Let $f:(D^2,S^1)\to(D^4,S^3)$ be a smooth embedding (so called a slice disk), and we set $M:=f(D^2)$. Then, is the restriction map $C^{\infty}(D^4)\to C^{\infty}(M)$ open map with relative to $C^2$ topology?
If we can prove it, I think the Slice-Ribbon conjecture will be solved, because we can approximate a height function of $D^4$ such that the restriction to $M$ is a Morse function, and after cancelling all 2-handles of $M$, we can say that $M$ is a ribbon.
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When thinking about this problem notice the following, there are plenty of proper smooth embeddings of disks which cannot be approximated by ribbon embeddings. Just take any disk and connect sum it with a non-trivial 2-knot in $B^4$.
Yes you can get the knot to be embedded so that the height function restricts to a Morse function on the disk but that doesn't seem to help. You can use cancellation to get a Morse function with only one critical point, but there is no way to make sense of this ambiently. In fact it should be somewhat obvious why the only knot bounding a disk with a single critical point after restricting a Morse function on $B^4$ is the unknot.
I think in Scharlemann's approach to the Shoenflies problem there is some handle cancellation being done ambiently, but this is an entirely different animal compared to the slice-ribbon conjecture.