Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 112 / 3 / e / Solution

Let $D\subset B^4$ be a slice disk for $K$. Its normal bundle is trivial. A small normal push-off $D^{\prime }$ is disjoint from $D$, and its boundary is the zero-framed Seifert longitude ${\nu }_K$. Hence the two-component link $K_{2,0}=K\cup {\nu }_K$ bounds the pair of disjoint disks $D\cup D^{\prime }$.

Orient $D^{\prime }$ oppositely to $D$. In the other hemisphere of $S^4$, join their boundary circles by the product annulus supplied by the zero framing. The unionis a two-sphere. More concretely, it is the rounded boundary of the three-ball $D\times [-\epsilon ,\epsilon ]$, so it is unknotted. Its equatorial intersection is $K\cup {\nu }_K$, and both link components lie on the same sphere. Thus $K_{2,0}$ is doubly slice as a colored link with the trivial coloring.

Solved by gpt-5.6-sol high.