The question is:
If $K_{1}$,$K_{2}$ are amphicherial knots. Show that $K_{1} \cong K_{2}$.
The definition of amphicheiral is as follows:
A knot $K$ is amphicheiral iff there exists an orientation preserving homeomorphism of $\mathbb{R^3}$ onto itself which maps $K$ onto its mirror image.
The definition of mirror image of a knot $K$ is as follows:
By the mirror image of a knot $K$ we mean the image of $K$ under the reflection $\mathscr{R}$ defined by $(x, y, z) \rightarrow (x, y, -z).$
The meaning of $K_{1} \cong K_{2}$, is given below:
The definition of equivalence of 2 knots according to Richard H. Crowell and Ralph H. Fox, edition 1963, is:
Assume that $K_{1}$,$K_{2}$ are 2 knots in $\mathbb{R^3}$, then they are equivalent , denoted by $K_{1} \cong K_{2}$, iff $\exists f: \mathbb{R^3} \rightarrow \mathbb{R^3}$, where $f$ is a homeomorphism and such that $f(K_{1}) = K_{2}.$
My thoughts:
By the definition of amphicherality there exists an orientation preserving homeomorphism of $\mathbb{R^3}$ onto itself which maps $K$ (take it to be $K_{1}$) onto its mirror image (take it to be $K_{2}$) and hence the equivalence definition is satisfied and we are done…… am I correct?
EDIT:
I think I am incorrect as the given definition of amphicherality deals with only one knot ….. am I correct?
Best Answer
This is just a long comment. Let $K$ be a knot and $-K$ be its mirror image. A knot $K$ is said to be amphichieral if it is equivalent to $-K$. So if you have two knots $K_1$ an $K_2$, they can only be amphichieral if they are already equivalent, but neither need be the mirror image of the other. Often, questions likes these are phrased in terms of their diagrams.