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Mar 25, 2016 at 20:56 comment added Blake Ok, so is this true in the case where $M$ is transverse to the faces of $C$?
Mar 25, 2016 at 12:07 comment added Sebastian Goette Then just use the word "transversely". Otherwise $M$ could intersect those faces in rather weird ways.
Mar 25, 2016 at 2:11 comment added Blake I don't believe that's true. Note that when I say "we can perturb $M$ so that it only intersects faces of $C$ of dimension $d−2$ or greater", I'm also saying that I assume $M$ intersects every face of $C$ transversally.
Mar 25, 2016 at 0:33 comment added Ryan Budney I think generally you will have to create new intersections when you smooth your map. Think of examples where your original map (of $M$) are highly not-transverse on the skeleta. Even for maps into the plane there appear to be problems.
Mar 24, 2016 at 22:39 history edited Blake
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Mar 24, 2016 at 3:58 review Close votes
Mar 24, 2016 at 10:33
Mar 24, 2016 at 3:49 comment added Blake I believe that M could be smoothed, which seems to be what the result you linked proves, but I don't see how this answer my question about the polyhedral complex.
Mar 24, 2016 at 3:45 comment added Ryan Budney Plenty. mathoverflow.net/questions/8789/…
Mar 24, 2016 at 3:43 comment added Blake Do you have a reference that might be useful?
Mar 24, 2016 at 3:41 comment added Ryan Budney Yes, this is a theorem of Whitney's.
Nov 26, 2015 at 8:49 history edited Blake
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Nov 17, 2015 at 5:24 history edited Blake
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Nov 15, 2015 at 19:09 history edited Blake
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Nov 14, 2015 at 3:01 review First posts
Nov 14, 2015 at 3:12
Nov 13, 2015 at 16:28 history asked Blake CC BY-SA 3.0