Given the unit sphere Sn, we prove the following theorem and several extensions: For any continuous function f : Sn - Sn, if f has no fixed point in Sn, then f must be surjective.

Sun, N.

Yang, Z.F.

English

Given the unit sphere Sn, we prove the following theorem and several extensions: For any continuous function f : Sn - Sn, if f has no fixed point in Sn, then f must be surjective

Tilburg University Repository

    Tilburg UniversitySurjective Function TheoremsSun, N.; Yang, Z.F.Publication date:1999Link to publication in Tilburg University Research PortalCitation for published version (APA):Sun, N., & Yang, Z. F. (1999). Surjective Function Theorems. (CentER Discussion Paper; Vol. 1999-75).Microeconomics.General rightsCopyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright ownersand it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights.            • Users may download and print one copy of any publication from the public portal for the purpose of private study or research.            • You may not further distribute the material or use it for any profit-making activity or commercial gain            • You may freely distribute the URL identifying the publication in the public portalTake down policyIf you believe that this document breaches copyright please contact us providing details, and we will remove access to the work immediatelyand investigate your claim.Download date: 12. May. 20211Surjective Function Theorems1Ning Sun 2 and Zaifu Yang3Abstract: Given the unit sphere Sn, we prove the following theorem andseveral extensions: For any continuous function f : Sn 7! Sn, if f has no xedpoint in Sn, then f must be surjective. Furthermore, there exists x 2 Snsuch that f(x) =  f( x).Keywords: Surjective function, xed point, antixed point, antipodal point,sphere, manifold.1 Main ResultsA function f : D 7! I is said to be surjective (or f is said to map D onto I) if everyelement of I is the image of some element of D under the function f , i.e., f(D) = I . fis said to have a xed point (an antipodal point) in D if there exists x 2 D such thatf(x) = x (f(x) =  f( x)). In this Note we present several surjective function theoremswhich are not only interesting on their own but also fundamental. Our approach isa topological one. Now we introduce some notation: Let n  2 denote any integernumber, IRn the n-dimensional Euclidean space, and x  y = Pi xiyi the inner productof vectors x and y. We write x 2 IRn+1 by x = (x0; x1;    ; xn) or x = (x0; x1;    ; xn)>.Furthermore, dene Bn+1 = fx 2 IRn+1 j x  x  1g (i.e. the (n + 1)-dimensional unitball), Bn = fx 2 Bn+1 j x0 = 0g, Sn = fx 2 IRn+1 j x x = 1g (i.e. the n-dimensional unitsphere), Sn 1 = fx 2 Sn j x0 = 0g, e0 = (1; 0;    ; 0) 2 IRn+1, and 0 = (0;    ; 0) 2 IRn+1.1The second author's research is partly supported by Grants-in-Aid of the Ministry of Education,Science, Sports and Culture of Japan, and partly by a grant of the Netherlands Organization for ScienticResearch (NWO). He also thanks CentER for Economic Research of Tilburg University for its hospitality.2N. Sun, Faculty of System Science and Technology, Akita Prefectural University, Honjo City,Akita 015-0055, Japan. E-mail: sun@akita-pu.ac.jp3Z. Yang, Faculty of Business Administration, Yokohama National University, Yokohama 240-8501,Japan. E-mail: zyang@business.ynu.ac.jp2Theorem 1.1 For any continuous function f : Sn 7! Sn, if f has no xed point in Sn,then f must be surjective. Furthermore, there exists x 2 Sn such that f(x) =  f( x).Proof: Suppose to the contrary that f is not surjective. Then there would exist somey 2 Sn which is not in f(Sn). Without loss of generality, we may assume that y = e0.Thus, it follows from the continuity of f on the compact set Sn that there exists apositive  such that f0(x)  1    for all x 2 Sn. Let C = fx 2 Sn j x0  1   g andCb = fx 2 Sn j x0 = 1  g. We have that f(Sn)  C. Dene the function g : C 7! Bnbyg(x) =12  s1 + x01  x0 (0; x1;    ; xn)>:Its inverse h = g 1 : Bn 7! C is given by for x 6= 0hi(x) =8><>:(2  )px  x  1; if i = 0;(2  )xir2 (2 )pxx(2 )pxx ; if i = 1;    ; n;with h(0) =  e0. It is easy to see that h(x) converges to  e0 as x goes to 0. Thus, bothg and h are continuous functions. It may be dicult to gure out how g is constructed.Geometrically, we can visualize the idea in the sphere S2 being imagined as the surfaceof the earth. Given any point P on the arctic circle Cb, there is a unique longitude linepassing through P which links both the north pole e0 and the south pole  e0. Thislongitude intersects the equator line Sn 1 uniquely at one point, say Q. Clearly, thesection of the longitude line between P and the south pole is homeomorphic to the straightline between the core 0 and Q. Function g maps P to Q and the south pole to the core.Consequently, we obtain a continuous function g  f h mapping from the convex andcompact set Bn into itself. By using Brouwer's xed point theorem, we know there existsz 2 Bn such that z = g  f h(z). Setting x = h(z), we obtain x = f(x) 2 C. Thiscontradicts the hypothesis that f has no xed point in Sn.Now we prove the last part. Since f has no xed point, it follows from Corollary 4(c) of Whittlesey [3] or Milnor [2] that f must have an antipodal point. We complete theproof. 23In term of equation theory, the above theorem states that for each y 2 Sn, the equationf(x) = y has a solution.Corollary 1.2 For any continuous function f : Sn 7! Sn, if x  f(x)  0 for everyx 2 Sn, then f must be surjective.Proof: Suppose to the contrary that f has a xed point x 2 Sn. Then 0 < x  x =x  f(x)  0 which is impossible. 2The above corollary is not so simple as it might appear. In fact we will show that itis at least as powerful as the classic Brouwer's xed point theorem is. We prove Brouwertheorem via Corollary 1.2: If f : Bn 7! Bn is a continuous function, then there existsx 2 Bn such that f(x) = x. Suppose to the contrary that Brouwer theorem is false.Then it holds f(x)  x 6= 0 for every x 2 Bn. Let g(x) = f(x)  x. Thenx  g(x) = x  (f(x)  x) = x  f(x)  x  x = x  f(x)  1 < 0for all x 2 Bn \ Sn = Sn 1. We dene the function h : Sn 7! Sn as follows: For x 2 Sn,let y(x) = (0; x1;    ; xn)>. Obviously, y(x) 2 Bn . If y(x)  g(y(x))  0, deneh(x) =g(y(x))pg(y(x))  g(y(x)):If y(x)  g(y(x)) > 0, it is easy to see that Pni=1 x2i < 1 and x0 6= 0. Denez0 =  y(x)  g(y(x))x0; z = (z0; 0;    ; 0)> 2 IRn+1;andh(x) =g(y(x)) + zp(g(y(x)) + z)  (g(y(x))+ z) :Since y(x)  g(y(x)) is a continuous function in x on a compact set, then there exists some > 0 such thaty(x)  g(y(x))  for all x 2 Sn 1. Furthermore, there exists a positive  such that y(x) g(y(x)) 0 impliesj x0 j . Now it is readily veried that h is a continuous function and x h(x)  0 for all4x 2 Sn. So all conditions of Corollary 1.2 are met. Then h must be surjective. But it isimpossible since there does not exist any x 2 Sn such that h(x) = e0. 2A function f : D 7! I is said to have an antixed point in D if there exists x 2 Dsuch that f(x) =  x.Corollary 1.3 For any continuous function f : Sn 7! Sn, if f has no antixedpoint in Sn, then f must be surjective. Furthermore, there exists x 2 Sn such thatf(x) =  f( x).Proof: Let g(x) =  f(x) for all x 2 Sn. Clearly, g is continuous on Sn, g(Sn)  Sn,and g has no xed point in Sn. It follows from Theorem 1.1 that g is surjective and henceso is f . 