How do we write a proof in knot theory?

Question

For example, show that any diagram of a link can be changed into a diagram of the unlink by suitable crossing changes.

I am reading some Knot Theory book, but I usually have no idea how to start a proof. For example, to prove the above question. How can I represent any diagram of a link? The statement is understandable, but what a rigorous proof looks like?

Answer 1

Very, very frequently, by induction. In this case, on the number of not-yet-unlinked link components. Can you show the result for a one component link? For the inductive step, can you pick an arbitrary component of a link and make it lie in front of the rest of the link, by switching a subset of its crossings?

Answer 2

A knot shadow is a knot diagram without the over/under information, and you can think of it as a particular embedding of a $4$-regular graph in the plane. The knot shadow can be given as an immersion $s:S^1\to \mathbb{R}^2$. Let $a\in S^1$ be a point where $s(a)$ is not a crossing. There is a (Morse) function $f:S^1\to \mathbb{R}$ where $f(a)=0$, $f(a-\epsilon)=1$ for an $\epsilon>0$ small enough so that $f([a-\epsilon,a])$ contains no crossings, and where $a$ and $a-\epsilon$ are the only two critical points. That is, $f$ is strictly increasing from $a$ to $a-\epsilon$, and then it quickly strictly decreases from $a-\epsilon$ to $a$.

Claim: the map $h:S^1\to \mathbb{R}^3$ defined by $h(t)=(s(t), f(t))$ is a knot, and it is unknotted.

First, it is a knot because if $h(t_1)=h(t_2)$ then $s(t_1)=s(t_2)$ and $f(t_1)=f(t_2)$. If $s(t_1)=s(t_2)$, then this is a crossing point in the knot shadow. But at crossing points we have guaranteed that $f(t_1)=f(t_2)$ implies $t_1=t_2$. Thus $h$ is an injective immersion, which since $S^1$ is compact means $h$ is an embedding.

Second, we can construct a reasonably explicit disk that this knot $h$ bounds. Notice that each cross-section $h(S^1)\cap (\mathbb{R}^2\times\{z\})$ for $0<z<1$ consists of two points, one from $h((a,a-\epsilon))$ and another from $h((a-\epsilon,a))$. So there is a straight-line path between these two points that does not intersect any other point of the knot. By taking the union of these paths for all $0<z<1$ along with all of $h(S^1)$, one gets an embedded disk in $\mathbb{R}^3$ whose boundary is $h(S^1)$. (I won't check this here.)

Therefore, every knot shadow is from the knot diagram of an unknot.

So: given a knot, take its knot shadow, and construct an unknot with the same shadow. Change all the crossings of the original knot to match the unknot's crossings. Therefore every knot can be unknotted by changing some crossings in any diagram of it.