I was considerin that counterexample too since the proof of dimension invariance by Borsuk-Ulam theorem uses same contradiction argument when homeomorphism is restricted to unit sphere. Seems to me that domain invariance is much deeper result.

]]>So the 1-dimensional case leads layman astray since homeomorphic image of a single point is just a single point, easy case, but homeomorphic image of a line can be pathological and one can’t immediately identify the connected domains?

A circle separates the plane into two but the homeomorphic image of that is the real question. So does n-dimensional Jordan curve theorem immediately solve dimension invariance, or is there some more pitfalls?

]]>Can you please explain, so that a layman can understand, why “invariance of dimension” is non-trivial? If a point separates a line but not a plane, then they are not homeomorphic, but why can’t we use that same argument again and argue that a line separates plane but not 3-dimensional space, and so on and so forth?

]]>dh1 ^ (g2) ^ dg3 ^ … (over the sphere). Now replace g2 by h2 (equal on the domain of integration). Repeat this process to replace all gi by hi, resulting in the volume form for the n-ball, which computes to a non-zero value. As Yakar Kannai points out in the Monthly selection, the same proof also applies to any reasonable n-manifold with boundary, embedded in Euclidean space of the same dimension (attributed to M. Hirsch amongst others). J Sjogren ]]>

I think the connection with degree theory comes in the fact that the degree of a smooth map can be expressed as the integral (normalised by the surface area of the sphere). An application of Stokes’ theorem shows that this quantity is invariant under smooth deformations of g; since the constant map has zero degree and the identity map has non-zero degree, one therefore cannot continuously deform one to the other. If one expresses your map in polar coordinates one sees that this argument is basically equivalent to Lima’s argument.

]]>As usual it’s enough to show that there is no continuous map which restricts to the identity on the boundary. As usual it’s enough to prove there is no such map. Suppose there were such a map. Consider : on the one hand this is zero as has less than full rank, and on the other hand it equals

by Stokes’ theorem.

But as is the identity on the boundary, we can replace by here (pause for thought) and reverse the argument, giving the alternate answer , the volune of the unit ball in . Hence no such exists. Is there degree theory lurking here?

A similar argument, combined with some ideas of Shchepin can be used to give a quite easy proof of the Borsuk–Ulam theorem. See http://www.maths.ed.ac.uk/~carbery/analysis/notes/bu3_public.pdf

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