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Jan 12, 2020 at 5:27 comment added niran90 No, you assumed correctly - I want to map the interiors to each other. I've now clarified the ambiguity in the post. Thanks.
Jan 11, 2020 at 19:15 comment added Robert Bryant Actually, I assumed that you wanted to map the interior of $B$ to the interior of $A'$, and that's what my answer addressed. Did you instead want to map the actual polyhedrons (i.e., the surfaces) conformally?
Jan 11, 2020 at 9:41 comment added niran90 Okay I see. Suppose I relax the requirement for conformality, and simply require that the mapping is smooth and bijectve, would that be feasible? I imagine there are infinitely many such mappings. But what's the most straightforward way to derive one of them? Actually for my specific application, conformality is not crucial. I just thought it would be a sufficient condition for a smooth bijective map.
Jan 11, 2020 at 9:15 history answered Robert Bryant CC BY-SA 4.0