# Proof that knot genus is a knot invariant

I have a proof of the the following fact concerning knot genus, but I'm not sure that it's correct.

If knot $J$ is isotopic to another knot $K$ then $J$ and $K$ have the same genus.

Proof. Let $f_t : \mathbb{R}^3 \times [0, 1] \to \mathbb{R}^3$ be the ambient isotopy such that $f(J,1) = K$. In particular, the map $f_1(x) = f(x,1)$ is a homeomorphism.

Let $g(J)$ denote the genus of $J$, and let $S$ be a Seifert surface of $J$ such that $S$ has genus $g(J)$. Then $f_1(S)$ is connected, orientable and has boundary $K$, so it is a Seifert surface of $K$. In this way we show that $g(J) \geq g(K)$.

Let $T$ be a Seifert surface of $K$ such that $T$ has genus $g(K)$. Then likewise $f_1^{-1}(T)$ is a Seifert surface of $J$, thus $g(J) \leq g(K)$. Therefore we conclude that $g(J) = g(K)$.

The proof seems a bit too simple; did I make a mistake somewhere?

## 1 Answer

This proof is fine. You may like to also mention why $f_1(S)$ is still an embedded surface, but this is an easy conclusion from the fact that $f_1$ is a homeomorphism. Other than that, I don't see anything missing.