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Statement 1: Knots of opposite chirality have ambient isotopy, but not regular isotopy.

Statement 2: We can then define two such knots to be equivalent if they are ambient isotopic, meaning that there exists an (orientation-preserving) homeomorphism $\mathbb{R}^3\to\mathbb{R}^3$ carrying one to the other.

Am wondering if these two statements contradict each other: How could an orientation-preserving homeomorphism $\mathbb{R}^3\to\mathbb{R}^3$ always carry one knot to its mirror image? Is it because ambient isotopy of knots are defined differently in the two statements? Which definition is "standard", or more commonly accepted?

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    $\begingroup$ I have no idea what Mathworld is talking about there. The equivalence relation defined by smooth isotopy of embeddings is the same thing as the equivalence relation defined by orientation-preserving homeomorphisms of $\mathbb R^3$. See my answer here. $\endgroup$
    – user98602
    Jul 17 '15 at 18:16
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The key reference on regular isotopy seems to be the following paper:

Kauffman, Louis H. “An invariant of regular isotopy.” Transactions of the American Mathematical Society 318, no. 2 (1990): 417–471.

Note that regular isotopy is an equivalence relation on knots diagrams, not the knots themselves.

The statements on MathWorld don't make a whole lot of sense, so I'm not quite sure how to respond to the rest of your question. It is possible for a knot to be ambient isotopic to its mirror image, e.g. the figure-eight knot-- see chiral knot on Wikipedia. There is only one commonly used definition of ambient isotopy of knots.

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