2Corollary 1.4 For any continuous function f : Sn 7! Sn, if x  f(x)  0 for everyx 2 Sn, then f must be surjective.Proof: Suppose to the contrary that f has an antixed point y 2 Sn. Then 0 >  y  y =f(y)  y  0 which is impossible. By Corollary 1.3 f is surjective. 2The coming result is Theorem 3 of Whittlesey [3]. Here we give an alternative proofwhich is conceptually simpler than Whittlesey's.Theorem 1.5 For any continuous function f : Sn 7! Sn, if f(x) =  f( x) for everyx 2 Sn, then f must be surjective.Proof: Suppose to the contrary that f is not surjective. Then there would exist somey 2 Sn which is not in f(Sn). Without loss of generality, we may assume that y = e0.By the assumption we know that there does not exist any x 2 Sn such that f(x) =  e0,either. Now we dene the function g : Sn 7! Bn bygi(x) =8>><>:0; if i = 0;fi(x)qPnj=1f2j (x); if i = 1;    ; n:5It follows from the assumption that g(x) =  g( x) for every x 2 Sn. Clearly, g is acontinuous function. This is impossible according to Borsuk-Ulam theorem (see [1, 4])which says that if l : Sn 7! Bn is a continuous function, then there exists x 2 Sn suchthat l(x) = l( x). 2It will be shown that Theorem 1.5 is actually equivalent to Borsuk-Ulam theorem.Suppose that the latter theorem is false. Let g(x) = f(x)   f( x). Then g(x) 6= 0 andg(x) =  g( x) for all x 2 Sn. Dene the function h : Sn 7! Sn byh(x) =g(x)pg(x)  g(x):Clearly, all conditions of Theorem 1.5 are satised. Then h must be surjective. But it isimpossible since h0(x) = 0 for all x 2 Sn. 2Now we proceed to prove one more result. DeneSn+ = fx 2 Sn j x0 > 0 g and Sn  = fx 2 Sn j x0 < 0 g:Theorem 1.6 For any continuous function f : Sn 7! Sn, if it satises thatf0(0; x1;    ; xn) = 0fi(0; x1;    ; xn) =  fi(0; x1;    ; xn); i = 1;    ; nfor all (0; x1;    ; xn) 2 Sn 1, then either Sn+  f(Sn) or Sn   f(Sn) or both.Proof: Suppose to the contrary that there exist two points y = (y0; y1;    ; yn) 2 Sn+ and~y = (~y0; ~y1;    ; ~yn) 2 Sn  neither of which belongs to f(Sn). We dene h : Bn 7! Sn byh(0; x1;    ; xn) = ( p1  x  x; x1;    ; xn);h 1 : Sn 7! Bn byh 1(x0; x1;    ; xn) = (0; x1;    ; xn);g : Bn n fh 1(y)g 7! Sn 1 with g(x) equal to the intersection point of the straight linegoing through x and h 1(y) on Sn 1;6bg : Bn n fh 1(~y)g 7! Sn 1 with bg(x) equal to the intersection point of the straight linegoing through x and h 1(~y) on Sn 1.It is easy to verify that h, h 1, g, and bg are continuous functions. Now we constructthe function F : Bn 7! Sn 1 byF (0; x1;    ; xn) =8>><>>>:g  h 1  f  h(0; x1;    ; xn); if f0  h(0; x1;    ; xn) > 0;bg  h 1  f  h(0; x1;    ; xn); if f0  h(0; x1;    ; xn) < 0;f  h(0; x1;    ; xn); if f0  h(0; x1;    ; xn) = 0:Note that h 1(x) = g(x) = bg(x) = x for all x 2 Sn 1. Then for any x 2 Bn withf0  h(x) = 0, it holds thatg  h 1  f  h(x) = bg  h 1  f  h(x) = f  h(x):From the continuity of h, h 1, g, and bg, F is a continuous function and furthermore itsatises thatF (0; x1;    ; xn) = f  h(0; x1;    ; xn) = f(0; x1;    ; xn) = f(0; x1;    ; xn) =  F (0; x1;    ; xn)for all (0; x1;    ; xn) 2 Sn 1. This contradicts an equivalent form of Borsuk-Ulam the-orem which says that there does not exist any continuous function l : Bn 7! Sn 1 suchthat l(x) =  l( x) for every x 2 Sn 1. 2References[1] V. Istratescu (1981), Fixed Point Theory: An Introduction, Reidel Publishing Company, Dordrecht.[2] J. Milnor (1965), Topology From The Dierentiable Viewpoint, The University Press of Virginia, Char-lottessville.[3] E.F. Whittlesey (1963), \Fixed points and antipodal points", American Mathematical Monthly 70, 807-821.[4] Z. Yang (1999), Computing Equilibria and Fixed Points, Kluwer Academic Publishers, Boston.

