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I'm looking for a intuitive proof of Tucker's lemma and/or the Borsuk-Ulam theorem. The proof should not make use of topology, cohomology etc. as it should be understandable by undergraduates.

Thanks in advance!

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This proof of Borsuk-Ulam seems to be elementary, I'm not sure it is intuitive (I haven't checked it in detail). –  Michael Greinecker Jul 29 '12 at 19:28
    
Thanks a lot for you help! –  user36773 Jul 29 '12 at 21:23

1 Answer 1

This is my intuition for the special case of $n=2$ (it cannot be easilly generalized to higher dimensions). If $f: S^2\to \mathbb{R}^2$ is antipodal and nowhere zero, than $g:=f/|f|$ is an antipodal map from $S^2$ to the circle $S^1$. Take two antipodal points in the equator, $A$ and $B$. Their images $f(A)$ and $f(B)$ are antipodal on the circle and the image of the half-circle $AB$ on the equator is mapped to some curve on the circle that winds around the circle $n+1/2$ times for some $n$. So, the image of the whole equator ($(AB)$+the antipodal halfcircle $(BA)$) winds around the circle $2n+1$ times, a nonzero number. The equator can be slipped towards the north pole and contracted in $S^2$. However, for its image, you cannot contract a curve in the circle that has a nontrivial winding number.

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