In mathematics, the Borsuk–Ulam theorem states that every continuous function from an n-sphere into Euclidean n-space maps some pair of antipodal points to the same point.
Here, two points on a sphere are called antipodal if they are in exactly opposite directions from the sphere's center.
Formally: if f: S^n \to \R^n is continuous then there exists an x\in S^n such that:  f(-x)=f(x).
The case n=1 can be illustrated by saying that there always exist a pair of opposite points on the Earth's equator with the same temperature.
The same is true for any circle.
This assumes the temperature varies continuously in space.
The case n=2 is often illustrated by saying that at any moment, there is always a pair of antipodal points on the Earth's surface with equal temperatures and equal barometric pressures, assuming that both parameters vary continuously in space.
The Borsuk–Ulam theorem has several equivalent statements in terms of odd functions.
Recall that S^n is the n-sphere and B^n is the n-ball:
If g : S^n \to \R^n is a continuous odd function, then there exists an x\in S^n such that:  g(x)=0.
If g : B^n \to \R^n is a continuous function which is odd on S^{n-1} (the boundary of B^n), then there exists an x\in B^n such that:  g(x)=0.
History
According to , the first historical mention of the statement of the Borsuk–Ulam theorem appears in .
The first proof was given by , where the formulation of the problem was attributed to Stanislaw Ulam.
Since then, many alternative proofs have been found by various authors, as collected by .
Equivalent statements
The following statements are equivalent to the Borsuk–Ulam theorem.
With odd functions
A function g is called odd (aka antipodal or antipode-preserving) if for every x: g(-x)=-g(x).
The Borsuk–Ulam theorem is equivalent to the following statement: A continuous odd function from an n-sphere into Euclidean n-space has a zero.
PROOF:
If the theorem is correct, then it is specifically correct for odd functions, and for an odd function, g(-x)=g(x) iff g(x)=0.
Hence every odd continuous function has a zero.
For every continuous function f, the following function is continuous and odd: g(x)=f(x)-f(-x).
If every odd continuous function has a zero, then g has a zero, and therefore, f(x)=f(-x).
Hence the theorem is correct.
With retractions
Define a retraction as a function h: S^n \to S^{n-1}.
The Borsuk–Ulam theorem is equivalent to the following claim: there is no continuous odd retraction.
Proof: If the theorem is correct, then every continuous odd function from S^n must include 0 in its range.
However,  0 \notin S^{n-1} so there cannot be a continuous odd function whose range is S^{n-1}.
Conversely, if it is incorrect, then there is a continuous odd function g: S^n \to \Bbb{R}^n with no zeroes.
Then we can construct another odd function h: S^n \to S^{n-1} by:
h(x)=\frac{g(x)}{|g(x)|}
since g has no zeroes, h is well-defined and continuous.
Thus we have a continuous odd retraction.
Proofs
1-dimensional case
The 1-dimensional case can easily be proved using the intermediate value theorem (IVT).
Let g be an odd real-valued continuous function on a circle.
Pick an arbitrary x.
If g(x)=0 then we are done.
Otherwise, without loss of generality, g(x)>0.
But g(-x)<0.
Hence, by the IVT, there is a point y between x and -x at which g(y)=0.
General case
Algebraic topological proof
Assume that h: S^n \to S^{n-1} is an odd continuous function with n > 2 (the case n = 1 is treated above, the case n = 2 can be handled using basic covering theory).
By passing to orbits under the antipodal action, we then get an induced continuous function h': \mathbb{RP}^n \to \mathbb{RP}^{n-1} between real projective spaces, which induces an isomorphism on fundamental groups.
By the Hurewicz theorem, the induced ring homomorphism on cohomology with \mathbb F_2 coefficients [where \mathbb F_2 denotes the field with two elements],
\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},
sends b to a.
But then we get that b^n = 0 is sent to a^n \neq 0, a contradiction.Joseph J. Rotman, An Introduction to Algebraic Topology (1988) Springer-Verlag  (See Chapter 12 for a full exposition.)
One can also show the stronger statement that any odd map S^{n-1} \to S^{n-1} has odd degree and then deduce the theorem from this result.
Combinatorial proof
The Borsuk–Ulam theorem can be proved from Tucker's lemma.
Let g : S^n \to \R^n be a continuous odd function.
Because g is continuous on a compact domain, it is uniformly continuous.
Therefore, for every \epsilon > 0, there is a \delta > 0 such that, for every two points of S_n which are within \delta of each other, their images under g are within \epsilon of each other.
Define a triangulation of S_n with edges of length at most \delta.
Label each vertex v of the triangulation with a label l(v)\in {\pm 1, \pm 2, \ldots, \pm n} in the following way:
The absolute value of the label is the index of the coordinate with the highest absolute value of g: |l(v)| = \arg\max_k (|g(v)_k|).
The sign of the label is the sign of g, so that: l(v) = \sgn (g(v)) |l(v)|.
Because g is odd, the labeling is also odd: l(-v) = -l(v).
Hence, by Tucker's lemma, there are two adjacent vertices u, v with opposite labels.
Assume w.l.o.g. that the labels are l(u)=1, l(v)=-1.
By the definition of l, this means that in both g(u) and g(v), coordinate #1 is the largest coordinate: in g(u) this coordinate is positive while in g(v) it is negative.
By the construction of the triangulation, the distance between g(u) and g(v) is at most \epsilon, so in particular |g(u)_1 - g(v)_1| = |g(u)_1| + |g(v)_1| \leq \epsilon  (since g(u)_1 and g(v)_1 have opposite signs) and so |g(u)_1| \leq \epsilon.
But since the largest coordinate of g(u) is coordinate #1, this means that  |g(u)_k| \leq \epsilon for each  1 \leq k \leq n.
So |g(u)| \leq c_n \epsilon, where c_n  is some constant depending on n  and the norm |\cdot|  which you have chosen.
The above is true for every \epsilon > 0; since  S_n is compact there must hence be a point u in which |g(u)|=0.
Corollaries
No subset of \R^n is homeomorphic to S^n
The ham sandwich theorem: For any compact sets A1, ..., An in \R^n we can always find a hyperplane dividing each of them into two subsets of equal measure.
Equivalent results
Above we showed how to prove the Borsuk–Ulam theorem from Tucker's lemma.
The converse is also true: it is possible to prove Tucker's lemma from the Borsuk–Ulam theorem.
Therefore, these two theorems are equivalent.
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 \R^n containing the origin (Here, symmetric means that if x is in the subset then -x is also in the subset).
Consider the function A which maps a point to its antipodal point: A(x) = -x.
Note that A(A(x))=x.
The original theorem claims that there is a point x in which f(A(x))=f(x).
In general, this is true also for every function A for which A(A(x))=x.
However, in general this is not true for other functions A.
See also
Topological combinatorics
Necklace splitting problem
Kakutani's theorem (geometry)
Imre Bárány
Notes
References
External links
