# From local to global diffeomorphism

Let $$M$$ be a manifold. Let $$B \cong S^1$$ be its boundary. Assume there exists a diffeomorphism $$\phi$$ from some neighbourhood $$U \subset M$$ of $$B$$ to $$S^1 \times [0,1)$$.

$$M$$ can be embedded in four-dimensional space. Does there exist a diffeomorphism from $$M$$ to $$f(M) \subset \mathbb{R}^4$$ so that $$f(U) = \{ (\cos(\theta), \sin(\theta), z,0) |\theta \in [0, 2 \pi], z \in [0,1) \}$$?

I am intuitively completely convinced that such a diffeomorphism must exist, but practically rather stumped how to actually construct it using $$\phi$$. I.e. I don't know if $$\phi$$ can be 'extended' to include all of $$M$$ while retaining its original image on $$U$$.

• Just to clarify, is $S^1$ the circle in this problem? Commented Jun 14, 2019 at 19:55
• @SamSkywalker, yes, it is the circle. I took this as an example, but the main property I need is for the boundary to be compact. Commented Jun 14, 2019 at 20:02
• Yes, this is indeed true since your $M$ is a surface. A proof requires some work though. One way to argue is to go through the step of the proof of Whitney embedding theorem (say, every surface embeds in $R^4$). Another is to argue that all 1-dimensional smooth framed knots in $R^4$ are smoothly ambiently isotopic. Commented Jun 14, 2019 at 22:54

It seems to be false, if I am not wrong. Regard the Möbius strip as a real projective plane with an open disc removed. This surface has $$S^1$$ as border and there is a diffeomorphism $$\phi$$ as the one you request.
However, the Möbius strip cannot be embedded into $$\mathbb R^3$$ as a cylinder because the former is non-orientable while the later is orientable. So there cannot be a diffeomorphism like the $$f$$ in the question.
However, I think we would have quite a different answer if we had more flexibility in the $$z$$ component of $$f$$. Were this the case, I am at this moment inclined to believe that all manifolds with $$S^1$$ as border can be embedded into $$\mathbb R^3$$, but at this moment I can think of no proof of this statement.
• Right, I mixed it up. What if we remove an open disc from the Möbius strip and we glue it with the $U$ from the problem? Commented Jun 14, 2019 at 20:59
• The manifold can be embedded in $\mathbb{R}^3$ by assumption. I had to choose a number to be able to properly define the parametrisation; however, as far as I'm concerned, manifolds embedded in \$\mathbb{R}^4} are fine too. I edited my original question to generalise. Commented Jun 14, 2019 at 21:50