Are the unit sphere and the unit cube in the n-dimensional Euclidean space homeomorphic? If so, can anyone give an explicit formula for the homeomorphism?
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1How about $v$ goes to $v/|v|$. Will that do it? – Gregory Grant Feb 22 '15 at 16:27
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That's for their boundaries. He might mean the solid ball? – Henno Brandsma Feb 22 '15 at 16:30
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Yes, I meant the solid cube and solid sphere. – user168826 Feb 22 '15 at 16:31
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Ah, in that case you have to work a bit harder. – Gregory Grant Feb 22 '15 at 16:35
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From the solid cube ($|x_i|\leq 1$) to the solid sphere ($||x|| \leq 1$), set $f(x)=max\{|x_1|,...,|x_n|\}*x/||x||$. It is clear from this formula that if $f(x)=f(y)$ then $x$ and $y$ are collinear, say $x=ty$. But then, it follows that $max(|x_1|,...,|x_n|)=max(t|x_1|,...,t|x_n|)$ so $t=1$, which proves injectivity. Surjectivity follows from the fact that on each line through the origin, the function restricted to that line gives you both points of intersection of the line with the sphere by plugging in the two points where the line intersects the cube, so by connectivity and continuity it must hit the entire segment in the sphere.
As per Julian's comment, I am technically done at this point. However, for an explicit inverse, set $g(y) = ||y||y/max(|y_1|,...,|y_n|)$, $g(0)=0$.
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5A continuous bijection between compact spaces is a homeomorphism. – Julián Aguirre Feb 22 '15 at 17:44