Note: One is able to do puzzles 2 and 3 without reading or understanding the text before them.
In knot theory, people define equivalence of knots using the concept of ambient isotopy. Two knots (embeddings of the unit circle into the 3-dimensional Euclidean space) are equivalent, intuitively, if they could be modified to look like each other if they were made of very elastic rope. Thus the intuitive idea is clear. However any of the common definitions of continuous deformations are not usually suitable here:
Any two knots are homeomorphic to each other (as the homeomorphic images of the circle).
Any two knots are homotopic.
Any two knots are isotopic!
Isotopy means the following: two mappings
[tex]f\colon X\to Y\text{ and }g\colon X\to Y[/tex]
are isotopic if there exists a homotopy [tex]H\colon X\times I\to Y[/tex] ([tex]I[/tex] is the unit interval) such that for all [tex]t\in I,\,H\restriction(X\times \{t\})[/tex] is an embedding from [tex]f[/tex] to [tex]g[/tex], i.e. [tex]H\restriction X\times\{0\}=f\circ i^{-1}[/tex] and [tex]H\restriction X\times\{1\}=g\circ j^{-1}[/tex], where: [tex]i,j[/tex] are (the obvious) inclusions [tex]X\hookrightarrow X\times \{0\}[/tex] and [tex]X\hookrightarrow X\times \{1\}[/tex].
Puzzle 1. Show that any two knots, i.e. embeddings [tex]f\colon S^1\to \mathbb{R}^3[/tex] and [tex]g\colon S^1\to \mathbb{R}^3[/tex] are isotopic. Hint: Use a heuristic argument. I mean, rigorous proof is not what is expected here.
We would like to define something which would be like isotopy, but still distinguish different knots! The solution is ambient isotopy. Instead of moving the knots alone, we move the whole space in which they sit and try to move one knot onto the other in this way. This leads to the definition equivalent to the intuitive definition given above.
Thus, two maps
[tex]f\colon X\to Y\text{ and }g\colon X\to Y[/tex]
are ambient isotopic, if there is a homeomorphism [tex]H\colon Y\to Y[/tex] which is isotopic (by the definition above) to the identity map [tex]id_Y[/tex] with the property that [tex]g=H\circ f.[/tex]
Now, knot theory is seeking ways to prove that two knots are not equivalent and algorithms to check whether or not / verify that two knots are equal. For instance knot theory can prove that the knots
are not equal.
Knots can be generalized to links and other embeddings: Embeddings of multiple circles at the same time (links), embeddings of graphs into [tex]\mathbb{R}^3[/tex] and embedding of two-manifolds to [tex]\mathbb{R}^3[/tex]. The following links are not equal:
and the following embeddings of graphs are not equal:
For the possible surprise of the reader, I state the following two puzzles concerning the last mentioned generalization, embedding of two-manifolds in [tex]\mathbb{R}^3[/tex]:
Puzzle 2. Show that the following two embeddings of the genus two surface are equivalent (ambient isotopic; continuously deformable into one another):
Puzzle 3. Show that the following two embeddings of the genus two surface together with a torus are equivalent (ambient isotopic; continuously deformable into one another):
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