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One can define a knot in two ways:

(1) A knot is a closed polygonal curve in $\mathbb R^3$

(2) A knot is an equivalence class of embeddings $S^1 \hookrightarrow \mathbb R^3$

And perhaps also:

(3) A knot is an equivalence class of smooth $1$-dimensional submanifolds of $S^3$

Question 1: Can I replace $S^3$ in (3) with $\mathbb R^3$?

Question 2: I would like to define what a regular projection of a knot is. Unfortunately, it depends on whether I use (1), (2) or (3). I would like to use (2) and I have the definition using (1), which goes as follows:

(Definition) A knot projection is called a regular if no three points on the knot project to the same point, and no vertex projects to the same point as any other point on the knot.

How can I define regular projection without (1), that is, how can I define it for embeddings instead of polygonal curves?

Thanks a lot!

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    @QiaochuYuan Reading the first paragraph [here](http://www.ams.org/staff/jackson/fea-nelson.pdf) suggests that one can either view knots as equivalence classes or as simple closed curves!2012-10-15

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  1. Yes, you can replace $\mathbf R^3$ with $S^3$ everywhere and it will essentially not make a difference. Your assumptions imply that the knot misses a point which you can add/remove to go back and forth between either one. This MO question might be stimulating: https://mathoverflow.net/questions/63158/

  2. A definition if the embedding is smooth (or just differentiable) is that a projection is regular if the composite map $\phi \colon S^1 \to \mathbf R^2$ has the property that no three points are mapped to the same point, and that if $\phi(p)=\phi(q)$, then nonzero tangent vectors at $p$ and $q$ are mapped to linearly independent tangent vectors under $d\phi$. In other words, the image of the knot crosses itself transversely.

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    Thank you! Would you point me to a book/notes or other reference where you got this definition from?2012-10-14