Surjective Function Theorems

    Tilburg UniversitySurjective Function TheoremsSun, N.; Yang, Z.F.Publication date:1999Link to publication in Tilburg University Research PortalCitation for published version (APA):Sun, N., & Yang, Z. F. (1999). Surjective Function Theorems. (CentER Discussion Paper; Vol. 1999-75).Microeconomics.General rightsCopyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright ownersand it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights.            • Users may download and print one copy of any publication from the public portal for the purpose of private study or research.            • You may not further distribute the material or use it for any profit-making activity or commercial gain            • You may freely distribute the URL identifying the publication in the public portalTake down policyIf you believe that this document breaches copyright please contact us providing details, and we will remove access to the work immediatelyand investigate your claim.Download date: 26. Sep. 20231Surjective Function Theorems1Ning Sun 2 and Zaifu Yang3Abstract: Given the unit sphere Sn, we prove the following theorem andseveral extensions: For any continuous function f : Sn 7! Sn, if f has no xedpoint in Sn, then f must be surjective. Furthermore, there exists x 2 Snsuch that f(x) =  f( x).Keywords: Surjective function, xed point, antixed point, antipodal point,sphere, manifold.1 Main ResultsA function f : D 7! I is said to be surjective (or f is said to map D onto I) if everyelement of I is the image of some element of D under the function f , i.e., f(D) = I . fis said to have a xed point (an antipodal point) in D if there exists x 2 D such thatf(x) = x (f(x) =  f( x)). In this Note we present several surjective function theoremswhich are not only interesting on their own but also fundamental. Our approach isa topological one. Now we introduce some notation: Let n  2 denote any integernumber, IRn the n-dimensional Euclidean space, and x  y = Pi xiyi the inner productof vectors x and y. We write x 2 IRn+1 by x = (x0; x1;    ; xn) or x = (x0; x1;    ; xn)>.Furthermore, dene Bn+1 = fx 2 IRn+1 j x  x  1g (i.e. the (n + 1)-dimensional unitball), Bn = fx 2 Bn+1 j x0 = 0g, Sn = fx 2 IRn+1 j x x = 1g (i.e. the n-dimensional unitsphere), Sn 1 = fx 2 Sn j x0 = 0g, e0 = (1; 0;    ; 0) 2 IRn+1, and 0 = (0;    ; 0) 2 IRn+1.1The second author's research is partly supported by Grants-in-Aid of the Ministry of Education,Science, Sports and Culture of Japan, and partly by a grant of the Netherlands Organization for ScienticResearch (NWO). He also thanks CentER for Economic Research of Tilburg University for its hospitality.2N. Sun, Faculty of System Science and Technology, Akita Prefectural University, Honjo City,Akita 015-0055, Japan. E-mail: sun@akita-pu.ac.jp3Z. Yang, Faculty of Business Administration, Yokohama National University, Yokohama 240-8501,Japan. E-mail: zyang@business.ynu.ac.jp2Theorem 1.1 For any continuous function f : Sn 7! Sn, if f has no xed point in Sn,then f must be surjective. Furthermore, there exists x 2 Sn such that f(x) =  f( x).Proof: Suppose to the contrary that f is not surjective. Then there would exist somey 2 Sn which is not in f(Sn). Without loss of generality, we may assume that y = e0.Thus, it follows from the continuity of f on the compact set Sn that there exists apositive  such that f0(x)  1    for all x 2 Sn. Let C = fx 2 Sn j x0  1   g andCb = fx 2 Sn j x0 = 1  g. We have that f(Sn)  C. Dene the function g : C 7! Bnbyg(x) =12  s1 + x01  x0 (0; x1;    ; xn)>:Its inverse h = g 