My understanding is that if the embeddings $f_0,f_1$ are tame knots then
$H(t,\theta) = (1-t)f_0(\theta) + t f_1(\theta)$
is a homotopy between them, thus all tame knots are homotopic. Is this the case?
To provide some context, I'm writing a paper on introductory knot theory and I want to use this to motivate why isotopy (where we additionally require $H(t,\theta)$ to be an embedding for all $t$) is used to define equivalence of tame knots. My supervisor thinks the above statement is wrong but after a quick exchange of emails I'm none the wiser as to how it is wrong. I'm meeting with her on monday but it would be nice to get this straightened out before then.
This follows from the fact that $\pi_{1}(\mathbb{R}^3) = 0$. In fact, any two knots are also isotopic to the un-knot by shrinking the knotted parts of the knots to a point (an embedding for all $t$). The correct idea is an ambient isotopy, since when you shrink a portion of Euclidean space to a point you do not have an embedding of Euclidean space into itself.