Knot Traces and Trace Lemma I'm currently reading the paper by Miller and Picorillo titled Knot Traces and Concordance. I have some confusion about the proof of theorem 1.8. In particular the 'if' direction.
I'll restate some of the definitions of objects they use. The trace $X_K$ of a knot $K$ is defined as the union of a 4-ball with a 2 handle attached along $K$. Then the 'if' statement claims that a smooth embedding of $X_K$ into $S^4$ means that $K$ is smooth slice I.e. bounds a smooth disc in $B^4$.
Consider the composition of the inclusion of the core of the 2 handle with the smooth embedding from the hypothesis: $D^2 \rightarrow X_K \rightarrow S^4$
Then this defines a smooth disc with  boundary $K$ in $S^4$ and thus $B^4$ since the image is not surjective.
I was wondering what is the problem with this approach? Is the 2 core not necessarily smoothly embedded?
The authors choose to consider a map from $S^2 \rightarrow X_K \rightarrow S^4$ and restrict it to a disc. Is this choice deliberate in order to avoid some problems with the argument above?
Thanks forgive my ignorance if i am asking something stupid, (below is an exempt from the paper)

 A: It is true that the existence of a smooth, nonsurjective embedding $f : X \hookrightarrow S^4$ implies the existence of a smooth embedding $g : X \hookrightarrow B^4$. The way that I would prove this implication is to choose a point $p$ not in the image of an embedding, choose a smooth tiny round 4-ball $B(p,\epsilon)$ disjoint from the image of that embedding, and then choose a diffeomorphism $h : S^4 \setminus B(p,\epsilon) \to B^4$, and define $g = h \circ f$.
However, the smooth slice hypothesis is different, it requires the existence of a smooth embedding $D^2 \hookrightarrow B^4$ which, when restricted to $S^1 = \partial D^2$, has image equal to $K$. The embedding $g : X \to B^4$ above can certainly be restricted to $D^2$ to give an embedding $g : D^2 \to B^4$, but when that is further restricted to $S^1$, its image is some random circle in the interior of $B^4$ that has nothing to do with the given knot $K$.
The point here is that when $f : X \hookrightarrow S^4$ is given as above, the exterior of $f(B^4)$ in $S^4$ is diffeomorphic to $B^4$. Composing $f$ with that diffeomorphism, and then restricting to $D^2$, one gets a smooth embedding of $D^2$ into $B^4$ which is still not quite what one needs but is close to what one needs to prove that $K$ is slice.
