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I am trying to teach myself about differentiable manifolds and diffeomorphisms, and I think I am coming to grips with the topic - but am struggling with examples like the following:

Consider the cylinder C = S^1*(-1,1) and the full-twist Möbius strip F (which like the cylinder I want to think of as being embedded in three-dimensional space, but I feel like giving a specific subset would be unnecessary).

I know that C and F are homeomorphic by preforming a cut, twisting, and gluing the cut back together, but I wanted to ask something stronger. Is it possible to find a homeomorphism between C and F which does not involve any cutting and respects their embeddings (i.e. some kind of homotopy between the two), and can this be done smoothly?

I understand that this cannot be done in three-dimensional space without self-intersection, but can this morphing between C and F be achieved if we consider embedding C and F in higher-dimensional spaces?

And if possible, can someone give me an explicit example of such a diffeomorphism (or explain why it cannot be done)?

EDIT: By full-twist Möbius strip I mean a 360 degree twist, whereas the standard Möbius strip has a 180 degree twist.

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    Yes, it can be done in higher dimensions. There are fancy reasons, but better to think about an explicit approach: instead of worrying about the strips, think about their boundaries. The $2n$-twist Mobius strip has boundary two circles, "linked together" $2n$ times. You can undo these links in 4-dimensional space rather easily. Can you picture how? – 2017-01-04
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    I am familiar with orientability. And I was actually thinking about the boundaries, which I understood can be "unlinked" in four-dimensions. I got to the point where I morphed F into a "figure-eight cylinder" (where the self-intersection was sorted by a four-dimensional bump). But anything I try to think of to unfold these figure-eight boundaries seem to lead to infinitely-tight folds, which I believe are not allowed in a diffeomorphism – 2017-01-04
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    Just to clarify, it is not the linking between the boundaries I am having a problem with - but once they are unlinked, I am struggling to morph them into circles (from figure-eights) in a diffeomorphic way – 2017-01-04
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    @juanarroyo Sorry I think there is confusion in what I'm asking. When I say "full-twist Möbius strip" I am talking about a 360 degree twist (whereas the standard Möbius strip is a 180 degree twist). The standard Möbius strip is non-orientable, but the full-twist Möbius strip is. – 2017-01-04
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    My mistake I just noticed. – 2017-01-04
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    The term for the property I think you want is a smooth "ambient isotopy" between these two embeddings of the cylinder in $\mathbb{R}^4$. – 2017-01-04
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    @JamesCameron I haven't heard the term "ambient isotopy" before, but after looking it up quickly I think it is. Is it possible prove that one exists between C and F in R^4? And I assume that ambient isotopies are smooth by definition? – 2017-01-04
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    @SamForster Ambient isotopies aren't necessarily smooth by definition, you need to add that condition. I think you can prove using the Whitney approximation theorem that if two smooth embeddings are ambient isotopic, then they are smoothly ambient isotopic. If this claim is true, then you don't have to worry about the infinitely tight folds. I think in your situation you can even find an explicit smooth isotopy, maybe with a twist at the crease. – 2017-01-04

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