Editing Borsuk–Ulam theorem (section)

=== With odd functions ===
A function <math>g</math> is called ''odd'' (aka ''antipodal'' or ''antipode-preserving'') if for every <math>x</math>, <math>g(-x)=-g(x)</math>.

The Borsuk–Ulam theorem is equivalent to each of the following statements:

(1) Each continuous odd function <math>S^n\to \R^n</math> has a zero.

(2) There is no continuous odd function <math> S^n \to S^{n-1}</math>.

Here is a proof that the Borsuk-Ulam theorem is equivalent to (1): 

(<math>\Longrightarrow</math>) If the theorem is correct, then it is specifically correct for odd functions, and for an odd function, <math>g(-x)=g(x)</math> iff <math>g(x)=0</math>. Hence every odd continuous function has a zero.

(<math>\Longleftarrow </math>) For every continuous function <math>f:S^n\to \R^n</math>, the following function is continuous and odd: <math>g(x)=f(x)-f(-x)</math>. If every odd continuous function has a zero, then <math>g</math> has a zero, and therefore, <math>f(x)=f(-x)</math>. 

To prove that (1) and (2) are equivalent, we use the following continuous odd maps:
* the obvious inclusion <math> i: S^{n-1}\to \R^n\setminus \{0\} </math>,
* and the radial projection map <math> p: \R^n\setminus \{0\} \to S^{n-1} </math> given by <math>x \mapsto \frac{x}{|x|}</math>.

The proof now writes itself.

<math>((1) \Longrightarrow (2))</math> We prove the contrapositive. If there exists a continuous odd function <math>f:S^n\to S^{n-1}</math>, then <math> i\circ f</math> is a continuous odd function <math>S^n\to \R^n\setminus \{0\}</math>.

<math>((1) \Longleftarrow (2)) </math> Again we prove the contrapositive. If there exists a continuous odd function <math>f:S^n\to \R^{n}\setminus\{0\}</math>, then <math> p\circ f</math> is a continuous odd function <math>S^n\to S^{n-1}</math>.