# Prove that every knot diagram with two crossings is equivalent to the unknot.

I am a beginner studying knot theory, and we covered the Reidemeister moves on link diagrams in class today. The question in the title is the one I am struggling with now. I attached an image of the solution to a similar problem (for diagrams with only one crossing) that was provided in class.

(I originally only had the first two diagrams drawn but was told in class that the correct answer included the additional two.)

The question I am looking to answer now: Prove that every knot diagram with two crossings is equivalent to the unknot.

So far, I have drawn four possible knot diagrams and shown they are equivalent to the unknot, but I am unsure how to proceed. The logic given for the solution in the image was that a ‘rotation is not a Reidemeister move,’ so by this logic, should I be including four additional diagrams? Any help is greatly appreciated. Thank you!

• Your instructor is not correct (although so are you). If you want to use only the first two diagrams in the 1-crossing example, you need to show that there is a PL-homeomorphism of $S^3$ that gives you the equivalent rotation of the diagram. If you have proved that then your solution is complete. Commented Oct 24, 2023 at 22:54
• For the 2-crossing version, same thing. Your instructor is presumably looking for a solution that use R2 to bring the two intersection to some "standard" position (say at $(\pm 1,0)$ with the usual diagonal directions) because he/she does not believe in PL-homeomorphism of the plane. After you have done that you just exhaust all possibilities of joining up the 4 segments. Commented Oct 24, 2023 at 22:59
• Hi, welcome to Mathematics Stack Exchange! Before posting a question, kindly read how to ask a good question and how to use LaTeX. I hope you enjoy your time being a part of this community! Commented Oct 24, 2023 at 23:37