We define a compressing disk of the knot $K$ (a smooth emmbedding from $S^1$ to $S^3$) to be a smooth map $f: D → S^3$ such that $f|∂D = K$ and such that $f| int(D)$ is transverse to $K$. Then $f| int(D)$ has only finitely many intersections with the knot $K$ .

Here: "$f| int(D)$ is transverse to $K$" means that $$ \forall p \in f^{-1}(K), f_*T_p(intD)+T_{f(p)}K=T_{f(p)}S^3 $$

So the Transversality theorem tells us that the intersetions $\{p\in (int D)| f(p)\in ImK\}$ is a 0-manifold. But why it's finite ?

  • 2
    $\begingroup$ This is not a bad question, but, as written, it is incomplete in several ways. In particular: You should specify what is $K$ (presumably, a smooth knot), degree of regularity of $f$, including its behavior near the boundary. Also, when revising the question, please include your thoughts on the problem, as well as a description of your differential/geometric topology background. For now, I am voting to close. Lastly, use MathJax to write math in your questions. $\endgroup$ Jul 25, 2021 at 22:56
  • $\begingroup$ Thanks for your comments, I agree with your points. I have edited it. $\endgroup$ Jul 30, 2021 at 3:28
  • $\begingroup$ With the current definition the claim is simply false. You have to assume that $f$ is an immersion along the boundary of the disk (or something similar to it). This assumption will eliminate the possibility that the subset $f^{-1}(K)$ accumulates to the boundary of the disk. $\endgroup$ Jul 30, 2021 at 6:14
  • $\begingroup$ Note that we have the fact that $ f|\partial D$ is an embedding. But I don't know why your assumption can eliminate that possibility. $\endgroup$ Aug 18, 2021 at 14:04


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