- #26

WWGD

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Yes, there's this interesting one, (not originally mine). It uses the degree of a map, specifically, homeomor-You can prove this (##\mathbb{R}^{2n+1}\ncong X\times X##) with algebraic topology (homology)- do you have a different argument in mind?

phisms have degree ## \pm 1## and satisfy ##def(f \circ g)=deg(f) \cdot deg(g)##, so that for any homeomorphism f, we have : ##deg(f \circ f) ==1 ##.

Assume then there is a homeo h ## X^2 \rightarrow \mathbb R^3 ##. Then we have a homeo

from ## X^4 \rightarrow \mathbb R^6 ##. Consider the homeo ## h: X^4 \rightarrow X^4 : (a,b,c,d) \rightarrow (d,a,b,c) ## with ## h \circ h : (a,b,c,d) \rightarrow (c,d,a,b) ## which gives us the map ##h' \circ h' \mathbb R^6 \rightarrow \mathbb R^6 (a,b,c,d,e,f) \rightarrow ##

Yes, I dont know if this applies to any other type of space though. And it seems like a novel proof.If ##X## is a smooth manifold then wouldn't ##X^2## be even dimensional?

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