I am currently doing a course in Knot Theory and after looking at different texts I have found many ways to define knots and knot equivalence. In our course we are given the following definitions:
A knot $K\subseteq\mathbb{R}^{3}$ is a subset of the form $K=f(\mathbb{R})$ where $f:\mathbb{R}\rightarrow\mathbb{R}^{3}$ is a map with the following properties.
- $f(x_{1})=f(x_{2})$ if and only $x_{1}-x_{2}\in\mathbb{Z}$
- The derivative $\frac{d^{n}}{dx^{n}}f$ exists for all $n\in\mathbb{N}$
- The derivative is not zero everywhere, $\frac{df}{dx}\neq 0$
Knots $K$ and $K^{\prime}$ are equivalent if there exists a collection of knots $K_{x}$ for each $x\in\left[ 0,1\right]$. such that $K_{0}=K$ and $K_{1}=K^{\prime}$, and such that there exists a smooth map $F:\left[ 0,1\right]\times \mathbb{R}\rightarrow\mathbb{R}^{3}$ with the property that $F(x,\cdot )$ represents $K_{x}$ in the sense of the previous definition.
I think I understand intuitively what is happening here and that this smooth map ensures that we make no illegal 'jumps'. I am at the same time taking a course in Algebraic Topology and for the sake of my note keeping I would like to give an alternative definition of equivalence more in line of homeomorphisms from Topology, which If im not mistaken is essentially what is going on here.
So I would say that two knots $K$ and $K^{\prime}$ are equivalent if there exists a homeomorphism $f:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ for which $f(K)=K^{\prime}$.
Would this definition be satisfactory, if not how could I change it? Is there any substantial difference from my definition and the one given by my lecturer?
As was pointed out in the comments, you need to differntiate mirror images to agree with your first definition. This is fixed by asking for an orientation-preserving homeomorphism $f: \mathbb{R}^3 \to \mathbb{R}^3$. See https://en.wikipedia.org/wiki/Knot_theory#Knot_equivalence, where it is also (briefly) explained why these two notions of equivalence are the same.