The trefoil sum with its reverse mirror is ribbon

Delete a small straight subarc of $T$ in a coordinate ball. Identify its complementary ball with $B^3$ and parametrize the remaining oriented arc by a smooth embedding $a:[0,1]\to B^3$ with endpoints on $S^2$, interior in the open ball, and transverse radial endpoint collars. These are the standard ball-arc data of [F2]. Move the arc slightly if necessary so that it misses the origin; this is possible by a small translation on its interior, fixed on endpoint collars, since a smooth compact arc in three-space has empty interior and its endpoint collars can be kept outside a small ball about the origin. Put $f(s)=\mathrm{\mid }a(s){\mathrm{\mid }}^2>0$, with $f(0)=f(1)=1$ and $f<1$ in the interior; transversality gives $f^{\mathrm{\prime }}(0)<0<f^{\mathrm{\prime }}(1)$.

Arrange that $f$ is Morse away from its endpoint collars, without changing the arc type. Here are finite perturbation details. The zeros of $f^{\mathrm{\prime }}$ lie in a compact subinterval. Cover that zero set by finitely many smaller open intervals on each of which a compactly supported smooth function ${\varphi }_i$ has derivative $1$. For the family $f_u=f+{\sum }_i{u}_i{\varphi }_i$, the map $(s,u)\mapsto f_u^{\mathrm{\prime }}(s)$ is a submersion near this zero set because one parameter derivative is $1$; on its compact complement $f^{\mathrm{\prime }}$ is bounded away from zero, so small parameters create no further zeros. Apply [F3] to this finite family and zero in $\mathbb{R}$. Choose arbitrarily small good $u$, so every zero of $f_u^{\mathrm{\prime }}$ is simple. Endpoint collars stay fixed and $0<f_u<1$ inside. Replace $a(s)$ by $\sqrt{f_u(s)\mathrm{/}f(s)}a(s)$. For sufficiently small parameters this is an isotopy of embeddings, since on a finite chart cover immersion and local injectivity persist in $C^1$, while compact pairs outside the chart diagonals have a positive separation margin. Its squared radius is $f_u$. Rename this arc $a$ and this function $f$; the finitely many simple critical points lie in a compact subinterval, so there are only finitely many.

In $B^4=\{(x,t)\in {\mathbb{R}}^3\times \mathbb{R}:\mathrm{\mid }x{\mathrm{\mid }}^2+t^2\le 1\}$ take the surface $D=\{(a(s),t):0\le s\le 1,t^2\le 1-f(s)\}$. The strip map $(s,t)\mapsto (a(s),t)$ is an embedding. Its parameter domain is explicitly a smooth disk: the ellipse $v^2\le s(1-s)$ maps onto it by $(s,v)\mapsto (s,v\sqrt{q(s)})$, where $q(s)=(1-f(s))\mathrm{/}(s(1-s))$ extends smoothly and positively to both endpoints because $f$ has simple transverse endpoint zeros of $1-f$. This map and its inverse are smooth, and every interior vertical fibre is an interval of positive length. Smoothness at those endpoints follows from the implicit function theorem applied to $f(s)+t^2=1$, because $f^{\mathrm{\prime }}$ is nonzero there; the domain boundary is locally $s=s(t)$ rather than a corner. Equivalently $D$ is the regular sublevel of $f(s)+t^2$ in a slightly extended embedded strip. Consequently it is a smooth compact properly embedded disk, its boundary lies in $S^3$, its interior lies in the open four-ball, and its intersection with $S^3$ is transverse.

The two boundary arcs of $D$ are $(a(s),\sqrt{1-f(s)})$ and $(a(s),-\sqrt{1-f(s)})$. The equatorial sphere $t=0$ meets their union only at $s=0,1$. The hemispheres are standard three-balls; projection to $x$ identifies their interiors with the interior of $B^3$, and their lifted arcs represent the original ball-arc pair by [F2]. To see this without a boundary-singular projection, first use transverse endpoint collars to straighten the arcs in smooth hemisphere collars, and then use projection on the compact remaining subarcs; the collar interpolation is an isotopy of the arcs fixing endpoints. Reflection $(x,t)\mapsto (x,-t)$ exchanges the hemispheres and reverses the boundary orientation of $S^3$. The orientation induced on $\mathrm{\partial }D$ runs along one lifted arc with increasing $s$ and along the other with decreasing $s$. Thus the capped oriented boundary arcs are $T$ and $-T^{\ast }$, with the fixed mirror reflection transported by positive coordinate frames as in [F2]. Gluing these standard hemispherical arc data gives $\mathrm{\partial }D=T\mathrm{\#}(-T^{\ast })$ as an oriented knot.

On $D$ the squared radial function is $\rho (s,t)=f(s)+t^2$. Its interior critical points are exactly $(s,0)$ with $f^{\mathrm{\prime }}(s)=0$, and their Hessians are $\mathrm{diag}(f^{\mathrm{\prime }\mathrm{\prime }}(s),2)$. They are nondegenerate, with index $0$ or $1$, never $2$. Since $\rho >0$, taking its square root preserves these critical indices. Hence the standard radial function makes $D$ a ribbon disk by [F1], proving the full conclusion.