The question is:
Let $K_{1}$,$K_{2}$ be two knots in $\mathbb{R^3}$. Let $T: \mathbb{R^3} \rightarrow \mathbb{R^3}$ be an invertible linear transformation such that $T(K_{1}) = K_{2}$. Show that $K_{1} \cong K_{2}$.
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:
I want to find a homeomorphism $f$, such that $f(K_{1}) = K_{2}$. I think the given linear transformation is that homeomorphism, and it is one-one because it is invertible, but I do not know why it should be onto (as the definition of homeomorphism requires the function to be bijection) could anyone explain this for me please?
Also I think that the given linear transformation is bicontinuous because cutting the ropes of the knotes are not allowed, am I correct?