Let $A,B\subset\mathbb{R}^n, n\geq 2,$ are two different shaped spindles. One is thick and one is thin. (Sorry for my unprofessional statements. Unsure about how to say it rigorously.) So there are homeomorphisms between $A$ and $B$.
My question is:
Is there such an homeomorphism $H$ that can be realized as
$$H(x_1,...,x_n)=(h_1(x_1),...,h_n(x_n)),$$ where
$$h_q:A_q \to B_q$$ and $A_q, B_q$ are the projection of $A, B$ to the $q$-th coordinate respectively?
Visually, $H$ maps coordinate-paralleling lines to paralleling lines and coordinate-paralleling sheets to paralleling sheets.
Thanks.
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1$\begingroup$ I might misunderstand the question. Consider "П" and "•". These are two compact, connected simply-connected sets. All the sections of the latter by horizontal lines are connected, but not so for the former. $\endgroup$– Boris BukhAug 26, 2014 at 18:07
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$\begingroup$ @BorisBukh Thanks for your example. I was trying to consider two "regular" subsets, such as squares and disks. I added the condition: nonempty interior on $A$ and $B$. $\endgroup$– LucyAug 26, 2014 at 18:22
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$\begingroup$ I'm probably being naive, but do you have a reference for your first statement? $\endgroup$– PVALAug 27, 2014 at 2:16
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$\begingroup$ A closed thickening of the 2-sphere and a closed 3-ball are a counterexample to your first statement as is. Should simply-connected be replaced with contractible? Even if so, I'm not convinced they're aren't counterexamples. $\endgroup$– PVALAug 27, 2014 at 2:25
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$\begingroup$ Do you simply mean self-homeomorphisms of an $X$ onto itself? $\endgroup$– Włodzimierz HolsztyńskiAug 27, 2014 at 9:13
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