this is elementary question about classical knot equivalence.
I know that just isotopy which need not to be ambient is not proper to define knot equivalence because bachelor's unknotting.
but this example is unsatisfied to me, because this unknotting isotopy is not differentiable.
If the isotopy is given "smooth" condition, it seems to define knot equivalence intuitively. i.e $ F(x,t): S^1 \times [0,1] \to S^3 $ s.t $F(x,t)$ is smooth map and level restriction $F(x,t_0)$ is embedding. Can it define knot equivalence?
I think it may be false.
Any reference is grateful to me.