Editing Borsuk–Ulam theorem (section)

== Generalizations ==
* In the original theorem, the domain of the function ''f'' is the unit ''n''-sphere (the boundary of the unit ''n''-ball). In general, it is true also when the domain of ''f'' is the boundary of any open bounded symmetric subset of <math>\R^n</math> containing the origin (Here, symmetric means that if ''x'' is in the subset then -''x'' is also in the subset).<ref>{{springer|title=Borsuk fixed-point theorem|id=p/b110770}}</ref>
* More generally, if <math>M</math> is a compact ''n''-dimensional [[Riemannian manifold]], and <math>f: M \rightarrow \mathbb{R}^n</math> is continuous, there exists a pair of points ''x'' and ''y'' in <math>M</math> such that <math>f(x) = f(y)</math> and ''x'' and ''y'' are joined by a geodesic of length <math>\delta</math>, for any prescribed <math>\delta > 0</math>.<ref>{{cite journal |last=Hopf |first=H. |year=1944 |title=Eine Verallgemeinerung bekannter Abbildungs-und Überdeckungssätze |journal=Portugaliae Mathematica}}</ref><ref>{{cite journal |last1=Malyutin |first1=A. V. |last2=Shirokov |first2=I. M. |year=2023 |title=Hopf-type theorems for f-neighbors |journal=Sib. Èlektron. Mat. Izv |volume=20 |issue=1 |pages=165–182}}</ref>
* Consider the function ''A'' which maps a point to its antipodal point: <math>A(x) = -x.</math> Note that <math>A(A(x))=x.</math> The original theorem claims that there is a point ''x'' in which <math>f(A(x))=f(x).</math> In general, this is true also for every function ''A'' for which <math>A(A(x))=x.</math><ref>{{cite journal|doi=10.2307/1969632 |author=Yang, Chung-Tao | year= 1954|volume=60 |issue=2 | title=On Theorems of Borsuk-Ulam, Kakutani-Yamabe-Yujobo and Dyson, I | journal=[[Annals of Mathematics]] | pages=262–282|jstor=1969632 }}</ref> However, in general this is not true for other functions ''A''.<ref>{{cite web | url=https://mathoverflow.net/q/61141 | title=Generalization of Borsuk-Ulam | publisher=Math Overflow | access-date=18 May 2015 | author=Jens Reinhold, Faisal | author2=Sergei Ivanov}}</ref>