1 : Bn 7! C is given by for x 6= 0hi(x) =8><>:(2  )px  x  1; if i = 0;(2  )xir2 (2 )pxx(2 )pxx ; if i = 1;    ; n;with h(0) =  e0. It is easy to see that h(x) converges to  e0 as x goes to 0. Thus, bothg and h are continuous functions. It may be dicult to gure out how g is constructed.Geometrically, we can visualize the idea in the sphere S2 being imagined as the surfaceof the earth. Given any point P on the arctic circle Cb, there is a unique longitude linepassing through P which links both the north pole e0 and the south pole  e0. Thislongitude intersects the equator line Sn 1 uniquely at one point, say Q. Clearly, thesection of the longitude line between P and the south pole is homeomorphic to the straightline between the core 0 and Q. Function g maps P to Q and the south pole to the core.Consequently, we obtain a continuous function g  f h mapping from the convex andcompact set Bn into itself. By using Brouwer's xed point theorem, we know there existsz 2 Bn such that z = g  f h(z). Setting x = h(z), we obtain x = f(x) 2 C. Thiscontradicts the hypothesis that f has no xed point in Sn.Now we prove the last part. Since f has no xed point, it follows from Corollary 4(c) of Whittlesey [3] or Milnor [2] that f must have an antipodal point. We complete theproof. 23In term of equation theory, the above theorem states that for each y 2 Sn, the equationf(x) = y has a solution.Corollary 1.2 For any continuous function f : Sn 7! Sn, if x  f(x)  0 for everyx 2 Sn, then f must be surjective.Proof: Suppose to the contrary that f has a xed point x 2 Sn. Then 0 < x  x =x  f(x)  0 which is impossible. 2The above corollary is not so simple as it might appear. In fact we will show that itis at least as powerful as the classic Brouwer's xed point theorem is. We prove Brouwertheorem via Corollary 1.2: If f : Bn 7! Bn is a continuous function, then there existsx 2 Bn such that f(x) = x. Suppose to the contrary that Brouwer theorem is false.Then it holds f(x)  x 6= 0 for every x 2 Bn. Let g(x) = f(x)  x. Thenx  g(x) = x  (f(x)  x) = x  f(x)  x  x = x  f(x)  1 < 0for all x 2 Bn \ Sn = Sn 1. We dene the function h : Sn 7! Sn as follows: For x 2 Sn,let y(x) = (0; x1;    ; xn)>. Obviously, y(x) 2 Bn . If y(x)  g(y(x))  0, deneh(x) =g(y(x))pg(y(x))  g(y(x)):If y(x)  g(y(x)) > 0, it is easy to see that Pni=1 x2i < 1 and x0 6= 0. Denez0 =  y(x)  g(y(x))x0; z = (z0; 0;    ; 0)> 2 IRn+1;andh(x) =g(y(x)) + zp(g(y(x)) + z)  (g(y(x))+ z) :Since y(x)  g(y(x)) is a continuous function in x on a compact set, then there exists some > 0 such thaty(x)  g(y(x))  for all x 2 Sn 1. Furthermore, there exists a positive  such that y(x) g(y(x)) 0 impliesj x0 j . Now it is readily veried that h is a continuous function and x h(x)  0 for all4x 2 Sn. So all conditions of Corollary 1.2 are met. Then h must be surjective. But it isimpossible since there does not exist any x 2 Sn such that h(x) = e0. 2A function f : D 7! I is said to have an antixed point in D if there exists x 2 Dsuch that f(x) =  x.Corollary 1.3 For any continuous function f : Sn 7! Sn, if f has no antixedpoint in Sn, then f must be surjective. Furthermore, there exists x 2 Sn such thatf(x) =  f( x).Proof: Let g(x) =  f(x) for all x 2 Sn. Clearly, g is continuous on Sn, g(Sn)  Sn,and g has no xed point in Sn. It follows from Theorem 1.1 that g is surjective and henceso is f . 2Corollary 1.4 For any continuous function f : Sn 7! Sn, if x  f(x)  0 for everyx 2 Sn, then f must be surjective.Proof: Suppose to the contrary that f has an antixed point y 2 Sn. Then 0 >  y  y =f(y)  y  0 which is impossible. By Corollary 1.3 f is surjective. 