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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?

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    $\begingroup$ For something like this, you can induct on the number of crossings. More generally, you might want to take a look at Lickorish's book. One method, for example, is to show that a knot invariant is well-defined is to show that it's invariant under Reidemeister moves. $\endgroup$ – anomaly Mar 23 '18 at 2:25
  • $\begingroup$ I would induct on the number of crossings. Reduce $k$ crossings of a link to $k-1$ crossings by using skein relations. $\endgroup$ – Mee Seong Im Mar 23 '18 at 2:30
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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?

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  • $\begingroup$ Are knots unlink? $\endgroup$ – user398843 Mar 23 '18 at 3:29
  • $\begingroup$ @user398843 : A knot can be self-linked. Can you show that there is some subset of its crossings that if switched would render it the unknot? This might be a good place to induct on the number of crossings. $\endgroup$ – Eric Towers Mar 23 '18 at 3:37
  • $\begingroup$ If given a specific link, I can do that. For example if we want to prove a one component link can be change into unlink, do we need to deal with all the knots? Because to change a trefoil and a 5-1 knot (katlas.math.toronto.edu/wiki/5_) to circles, we need to change different numbers of crossings. $\endgroup$ – user398843 Mar 23 '18 at 4:01
  • $\begingroup$ @user398843 : It's unclear to me that you are distinguishing between popping each link component out of the link from un-self-linking each component after having done so. To what I think your question is: Can you un-self-link all 0-crossing knots? Given an $n>0$-crossing knot, can you turn it into an $n-1$-crossing knot? $\endgroup$ – Eric Towers Mar 23 '18 at 14:27
  • $\begingroup$ Given an $n>0$-crossing knot, we can use R1 move or change one crossing, then use R2 move to turn it into an $n−1$-crossing knot. So you want to say the crossing is finite? then we can do this finitely many times and get an unknot? Would you mind elaborating in moire detail? $\endgroup$ – user398843 Mar 24 '18 at 1:23
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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.

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