When you cut off that alleged lower dimensional “topological sphere” is the remaining part path-connected because R^n is a vector space? Are there any subleties? This can be a silly question, but I haven’t seen that direct & easy proof before…

]]>Thank you professor for clarifying this issue.

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.

]]>Yes, I believe the Jordan-Brouwer separation theorem implies that is not homeomorphic to for (otherwise the image of the unit sphere in would be a counterexample to that theorem). But I think the standard proof of that theorem requires at least as much topological machinery as would be needed to establish invariance of domain.

]]>Thanks for your answer!

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?

]]>While a line clearly cannot separate a 3-dimensional space, it is not immediately obvious that every homeomorphic copy of a line (which includes fractals such as the Koch snowflake) also cannot separate a 3-dimensional space, as one can already see from to the existence of space-filling curves. Note that these space-filling curves cross each other and are thus not homeomorphic to a line, but they do indicate that there is something non-trivial to be done in order to rigorously justify the intuition here (much as the Jordan curve theorem, though similarly “obvious”, must also require rigorous proof).

]]>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?

]]>Greetings Terry Tao from af friends esp Robert Bonneau. Also greeting to A. Carberry from Prof. Wm Moran of Melbourne. I would not be hasty about attributing the proof of Brouwer FPT (using differential forms) to “E. Lima”. I know that reference comes from a book of do Carmo. On the other hand, this exact proof is found in Amer Math Monthly article April 1981, 264-268, by Yakar Kannai. He credits discussions with Prof. H.Scarf. First Dr. Kannai goes through the proof from the point of view of classical divergence theorem. However he also covers the proof again, in seven lines, using Stokes’ theorem in its formulation using exterior forms. A refinement that could be suggested to both “differential forms” proofs, is two consider two mappings g, h from the n-ball to (n-1) sphere both fixing every boundary point. In the integral of dg1 ^ dg2 ^ … ^ dgn over the ball, you want to replace each “g” by and “h”. You finally replace the retraction g by the identity h. First apply Stokes’ to equate the first integral with g1 dg2 ^ dg3 ^ … , integrated over the sphere, and note that now g1 may be replaced by h1 . Perform Stokes’ now to the new expression (“in reverse”) to obtain dh1 ^ dg2 ^ dg3 ^ … ^ dgn (over the ball), and now Stokes’ again to get

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

Dear Tony,

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.

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