# Equivalence of knots

It's intuitively clear what it means that two knots $K,K'$ are essentially the same, but it can be termed and defined more precisely in different ways. Are all of them equivalent?

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The knots themselves are always homotopy equivalent - indeed, homeomorphic - because they are all homeomorphic to $\mathbb{S}^1$. Perhaps you want to ask if the knot complements are homotopy equivalent? –  Zev Chonoles Feb 28 '13 at 17:50
@Zev: You see, I did not understand the concept of homotopy (equivalence) fully, yet. I believed that two knots are h-equivalent if they can be continuously deformed into each other. –  Hans Stricker Feb 28 '13 at 17:55
It depends on how you interpret "continuously deformed into each other". For example, the circle $\mathbb{S}^1$, the cylinder $\mathbb{S}^1\times\mathbb{R}^1$, and the solid torus $\mathbb{S}^1\times\mathbb{D}^2$ are all homotopy equivalent to each other; homotopy equivalence does not detect "contractible thickness", in other words. I think the intuitive idea you are referring to with the phrase "continuously deformed into each other" is the one captured by the notion of ambient isotopy, because it requires that the homotopy "keep track of" how the spaces are embedded in the ambient space. –  Zev Chonoles Feb 28 '13 at 18:00
@HansStricker As an aside - I think "being continuously deformed into each other" is a perfectly good way of thinking about homotopy, providing you remember that in this continuous deformation the object is allowed to pass through itself, so all knots can be untied. When you don't permit the object to self-intersect, you get an intuitive notion of isotopy. –  Matthew Pressland Feb 28 '13 at 18:10
Knots are ambient isotopy classes of embeddings of $S^1$ into things. –  Qiaochu Yuan Feb 28 '13 at 18:10

Isotopy (as opposed to ambient isotopy) is not a good equivalence relation on knots, because any knot is isotopic to an unknot. The idea is to pull the knot tighter and tighter until it shrinks to a point. More precisely (but still not completely precisely, because that would be a mess to write down), the isotopy would start with the given knot at time $t=0$. At a later time $t$ (in $(0,1)$), it would look like a $t$ fraction of a circle (unknotted) with a copy of the original knot compressed into a $(1-t)$-sized part. Finally, at time $t=0$, it's the whole unknotted circle.
That's obviously consistent with your other comment: X = knot, Y = ambient isotopy. But that's for embeddings of $S^1$. In my comment to Andreas' answer, I was drifting away to embeddings of surfaces. –  Hans Stricker Mar 1 '13 at 22:23