Smooth sliceness is equivalent to concordance to the unknot

Suppose first that a smooth proper disk $D\subset B^4$ has boundary $K$. Choose an interior point $p\in D$ and a sufficiently small round four-ball $E$ about $p$ contained in $\mathrm{int}{B}^4$. In orthonormal coordinates at $p$, the local graph theorem writes $D$ as $z=g(u)$ over a disk in its tangent plane, with $g(0)=0$ and $Dg(0)=0$. For small radius $R$, the function $r\mapsto r^2+g(r\theta {)}^2$ has derivative $2r+O(r^3)>0$ for every unit direction $\theta$ and $0<r\le R$. Hence $D\cap \mathrm{\partial }E$ is one radial graph over the tangent circle. Replacing $g$ by $\lambda g$, $0\le \lambda \le 1$, and solving the same strictly increasing radial equation gives an isotopy through smooth embeddings of $S^1$ in $\mathrm{\partial }E\cong S^3$ to the equatorial circle. By [F4], with target $\mathrm{\partial }E\cong S^3$, this isotopy extends to an ambient isotopy of $S^3$; hence $D\cap \mathrm{\partial }E$ is ambient isotopic to the standard unknot. The family of round balls moving the center of $E$ to $0$ along a path in $\mathrm{int}{B}^4$ is an isotopy of compact embeddings supported in $\mathrm{int}{B}^4$; apply [F4] with target ${\mathbb{R}}^4$ to extend it. Hence $B^4\setminus \mathrm{int}E$ is diffeomorphic, with boundary collars, to the standard shell $S^3\times I$. Removing $\mathrm{int}(D\cap E)$ from $D$ gives an annulus in that shell with boundary $K$ and a standard unknot.

Conversely, suppose $K$ is smoothly concordant to the standard unknot. Reverse the interval coordinate if needed so the unknot is the lower boundary. Glue a standard four-ball along the lower $S^3$ end of $S^3\times I$. The union is a standard four-ball with an added boundary collar: explicitly, the collar extends the radial coordinate of the unit ball from radius $1$ to radius $2$, and radial rescaling identifies it with the unit ball. Cap the lower unknot boundary of the annulus by the standard disk from [F3]. The annulus and cap meet in matching product collars, so their union is a smooth properly embedded disk with boundary $K$.

The product collars at the two boundary circles identify the annulus from step 1.1 with a smooth concordance, after orienting it so the lower boundary is the negative of its induced knot orientation and the upper boundary is positive. If necessary, reverse the interval coordinate to place the unknot at the lower end; a half-turn of the standard round circle about a diameter is an orientation-preserving ambient rotation that reverses its circle orientation, so the endpoint still has the standard oriented unknot type. Reversing the annulus orientation when required gives the boundary convention in [F1]. Therefore $K$ is smoothly concordant to the standard unknot.

Steps 1.1 and 2.1 prove that a smooth slice disk gives smooth concordance to the unknot, and step 1.2 proves the converse.