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Borsuk–Ulam theorem
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== Proofs == ===1-dimensional case=== The 1-dimensional case can easily be proved using the [[intermediate value theorem]] (IVT). Let <math>g</math> be the odd real-valued continuous function on a circle defined by <math>g(x)=f(x)-f(-x)</math>. Pick an arbitrary <math>x</math>. If <math>g(x)=0</math> then we are done. Otherwise, without loss of generality, <math>g(x)>0.</math> But <math>g(-x)<0.</math> Hence, by the IVT, there is a point <math>y</math> at which <math>g(y)=0</math>. ===General case=== ====Algebraic topological proof==== Assume that <math>h: S^n \to S^{n-1}</math> is an odd continuous function with <math>n > 2</math> (the case <math>n = 1</math> is treated above, the case <math>n = 2</math> can be handled using basic [[Covering space|covering theory]]). By passing to orbits under the antipodal action, we then get an induced continuous function <math>h': \mathbb{RP}^n \to \mathbb{RP}^{n-1}</math> between [[Real projective space|real projective spaces]], which induces an isomorphism on [[Fundamental group|fundamental groups]]. By the [[Hurewicz theorem]], the induced [[ring homomorphism]] on [[cohomology]] with <math>\mathbb F_2</math> coefficients [where <math>\mathbb F_2</math> denotes the [[GF(2)|field with two elements]]], :<math> \mathbb F_2[a]/a^{n+1} = H^*\left(\mathbb{RP}^n; \mathbb{F}_2\right) \leftarrow H^*\left(\mathbb{RP}^{n-1}; \mathbb F_2\right) = \mathbb F_2[b]/b^{n},</math> sends <math>b</math> to <math>a</math>. But then we get that <math>b^n = 0</math> is sent to <math>a^n \neq 0</math>, a contradiction.<ref name=rotman>Joseph J. Rotman, ''An Introduction to Algebraic Topology'' (1988) Springer-Verlag {{ISBN|0-387-96678-1}} ''(See Chapter 12 for a full exposition.)''</ref> One can also show the stronger statement that any odd map <math>S^{n-1} \to S^{n-1}</math> has odd [[degree of a continuous mapping|degree]] and then deduce the theorem from this result. ====Combinatorial proof==== The Borsuk–Ulam theorem can be proved from [[Tucker's lemma]].<ref name=prescott2002/><ref name=FreundTodd1982>{{Cite journal|doi=10.1016/0097-3165(81)90027-3 | year=1982| volume=30 | issue=3| title=A constructive proof of Tucker's combinatorial lemma | journal=[[Journal of Combinatorial Theory]] | series=Series A | pages=321–325|author1=Freund, Robert M.|author2=Todd, Michael J.| doi-access=free}}</ref><ref name=SimmonsSu2003>{{Cite journal|author1 = Simmons, Forest W.| author2=Su, Francis Edward | doi=10.1016/s0165-4896(02)00087-2 |year=2003| volume=45 | title=Consensus-halving via theorems of Borsuk–Ulam and Tucker | journal=Mathematical Social Sciences | pages=15–25| hdl=10419/94656 | url=https://scholarship.claremont.edu/hmc_fac_pub/677 | hdl-access=free }}</ref> Let <math>g : S^n \to \R^n</math> be a continuous odd function. Because ''g'' is continuous on a [[Compact space|compact]] domain, it is [[uniformly continuous]]. Therefore, for every <math>\epsilon > 0</math>, there is a <math>\delta > 0</math> such that, for every two points of <math>S_n</math> which are within <math>\delta</math> of each other, their images under ''g'' are within <math>\epsilon</math> of each other. Define a triangulation of <math>S_n</math> with edges of length at most <math>\delta</math>. Label each vertex <math>v</math> of the triangulation with a label <math>l(v)\in {\pm 1, \pm 2, \ldots, \pm n}</math> in the following way: * The absolute value of the label is the ''index'' of the coordinate with the highest absolute value of ''g'': <math>|l(v)| = \arg\max_k (|g(v)_k|)</math>. * The sign of the label is the sign of ''g'' at the above coordinate, so that: <math>l(v) = \sgn (g(v)_{|l(v)|}) |l(v)|</math>. Because ''g'' is odd, the labeling is also odd: <math>l(-v) = -l(v)</math>. Hence, by Tucker's lemma, there are two adjacent vertices <math>u, v</math> with opposite labels. Assume w.l.o.g. that the labels are <math>l(u)=1, l(v)=-1</math>. By the definition of ''l'', this means that in both <math>g(u)</math> and <math>g(v)</math>, coordinate #1 is the largest coordinate: in <math>g(u)</math> this coordinate is positive while in <math>g(v)</math> it is negative. By the construction of the triangulation, the distance between <math>g(u)</math> and <math>g(v)</math> is at most <math>\epsilon</math>, so in particular <math>|g(u)_1 - g(v)_1| = |g(u)_1| + |g(v)_1| \leq \epsilon </math> (since <math>g(u)_1</math> and <math>g(v)_1</math> have opposite signs) and so <math>|g(u)_1| \leq \epsilon</math>. But since the largest coordinate of <math>g(u)</math> is coordinate #1, this means that <math>|g(u)_k| \leq \epsilon</math> for each <math>1 \leq k \leq n</math>. So <math>|g(u)| \leq c_n \epsilon</math>, where <math>c_n </math> is some constant depending on <math>n </math> and the norm <math>|\cdot| </math> which you have chosen. The above is true for every <math>\epsilon > 0</math>; since <math>S_n</math> is compact there must hence be a point ''u'' in which <math>|g(u)|=0</math>.
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