I really recommend that you just accept this as obvious and continue, since the proof is long, nowhere near complete, and is not referred to anywhere else. I never found a reference to cite for this, although I can't imagine how one could prove without this that the three Reidemeister moves are sufficient for describing legal transformations to link diagrams.

First, one needs definitions.

- A
**knot**is a polygon embedded in threespace, that is, a set of line segments forming a cycle with line segments overlapping only at endpoints, and exactly two line segments meeting at each endpoint. - A
**directed knot**is a knot where the cycle is directed, that is, each endpoint has one input and one output. - A
**link**is a set of disjoint knots. - A
**directed link**is a link in which all the knots are directed. - link1 and link2 are
**equivalent**if there is a continuous mapping f:R^3x[0,1] -\gt R^3 such that f(link1,0) = link1 and f(link1,1) = link2 and f(*,x):R^3 -\gt R^3 is bijective for all x in [0,1]. - A
**projection**is, um, the standard mathematical projection of 3-space onto 2-space, except we require that no more than two line segments be projected to any single point in 2-space. This is possible because any three line segments form a one-dimensional set of projections that maps some point of all three to a single point; there are a finite number of such one-dimensional sets but a 2-space of projections to choose from. - A
**crossing**in a projection is a point in 2-space that two points in the link map to. Note that the two points will be on the interiors of two distinct line segments. - An
**overpass**, first, choose "up" and "down" in the direction that was eliminated in the projection. Given a crossing in a projection, the**overpass**is the string segment in the link with the top point, the**underpass**is the string segment in the link with the bottom point.

Claim: Any two links with the same projection and the same overpass information are equivalent.

- Given the common projection, at each crossing, choose a point in 3-space at height 1 for the overpass and height 0 for the underpass.
- Given the string segment from the projection connecting crossings x and y with projected length s, for a point at distance t from x along the projected string segment, choose the point at height height(x)+((height(y)-height(x))*(t/s)), where height(x) is 1 if the string segment is an overpass at x, 0 otherwise, similarly for y.
- You now have a third link with the same projection (and link diagram) as the first two.
- Define f(link1,0)=link1, f(link1,.5) to be this third link, and f(link1,1)=link2. Bleah, I need to specify the smooth interpolation of the link edges and also of the surrounding 3-space. There is some minimal distance that all line segments are apart. This can all be done, it's just ugly.

Claim: If two links have projections with the same crossings, the same overpass information, and crossings are connected the same way, then the two links are equivalent.

This is the hardest part. The connecting arcs could be all over the plane, mixed up like kneaded dough. I believe a reasonable strategy is to unwind both knots until the connecting arcs are going more or less directly to where they ought to go, then to transform one knot into the other. I've never spelled out this part of the proof.

Claim: If two links have the same link diagram, then they can be converted to having projections with the same crossings, the same overpass information, and the crossings connected in the same way, so they are equivalent.

This requires arguing about embedding planar graphs in the plane, since the link diagram is a planar graph (plus crossing information), but it is not an embedded planar graph. It also requires a definition of a link diagram.

Claim: The three Reidemeister moves are sufficient.

This is similar to the argument that you can choose projections where no more than two line segments are mapped to any point. The Reidemeister moves are a list of ways that the link diagram can change by moving through a projection that has one point with three line segments mapping to it. This can happen in only a few ways:

- One line segment being on end (type I)
- The endpoint of two line segments colliding with the middle of another line segment (type I or II)
- Two endpoints colliding (type I or II or nothing)
- The interiors of three line segments colliding (type III)