Fun tions and the Gradient Flow De(cid:12)nition 6. Let M be a smooth manifold. A smooth one-parameter group of di(cid:11)eomor- phisms on M is a smooth map (cid:30): R(cid:2) M ! M de(cid:12)ned as (t;x) 7! (cid:30)t(x) satisfying the following properties: (cid:15) (cid:30)0 = idM (cid:15) For ea h t 2 Rthe map (cid:30)t: M !M as t 7! (cid:30)t(x) is a di(cid:11)eomorphism. (cid:15) For r;s2 R;(cid:30)r+s(x)= (cid:30)r Æ(cid:30)s(x) for all x 2 M. Sin e the map (cid:30) is smooth, for ea h x 2 Rthe map t 7! (cid:30)t(x) is a smooth urve in M. Hen e d ((cid:30)t(x))jt=0= X(x)2TxM: dt d That is, the map X: M ! TM de(cid:12)ned as x 7! dt((cid:30)t(x)) jt=0 is a smooth ve tor (cid:12)eld on M. This ve tor (cid:12)eld is said to be generated by the smooth one-parameter group of di(cid:11)eomorphism (cid:30) on M. Conversely, any smooth one-parameter group of di(cid:11)eomorphism arises this way on any ompa t manifold. More pre isely: Theorem 4. Let M be a smooth ompa t manifold. Let X 2 (cid:31)(M) be a smooth ve tor (cid:12)eld on M. Then there exists a unique smooth one-parameter group of di(cid:11)eomorphism (cid:30): R(cid:2)M !M whi h generates X. That is, d ((cid:30)t(x))jt=0= X(x) and (cid:30)0(x)= x: dt 7 This theorem follows from a well known basi theorem in ODE. n n Theorem 5. Let U (cid:26) R be open. Let X: U !R be a smooth map. Given x0 2 U there exists an open neighbourhood (cid:10) of x0 in U, an (cid:15) > 0 and a smooth map F: ((cid:15);(cid:15))(cid:2)(cid:10) !U d su h that F(y;0)= y for all y 2 (cid:10) and dt(F(t;x)jt=0= X(x). If (Æ;V;G) is another solution satisfying onditions similar to above then F = G on ((cid:17);(cid:17))(cid:2)V where (cid:17) :=minfÆ;(cid:15)g. Proof. Refer to [1℄ or [5℄. We will use one-parameter group of di(cid:11)eomorphisms to prove many important results in this arti le. De(cid:12)nition 7. Let f be a smooth fun tion on a surfa e M. Then f is said to be a Morse fun tion if all its riti al points are non degenerate. Example 8. 1. All model fun tions are Morse fun tions. 2 2. Height fun tions on the sphere S and the torus are Morse fun tions. 3. The fun tion de(cid:12)ned in Example 7 is a Morse fun tion. There always exist Morse fun tions on any losed manifold. (We will not get into the te hni alities of this result. Interested readers an refer [7℄.) Let f be a smooth fun tion on any surfa e M. Let a and b be real numbers su h that a < b. We will use the following notations: (cid:0)1 M(a) = fx2 M :f(x)(cid:20) ag = f ((cid:0)1;a℄; 0 (cid:0)1 M (a) = fx2 M :f(x)(cid:21) ag = f [a;1); (cid:0)1 V(a) = fx2 M :f(x)= ag = f (a); (cid:0)1 W(a;b) = fx2 M :a(cid:20) f(x) (cid:20) bg = f [a;b℄: These sets are illustrated in Fig. 4. Theorem 6. LetM andN bendimensionalsmooth manifoldswith boundary. Letf: M ! N be a di(cid:11)eomorphism. Them f maps interior of M onto interior of N and the boundary of M onto the boundary of N. Theorem 7. Let M be a ompa t surfa e and f be a smooth fun tion on M. Let a and b be regular values of f. Then M(a) and W(a;b) are ompa t surfa es with V(a) as the boundary of M(a) and disjoint union of V(a) and (b) as the boundary of W(a;b). Also V(a) is a losed 1-manifold. The proofs of theorem 6 and theorem 7 are simple appli ations of Impli it Fun tion Theorem and Inverse Mapping Theorem. N Any losed manifold M an be embedded in R for some large N. Hen e for ea h N x 2 M, the tangent spa e TxM inherits an inner produ t from R . 