2The coming result is Theorem 3 of Whittlesey [3]. Here we give an alternative proofwhich is conceptually simpler than Whittlesey's.Theorem 1.5 For any continuous function f : Sn 7! Sn, if f(x) =  f( x) for everyx 2 Sn, then f must be surjective.Proof: Suppose to the contrary that f is not surjective. Then there would exist somey 2 Sn which is not in f(Sn). Without loss of generality, we may assume that y = e0.By the assumption we know that there does not exist any x 2 Sn such that f(x) =  e0,either. Now we dene the function g : Sn 7! Bn bygi(x) =8>><>:0; if i = 0;fi(x)qPnj=1f2j (x); if i = 1;    ; n:5It follows from the assumption that g(x) =  g( x) for every x 2 Sn. Clearly, g is acontinuous function. This is impossible according to Borsuk-Ulam theorem (see [1, 4])which says that if l : Sn 7! Bn is a continuous function, then there exists x 2 Sn suchthat l(x) = l( x). 2It will be shown that Theorem 1.5 is actually equivalent to Borsuk-Ulam theorem.Suppose that the latter theorem is false. Let g(x) = f(x)   f( x). Then g(x) 6= 0 andg(x) =  g( x) for all x 2 Sn. Dene the function h : Sn 7! Sn byh(x) =g(x)pg(x)  g(x):Clearly, all conditions of Theorem 1.5 are satised. Then h must be surjective. But it isimpossible since h0(x) = 0 for all x 2 Sn. 2Now we proceed to prove one more result. DeneSn+ = fx 2 Sn j x0 > 0 g and Sn  = fx 2 Sn j x0 < 0 g:Theorem 1.6 For any continuous function f : Sn 7! Sn, if it satises thatf0(0; x1;    ; xn) = 0fi(0; x1;    ; xn) =  fi(0; x1;    ; xn); i = 1;    ; nfor all (0; x1;    ; xn) 2 Sn 1, then either Sn+  f(Sn) or Sn   f(Sn) or both.Proof: Suppose to the contrary that there exist two points y = (y0; y1;    ; yn) 2 Sn+ and~y = (~y0; ~y1;    ; ~yn) 2 Sn  neither of which belongs to f(Sn). We dene h : Bn 7! Sn byh(0; x1;    ; xn) = ( p1  x  x; x1;    ; xn);h 1 : Sn 7! Bn byh 1(x0; x1;    ; xn) = (0; x1;    ; xn);g : Bn n fh 1(y)g 7! Sn 1 with g(x) equal to the intersection point of the straight linegoing through x and h 1(y) on Sn 1;6bg : Bn n fh 1(~y)g 7! Sn 1 with bg(x) equal to the intersection point of the straight linegoing through x and h 1(~y) on Sn 1.It is easy to verify that h, h 1, g, and bg are continuous functions. Now we constructthe function F : Bn 7! Sn 1 byF (0; x1;    ; xn) =8>><>>>:g  h 1  f  h(0; x1;    ; xn); if f0  h(0; x1;    ; xn) > 0;bg  h 1  f  h(0; x1;    ; xn); if f0  h(0; x1;    ; xn) < 0;f  h(0; x1;    ; xn); if f0  h(0; x1;    ; xn) = 0:Note that h 1(x) = g(x) = bg(x) = x for all x 2 Sn 1. Then for any x 2 Bn withf0  h(x) = 0, it holds thatg  h 1  f  h(x) = bg  h 1  f  h(x) = f  h(x):From the continuity of h, h 1, g, and bg, F is a continuous function and furthermore itsatises thatF (0; x1;    ; xn) = f  h(0; x1;    ; xn) = f(0; x1;    ; xn) = f(0; x1;    ; xn) =  F (0; x1;    ; xn)for all (0; x1;    ; xn) 2 Sn 1. This contradicts an equivalent form of Borsuk-Ulam the-orem which says that there does not exist any continuous function l : Bn 7! Sn 1 suchthat l(x) =  l( x) for every x 2 Sn 1. 2References[1] V. Istratescu (1981), Fixed Point Theory: An Introduction, Reidel Publishing Company, Dordrecht.[2] J. Milnor (1965), Topology From The Dierentiable Viewpoint, The University Press of Virginia, Char-lottessville.[3] E.F. Whittlesey (1963), \Fixed points and antipodal points", American Mathematical Monthly 70, 807-821.[4] Z. Yang (1999), Computing Equilibria and Fixed Points, Kluwer Academic Publishers, Boston.

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Surjective Function Theorems

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