# The definition of a knot

The definition of a knot is an injective piecewise linear map from $S^{1}$ to $\mathbb{R}^{3}$. Isn't that equivalent to a subset of $\mathbb{R}^{3}$ homeomorphic to $S^{1}$ that is piecewise linear?

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What does it mean for a subset of $\mathbb R^3$ to be piecewise linear? –  joriki Dec 2 '12 at 12:28
This is not the definition of a knot. A knot is an equivalence class of such things up to ambient isotopy. –  Qiaochu Yuan Dec 2 '12 at 23:14
You are asking whether a function is equivalent its range. One can often get away with confusing the two, that is, one can often refer to the one when really meaning the other and not get into any difficulties, but at bottom they are not the same thing. Many different functions can have the same range, and the flexibility available in choosing the function can be useful. –  Gerry Myerson Dec 3 '12 at 11:27

Let us rephrase your question in the following way: "The definition of a knot is an embedding of $S^1$ in $\mathbb{R}^3$. Isn't that equivalent to a submanifold of $\mathbb{R}^3$ isomorphic to $S^1$?"

The answer is no. A submanifold can be embedded in different ways: Compose a given embedding with any isomorphism of $S^1$ that isn't the identity and you get another embedding.

Note however, that two knots $a, b: S^1 \to \mathbb{R}^3$ are being considered equivalent (with some mathematicians even equal) if there is an isotopy $H: \mathbb{R}^3 \times [0, 1] \to \mathbb{R}^3$, such that $a = H_1 \circ b$.

We could try to recreate this notion of equivalence for your second definition: Two "knots" $a, b \subset S^1$ are equivalent if there is an isotopy $H: \mathbb{R}^3 \times [0, 1] \to \mathbb{R}^3$, such that $H_1(a) = b$.

So if two knots are equivalent in the first definition, they are equivalent in the second definition. I'm not sure if the converse is true, i.e. whether there is an ambient isotopy for every automorphism of the embedded manifold, but I would believe so. So if you consider equivalence classes of knots, it shouldn't matter which definition you use, although I recommend the first one.

• Often, knots are considered as embeddings of $S^1$ into $S^3$, which is the one-point-compactification of $\mathbb{R}^3$. My answer applies to that definition equally.