Unknot: embedded vs immersed bounding disk

Question

Suppose that we have a knot $K\subset \mathbb{S}^3$. If $K$ bounds* an embedded 2-disk then $K$ is the unknot. But what happens if $K$ bounds an immersed 2-disk?

The immersed disk generically will have two kind of singularities, double points and triple points. The double points will be closed curves while the triple points a finite number of points.

1) What is an example when $K$ bounds an immersed disk but not an embedded one?

2) Can we remove the double point singularities?

3)Are the triple points related to the genus of a Seifert surface for $K$?

*Formally, $K$ bounds an embedded (immersed) 2-disk if $\exists f:(\mathbb{D}^2,\partial \mathbb{D}^2)\to (\mathbb{S}^3,K)$ that is a smooth embedding (immersion). We can assume that in both cases $f|_{\partial \mathbb{D}^2}$ is a diffeomorphism onto the image if necessary.

Answer 1

1. Every (smooth) knot in $S^3$ bounds an immersed smooth disk in $S^3$, see e.g.

Bruce Trace, A general position theorem for surfaces in Euclidean 4-space. Combinatorial methods in topology and algebraic geometry (Rochester, N.Y., 1982), 123–137, Contemp. Math., 44, Amer. Math. Soc., Providence, RI, 1985.

2. If A knot $K\subset S^3$ is a "ribbon knot" if it bounds an immersed disk in $S^3$ with only double singularities. There exist nontrivial ribbon knots in $S^3$, in particular, you cannot remove their double intersections. In particular, there are knots which bound immersed disks with double self-intersections, which do not bound embedded disks.

Here is an example taken from Wikipedia:

Every ribbon knot bounds a smooth embedded disk in $B^4$, the unit 4-ball. In other words, every ribbon knot is "slice." It is a famous open problem if every slice knot is ribbon.

In general, triple self-intersections are more directly related to the 4-ball genus rather than the Seifert genus of a knot.