# Is unknot a composite knot?

I am reading "The-knot-book-an-elementary-introduction-to-the-mathematical-theory-of-knots" by Collins C. Adams. When I was reading in introduction chapter in composite knot subchapter, I came up with idea that if we somewhat composite 2 or more factor knots each another under rules of composition, we can end up with unknot knot, that is you composite knots in such a way that composition of these knot remove each other to become unknot knot. However, in the book, after 2 page where I thought this idea there is written that there is no way to composite knots and end up with unknot knot.

I did not understand the justification in the book, can someone explain the main idea and show that the given explanation is enought to see there is no composition to make unknot from un-unknots [$*$]

How can we show(prove, or understand intiutively) there is no composition of un-unknots end up with unknot?[My Question]

$*$(the explanation from the book,"The-knot-book-an-elementary-introduction-to-the-mathematical-theory-of-knots" by Collins C. Adams, page 9 ):

As noted in the comments, the fact that the unknot is not composite follows from additivity of the Seifert genus of knots. Here's an overview of how this goes.

A Seifert surface of a knot $K$ is a compact, connected, oriented surface $F$ with boundary $K$. The Seifert genus $g(K)$ of a knot $K$ is the minimum genus of any Seifert surface of a knot. The only knot with Seifert genus zero is the unknot.

Theorem. Let $K_1$ and $K_2$ be two knots, and let $K_1\# K_2$ be their connected sum. Then $$g(K_1 \# K_2) = g(K_1) + g(K_2).$$

For now, let's assume the theorem.

Corollary. The unknot is not composite, that is, if $U$ is the unknot and $U=K_1\# K_2$, then $K_1=K_2=U$.

Proof of Corollary. Suppose that $U=K_1 \# K_2$. Then by the theorem $$0=g(U) = g(K_1\# K_2) = g(K_1) + g(K_2).$$ Thus $g(K_1) = g(K_2) = 0$, and since the unknot is the only genus zero knot, it follows that $K_1=K_2=U$. $$\tag*{\blacksquare}$$

Now let's give a sketch of the proof of the theorem that Seifert genus is additive.

Sketch of proof of Theorem. Let $K_1$ and $K_2$ be two knots with minimal genus Seifert surfaces $F_1$ and $F_2$ respectively.

First, we show that $g(K_1) + g(K_2) \geq g(K_1\# K_2)$. Embed the knots $K_1$ and $K_2$ and their Seifert surfaces $F_1$ and $F_2$ into $S^3$ so that there is a $2$-sphere $\Sigma$ with $K_1$ and $F_1$ on one side and $K_2$ and $F_2$ on the other. Draw an arc between $K_1$ and $K_2$ that misses the Seifert surfaces. Fattening that arc into a band yields a single surface $F$ whose boundary is $K_1\# K_2$. Moreover, the genus of $F$ is the sum of the genera of $F_1$ and $F_2$. So $K_1\# K_2$ has a Seifert surface of genus $g(K_1) + g(K_2)$, and hence $g(K_1) + g(K_2) \geq g(K_1\# K_2)$.

Now we show that $g(K_1) + g(K_2) \leq g(K_1\# K_2)$. Let $F$ be a minimal genus Seifert surface for $K_1\# K_2$, and let $\Sigma$ be a $2$-sphere intersecting $K_1 \# K_2$ in exactly two points $p$ and $q$ with $K_1$ on one side of $\Sigma$ and $K_2$ on the other, as below.

If we put $\Sigma$ into general position, then $F\cap \Sigma$ is a finite collection of simple closed curves and an arc $\beta$ from $p$ to $q$.

Choose an innermost simple closed curve $C$ in $F\cap \Sigma$. The curve $C$ must bound a disk in the Seifert surface $F$. If it did not bound a disk, then one could chop $F$ along a small cylinder with $C$ as a core and cap off the new boundary components of $F$ to form a surface $F'$. If $C$ does not bound a disk, then $F'$ is connected and has genus $g(K_1\# K_2) - 1$, contradicting that $F$ has minimal genus. So a neighborhood of $C$ in $F$ must look like below.

Since $C$ bounds a disk, we can push that disk to one side of $\Sigma$ removing that circle of intersection. Continue this process until the only thing in the intersection $F\cap \Sigma$ is the arc $\beta$ from $p$ to $q$. Following the knot $K_1 \# K_2$ from $p$ to $q$ and then following the arc $\beta$ from $q$ to $p$ yields the knot $K_1$, and following the knot $K_1 \# K_2$ from $q$ to $p$ and the arc $\beta$ from $p$ to $q$ yields the knot $K_2$. Moreover, the arc $\beta$ chops the Seifert surface $F$ into Seifert surfaces $F_1$ and $F_2$ for $K_1$ and $K_2$ respectively. The genus of $F$ is the sum of the genera of $F_1$ and $F_2$. Hence $g(K_1) + g(K_2) \leq g(K_1 \# K_2)$, as desired. $$\tag*{\blacksquare}$$