0
$\begingroup$

Let $Z\subset \mathbb R^N$ be a embedded submanifold (or just a concrete manifold).

For a smooth immersion $f:Z\to \mathbb R^N$, can we find a open set $U\subset \mathbb R^N$ containing $Z$ such that $f$ has extension $\tilde f:U\to $, which is also a immersion.

The original goal is to prove that the diffeomorphism between two submanifolds $f: Z_1\subset \mathbb R^N\to Z_2\subset \mathbb R^N$ can extend to a diffeomorphism between two open sets $U_1\supset Z_1,U_2\supset Z_2$. Before using inverse function theorem, I stuck in constracting a extension $\tilde f:U_1\to \mathbb R^N$ st $df_p$ is isomorphism for all $p\in Z_1$.

$\endgroup$
  • $\begingroup$ What is $X$? What is the relation between $N$ et $M$ ? $\endgroup$ – R. Alexandre Apr 17 '17 at 17:47
  • $\begingroup$ @R.Alexandre M, sorry $\endgroup$ – yaoliding Apr 18 '17 at 2:12
  • $\begingroup$ In terms of your "original goal" - it doesn't seem to be true in this level of generality. First, if $M$ is disconnected with components of different dimensions, then this is hopeless. Even for connected $M$, this seems not true. For example, if $M = \mathbb{R}P^2$, let $Z_1$ denote the "equator" in the hemisphere model of $\mathbb{R}P^2$ (which generates $\pi_1$), and let $Z_2$ denote a small circle enclosing a disc. Then, removing $Z_1$ from any open set around it leaves a connected thing, while removing $Z_2$ from any open set around it leaves a disconnected thing. $\endgroup$ – Jason DeVito Apr 18 '17 at 2:33
  • $\begingroup$ @JasonDeVito I see, so the conclusion should not generalize too much. Book in Guillemin & Pollack define diffeomorphism as a bijection that can be extend to a open subset in $\mathbb R^N$, but I don't know how to prove it consistent to the classical definition. $\endgroup$ – yaoliding Apr 18 '17 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.