Example of knot that "becomes unknotted in a different 3-manifold"?

Question

Suppose I had a knot $K \subset \mathbb{R}^3$, and an open ball $U$ inside a $3$-manifold $M$. If I now consider $K \subset U$ (via some homeomorphism of $\mathbb{R}^3$ with $U$), could it be possible that $K$ is unknotted in $M$? That is to say, $K$ becomes isotopic to the unknot in $U$; somehow "using the additional geometry that $M$ provides to make unknotting easier"? If so, are there any easy examples I can write down?

Answer 1

No, this isn't possible. Let $S$ be the boundary of the closure of $U$, which is a $2$-sphere in $M$. If $K$ is unknotted, that means there is an embedded disk $D\subset M$ whose boundary is $K$. We can isotope $D$ so that it intersects $S$ transversely. The intersection $D\cap S$ is a disjoint union of circles. Let's assume $D$ is chosen to minimize the number of circles in this intersection. If there is a circle, then there is one that bounds a disk $D'\subset S$ whose interior contains none of these circles. Compressing $D$ along $D'$ (and then throwing away any $2$-sphere components) yields a new disk whose boundary is $K$ and which intersects $S$ in fewer circles, which is a contradiction. Therefore, $D\subset U$, and $K$ is an "unknot in $\mathbb{R}^3$."