# Boundary of slice disk exterior is the zero surgery of slice knot

I couldn't exactly guess the level of question. I asked in Math Stack Exchange. (Depending on the situation, I will remove it from here.)

I'm trying to understand a sketch of proof of Livingston and Naik's draft book Introduction to Knot Concordance which is displayed in the following picture. Using some general topological identities, I couldn't follow why

$$\partial X(D) = (\partial B^4 - (Int(N(D)) \cap \partial B^4)) \cup (\partial N(D) \cap Int(B^4))$$ ?

$$\partial N(D) = (D \times \partial B^2) \cup (\partial D \times B^2)$$ implies $$\overline{\partial N(D) \cap Int(B^4)} = D \times B^2$$ ?

Any help will be appreciated.

• When you are taking out a neighbourhood of a Disk which intersecting the the boundary along a knot, then you are taking out a nbd of the knot from the bounday, but the interior of the disk helps you to add some more part in the torus boundary of the knot(which is 0 surgery essentially). Try to draw two dimensional analogous pic,.ie, start with a square and see what happen if you take out a nbd of an arc which intersect the bounday at exactly two points. Also if you are famalier with Kirby calculus, try to recall cancelling 1-2 pairs and dotted circle notation.(Gompf and Stipsicz page 168) – Anubhav Mukherjee Nov 12 '19 at 17:08
• I’m familiar with handlebody analogue of the proof. I’m just curious about this point-set topological proof. – Diego Hernández Rodríguez Nov 12 '19 at 19:04