1

Is it canonical immersion when it appears on Page 15, and cannonical submersion on Page 20? I never really see where it is defined, except for Page 15:

"Define $G$ s that $g = G \circ$ (canonical immersion)" - which does not really defined $g$ neither.

In both case, is $g$ an open map?

Thank you~

WishingFish
  • 2,412
  • $g$ is defined in the diagram on the same page; it's a "local/coordinate representation" of $g$ around $x$. The sentence you quote does not define $G$; rather, it's a consequence of the definition $G(x, z) = g(x) + (0, z)$. The composition sends $x \in \mathbb{R}^k$ to $G(x, 0)$, which is just $g(x)$. – TTS Jul 04 '13 at 19:19
  • 2
    Giving a page number only helps if you tells us which book it's from(!) – Fly by Night Jul 04 '13 at 19:23
  • Thanks @TTS, you mean "local/coordinate representation" of $f$? – WishingFish Jul 04 '13 at 19:23
  • @FlybyNight there there... – WishingFish Jul 04 '13 at 19:25
  • @WishingFish Yes. Can't edit it now, unfortunately. Also, $g$ is probably not open: the point of the theorem is that you can change coordinates to get a coordinate representation $x \mapsto (x, 0)$ and this isn't an open map if the dimensions aren't the same. [The image is much too "thin".] – TTS Jul 04 '13 at 19:25
  • No problem @TTS, I got it. :-B "By the denition of $g$, it also maps open sets to open sets": http://math.stanford.edu/~lhhuang/147/147_hw2.pdf. That's why I ask "the denition of g" and "its openness".. :-P – WishingFish Jul 04 '13 at 19:27
  • @WishingFish Submersions are different! Projection maps are open. – TTS Jul 04 '13 at 19:28
  • How about canonical submersion @TTS? Are they open..? – WishingFish Jul 04 '13 at 19:30
  • @TTS, locally, canonical submersion is just projection, right? – WishingFish Jul 04 '13 at 19:34
  • 1
    Yes, it seems like you've got the right idea. You could prove it directly in this case with balls, if you like. – TTS Jul 05 '13 at 16:47

0 Answers0