# Trying to explain the unknotting number of $7_5$

I am trying to proof that the unknotting number of prime knot $$7_5$$ is $$2$$.

For this purpose, I am studying the minimal number of crossings that need to be changed in the knot in order to get a diagram of the trivial knot.

However, I am stuck when I change the following crossing (circled in red):

I know that I get a knot with five crossings, but I am not capable of distinguish which of the following is:

How do I proceed?

We can cheat by using a computer tool. I'm using KnotFolio. I took your drawing of the knot and input it:

It recognizes this as $$7_5$$ by calculating some knot invariants and using the fact that no other knot with at most 7 crossings has the same invariants.

Flipping the crossing in question,

by a similar process it recognizes it as $$5_2$$. Unfortunately, KnotFolio doesn't have any Reidemeister move tool yet, so you have to edit the diagram manually, but after applying Reidemeister II we get this:

It's still recognized as $$5_2$$. The "Beautify" tool smooths out the knot diagram (an isotopy of the diagram -- note that it does this isotopy on a sphere, so it might choose a different face to be the outer face). In this form, it's more obvious that it's $$5_2$$.

By the way, with two flipped crossings we can get the unknot:

I'm guessing you already know this and that you're systematically checking that flipping each individual crossing in your diagram of $$7_5$$ doesn't give the unknot. Beware that this alone is not sufficient in proving that the knot $$7_5$$ has unknotting number $$2$$ -- you'd only be showing that the diagram in question has unknotting number $$2$$. There might still be some diagram for $$7_5$$ that has unknotting number $$1$$!

Do the obvious Reidemeister move. Now treat it as a planar graph (with the crossings as vertices of valency $$4$$). Label each region with the number of edges that surround it.

Which of the knots with $$5$$ crossings has the same "region structure" ?