I was reading this website: http://www.maa.org/sites/default/files/images/upload_library/23/stemkoski/knots/page1.html and I was confused by the definition of knot provided. The definition is as follows.
A $\bf{knot}$ $K$ is the image of a continuous function $f : \left[0, 1\right] \rightarrow \mathbb{R}^3$ such that $f(0) = f(1)$, $f$ restricted to $[0, 1)$ is injective and $\frac{df}{dt} \neq \vec{0}$ for any $t \in \left[0, 1\right]$. My issue with this definition of a knot is that knots are equivalent if they are ambient isotopic. The definition of ambient isotopy from the wikipedia page starts: "Let $N$ and $M$ be manifolds and $g$ and $h$ be embeddings of $N$ in $M$." Well any knot as defined above cannot be an embedding since the map $f$ is not injective since $f(0) = f(1)$ by definition.
So if we define knots as above:
- Is the definition incorrect for a knot from this website?
- Is there a different way that we should define equivalence?
- Parameterizations of knots seem like they could be nice to work with so is there a way to make this definition work?
A knot isn't an embedding of $[0,1]$ in $\mathbb R^3$, it's an embedding of $[0,1]/(0\sim 1)\cong\mathbb S^1$ in $\mathbb R^3$. Given a continuous map $f:[0,1]\to\mathbb R^3$ with $f(0)=f(1)$, there is a continuous map $\tilde f:\mathbb S^1\to\mathbb R^3$ with $f=\tilde f\circ\pi$ where $\pi:[0,1]\to\mathbb S^1$ is the quotient map. This map $\tilde f$ gives the embedding, not $f$. Maybe this is what confused you?