8 b 0 M (a) W(a;b) a PSfrag repla ements M(a) 0 Figure 4: The sets M(a), M (a), W(a;b) and V(a) Theorem 8. Let M be a surfa e and f: M !R,a smooth fun tion on M. Assume that a and b are regular values of f with a< b su h that W(a;b)does not ontain any riti al point of f. Then M(b) is di(cid:11)eomorphi to M(a), V(b) is di(cid:11)eomorphi to V(a) and W(a;b) is di(cid:11)eomorphi to V(a)(cid:2)[a;b℄. Proof. The idea of the proof is to push M(a) to M(b) using the one-parameter group of di(cid:11)eomorphisms. Sin e we want V(a) to be mapped di(cid:11)eomorphi ally to V(b), the integral urves of the ve tor (cid:12)eld should be transversal to the level urves of f. In parti ular the modi(cid:12)ed gradient ve tor (cid:12)eld of f may do the job. The details are given below. (cid:0)1 Let (cid:15) > 0 be small enough su h thatf (a(cid:0)(cid:15);b+(cid:15))does not ontain any riti al points of f. Let (cid:11): M !Rbe a non negative smooth fun tion su h that (cid:11) is 1 on W(a;b) and it is 0 on the omplement of W(a(cid:0)(cid:15);b+(cid:15)). See Fig. 5. De(cid:12)ne a ve tor (cid:12)eld ( (cid:11)(x) 2grad(f(x)) x2 W(a(cid:0)(cid:15);b+(cid:15)); kgrad(f(x))k Y(x)= 0 otherwise. This is a smooth ve tor (cid:12)eld transversal to the level urves of f in W(a;b). Let (cid:30) be the one-parameter group of di(cid:11)eomorphisms asso iated with the ve tor (cid:12)eld Y. For ea h x 2 M onsider the map t 7! f((cid:30)t(x)) = (t). This is a smooth fun tion from Rto R. Its derivative d d (t)= Df((cid:30)t(x)) (cid:30)t(x)= Df((cid:30)t(x))Y((cid:30)t(x))= (cid:11)((cid:30)t(x)): dt dt d That is, if (cid:30)t(x) 2 W(a;b) then dt (t) = 1. Hen e is linear as long as (cid:30)t(x) lies in 9 b+(cid:15) b a a(cid:0)(cid:15) PSfrag repla ements Figure 5: M(b) is di(cid:11)eomorphi to M(a) W(a;b). This means that (t)= t+A for some onstant A. But (0)= f((cid:30)0(x))= f(x). Hen e (t)= t+f(x) for all x su h that (cid:30)t(x)2 W(a;b). We laim that (cid:30)b(cid:0)a maps M(a) di(cid:11)eomorphi ally onto M(b). Let us (cid:12)x x 2 M(a). By Mean-Value Theorem f((cid:30)b(cid:0)a(x))(cid:0)f((cid:30)0(x)) (b(cid:0)a)(cid:0) (0) 0 j j= j j(cid:20) sup j (t)j(cid:20) 1: b(cid:0)a b(cid:0)a t2[0;b(cid:0)a℄ Thisimplies thatjf((cid:30)b(cid:0)a(x))(cid:0)f(x)j(cid:20)b(cid:0)a. Thatis,f((cid:30)b(cid:0)a(x))(cid:20) b(cid:0)a+f(x)(cid:20) b(cid:0)a+a= b. Thus (cid:30)b(cid:0)a maps M(a) into M(b). To prove that (cid:30)b(cid:0)a is onto as a map from M(a) to M(b), it is enough to prove that if f(x) > a then f((cid:30)b(cid:0)a(x)) > b. Let us assume this and prove that (cid:30)b(cid:0)a is onto. Let x 2 M(b). Sin e (cid:30)b(cid:0)a is a di(cid:11)eomorphism of M, there exists y 2M su h that (cid:30)b(cid:0)a(y)= x. If f(y) > a, then by assumption f((cid:30)b(cid:0)a(y)) = f(x) > b. Whi h is a ontradi tion to the fa t that x 2 M(b). Hen e f(y) (cid:20) a. This proves that (cid:30)b(cid:0)a is onto as a map from M(a) to M(b). Also (cid:30)b(cid:0)a is one-one and hen e (cid:30)b(cid:0)a is a di(cid:11)eomorphism as a map from M(a) to M(b). 0 Now let us prove the laim. Let f((cid:30)b(cid:0)a(x)) (cid:20) b for f(x) > a. Sin e (t) (cid:21) 0, is in reasing. That is, (b(cid:0) a) (cid:21) (0). This implies that for all x 2 M;f((cid:30)b(cid:0)a(x)) (cid:21) f((cid:30)0(x))= f(x) > a. Thus a (cid:20) f((cid:30)t(x)) (cid:20) b for all t 2 [0;b(cid:0)a℄. Hen e f Æ(cid:30)t(x) is linear in [0;b(cid:0)a℄. Hen e f((cid:30)b(cid:0)a(x))= b(cid:0)a+f(x)> b(cid:0)a+a> b. This is a ontradi tion. (cid:30)b(cid:0)a maps V(a) di(cid:11)eomorphi ally onto V(b) by theorem 6. For the last part we de(cid:12)ne a map (cid:18): V(a)(cid:2)[a;b℄!W(a;b) as (cid:18)(x;t)= (cid:30)t(cid:0)a(x): 10