Timeline for Is the preimage of a face under an affine map a face?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 12, 2018 at 7:26 | comment | added | Tom Werner | As I think of it, it cannot. Maybe that's the mistake. It is surely true of facets, following from Fourier-Motzkin elimination (see, eg, mathoverflow.net/questions/260107/…), but apparently not of faces. My argument above thus only proves that facets not always back-project to facets - which is obvious. | |
| Aug 12, 2018 at 4:03 | comment | added | Robert Furber | @TomWerner Can you show me an explicit example where the projection has more faces? I can't think of one. | |
| Aug 11, 2018 at 11:23 | comment | added | Tom Werner | Still there is something I do not understand. Suppose $F$ is a face of $Y$ that back-projects to a face $E=f^{-1}(F)$ of $X$. Then, obviously, $f(E)=F$. But it is known that a projection of a polytope can have more faces than the original polytope. So sometimes it must happen that two different faces of $Y$, say $F$ and $F'$, back-project to a single face of $X$, i.e., $f^{-1}(F)=E=f^{-1}(F')$. But this contradicts the fact that $f$ is a function, because $f(E)$ would be ambiguous. Where am I doing a mistake? | |
| Aug 10, 2018 at 19:24 | comment | added | Robert Furber | @TomWerner Quite right, I've just made the edit. | |
| Aug 10, 2018 at 19:24 | history | edited | Robert Furber | CC BY-SA 4.0 |
Fixed typo, as suggested by Tom Werner
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| Aug 10, 2018 at 19:20 | comment | added | Tom Werner | Many thanks, I did not expect the proof would be so simple. I rephrased the title as you suggest. Note the typo in your proof: in the last sentence there should be "$x,x'\in f^{-1}(F)$, proving that $f^{-1}(F)$ is a face". | |
| Aug 10, 2018 at 19:18 | vote | accept | Tom Werner | ||
| Aug 10, 2018 at 16:08 | history | answered | Robert Furber | CC BY-SA 4.0 |