TITLE: Equivalence of knots: ambient isotopy vs. homeomorphism
QUESTION [17 upvotes]: I am looking into knot theory and have found two different definitions stating that two knots $K_1$ and $K_2$ are equivalent, namely the concept of an ambient isotopy:
These two knots are ambient isotopic if there is an isotopy $h:\mathbb{R}^3 \times [0,1]\to \mathbb{R}^3$ such that $h(K_1,0)=K_1$ and $h(K_1,1)=K_2$.
Or the other definition takes a homeomorphism $g:\mathbb{R}^3\to\mathbb{R}^3$ that maps $K_1$ to $K_2$ (ie. $h(K_1) = K_2$).
I guess these definitions are not identical and I suppose this has something to do with orientations of your knots. However I find it a bit hard to get my head around it. So I hoped someone could either prove I'm right by giving a counterexample or prove I'm wrong (ie. prove these concept are identical)

REPLY [5 votes]: Milnor's Topology from the Differentiable Viewpoint (page 34) has a very elementary proof: 
Filling in the blanks:

We may assume that $f(0) = 0$: Translation is an isotopy.
We write $f$ in the form $f(x) = x_1g_1(x) + \cdots + x_mg_m(x)$: Hadamard's Lemma.
...the linear mapping $df_0$, which is clearly isotopic to the identity: $GL^+(m)$ is connected.

In symbols, the lemma says $\operatorname{Diff}^+(\mathbf R^m) \subseteq \operatorname{Diff}_0(\mathbf R^m)$. But it's easy to see that the reverse inclusion also holds. Thus, the main result here is $\operatorname{Diff}^+(\mathbf R^m) = \operatorname{Diff}_0(\mathbf R^m)$.