A Borsuk theorem Let $M$ and $L$ be two subspaces of Banach space $X$ such that $\dim L<\dim M<\infty$.
Let $S=\{m\in M : \|m\|=1\}$ and let $g$ be a continuous function from  $S$ to $L$ such that $g(-m)=-g(m)$ for all $m\in S$. I need to prove that there exists an $m \in S$ such that $g(m)=0$.
 A: There are various statements. Here are a few:
(1) Suppose $f\colon S^n\to \mathbb R^n$ is continuous and $f(-x)=-f(x)$. Then there is some $x\in S^n$ with $f(x)=0$.
(2) There is no continuous odd function $f\colon S^n\to S^m$ for $m<n$. 
(3) Let $f\colon S^n\to \mathbb R^n$ be continuous. Then there is some $x\in S^n$ so that $f(x)=f(-x)$. [Note that this works for $f\colon S^n\to \mathbb R^m$ for any $m\le n$.]
(4) An odd map $f\colon S^n\to S^n$ must have odd degree.
The standard proof uses some serious algebraic topology (cup product structure on $H^*(\mathbb RP^n)$). My favorite proof (see Guillemin and Pollack) comes from using differential topology (assuming the maps are smooth) and using an inductive argument together with the following generalization of the argument principle in complex analysis: If $D\subset\mathbb R^n$ is the closed unit ball and $f\colon D\to\mathbb R^n$ is smooth and nonzero on $\partial D$, then the degree of the map $f/\|f\|\colon \partial D\to S^{n-1}$ [this is the winding number of $f|_{\partial D}$ around $0$] computes the number of roots of $f$ (with multiplicity) in $D$.
See also Intuitive proof of Tucker's lemma or Borsuk-Ulam theorem and http://planetmath.org/proofofborsukulamtheorem.
