Write the 4-ball as $\mathbb{D}^4=\mathbb{D}^2\times \mathbb{D}^2$. Then its boundary $\mathbb{S}^3\simeq \mathbb{S}^1\times \mathbb{D}^2\cup \mathbb{D}^2\times \mathbb{S}^1$. We will use implicitely this homeomorphism.
Consider the unknot $K=\mathbb{S}^1\times\{0\}$ in the boundary $\partial \mathbb{D}^4\simeq\mathbb{S}^3$ of the four ball.
Notice that the disk $D = \mathbb{D}^2\times \{0\}$ is a smooth slice disk for $K$.
Is $D$ boundary parallel? I.e. is it obtained by pushing an unknotting disk $D'\subset \mathbb{S}^3,$ $\partial D' = K$ inside $int(\mathbb{D}^4)?$
A possible approach: take as $D'\subset \partial \mathbb{D}^4$ the PL disk obtained by gluing the annulus $\mathbb{S}^1\times ([-1,1]\times \{0\})\in \mathbb{S}^1\times \mathbb{D}^2\subset\partial \mathbb{D}^4$ to the disk $\mathbb{D}^2\times (\{-1\}\times \{0\})\in \mathbb{D}^2\times \mathbb{S}^1\subset\partial \mathbb{D}^4$. Then gluing $D'$ to $D$ we get an embedded 2-sphere. If we manage to prove that this sphere is unknotted in $\mathbb{D}^4$, i.e. it bounds a 3-ball then we can use the latter to push $D'$ to $D$.
Relevance of this problem
In studying Kirby calculus, one finds often the claim that when you attach a $2$-handle, the cocore of a $2$-handle is an unknotted $2$-disk, i.e. boundary parallel.
The above problem is a possible way to prove this.
