Existence of Dowker Notation

Question

If a knot diagram is oriented, and we follow this orientation, labelling each crossing with consecutive integers, then each crossing will be numbered with an even and an odd number. From this labelling we get Dowker's notation. However I am having trouble trying to prove that we are guaranteed an even and odd pair at each crossing. Why is this guaranteed?

Answer 1

Given the shadow of a knot diagram (an immersion of a curve in the plane with normal crossings), then you can checkerboard color the regions in the complement of the curve, meaning each region is colored white or black such that for every arc, the two regions on either side are given different colors:

Consider a crossing along with an interval of the curve that starts and ends at that crossing, denoted in blue here:

If we record the colors appearing on one side of the curve as we walk along it, then we will see (like in this example) white, black, white, ..., black, white. The first color is the same as the last color since they involve the two incident arcs on the same region. Hence, every such interval is composed of an odd-length sequence of arcs.

The integer labeling for Dowker-Thistlethwaite notation starts with $1$, and the integers(s) you put down at each crossing correspond to the number of arcs you've so far traveled along. Thus, the second number you put down is the first number plus an odd number, which is odd if the first number was even and even if the first number was odd.