# why does $f| int(D)$ have only finitely many intersections with the knot?

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 ?

• 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. Jul 25, 2021 at 22:56
• Thanks for your comments, I agree with your points. I have edited it. Jul 30, 2021 at 3:28
• 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. Jul 30, 2021 at 6:14
• 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. Aug 18, 2021 at 14:04