A reverse-mirror sum admits an explicit ribbon immersion

Choose $p\in S^3\setminus K$ and a smooth path $p_t$ from $p$ to $\mathrm{\infty }$, constant near both endpoints of $I$. This is an isotopy of embeddings of the compact $0$-manifold consisting of one point. By [F2] it extends to an ambient isotopy $H_t$ of $S^3$. Set $K^{\mathrm{\prime }}:=H_1(K)$, so $K^{\mathrm{\prime }}\subset S^3\setminus \{\mathrm{\infty }\}={\mathbb{R}}^3$, and take a regular diagram of $K^{\mathrm{\prime }}$ by [F1]. Delete a short open subarc of an edge incident to the unbounded complementary region. Its remaining projected arc can be placed in a closed planar disk with its two endpoints on the disk boundary, all its other points in the disk interior, and the deleted closing arc outside that disk: draw the boundary through the deleted edge along the adjacent unbounded region and round it away from the diagram. Parameterize the remaining immersed planar arc by $\gamma :[0,1]\to {\mathbb{R}}^2$, straight near its endpoints. It has nonzero derivative, finitely many transverse crossing pairs $(u_i,v_i)\in (0,1{)}^2$, no triples, and no endpoint multiple points. The closing arc has no crossing. At each crossing mark which of $u_i,v_i$ is over according to the diagram of $K^{\mathrm{\prime }}$. Since $H_t$ is orientation-preserving, the conjugate isotopy $r{H}_t{r}^{-1}$ shows $K^{\mathrm{\prime }†}$ is oriented ambient-isotopic to $K^{†}$. By [F3], the oriented connected-sum type $K^{\mathrm{\prime }}\mathrm{\#}{K}^{\mathrm{\prime }†}$ is therefore $K\mathrm{\#}{K}^{†}$ under the fixed gluing convention.

Choose a smooth function $q:[0,1]\to (0,\mathrm{\infty })$ so that $h(s)=\sqrt{s(1-s)q(s)}$ has the prescribed strict over-minus-under sign at every crossing pair. This is a finite interpolation: assign positive heights $2$ and $1$ at the over and under parameters, respectively, and prescribe $q(s)=h(s{)}^2\mathrm{/}(s(1-s))$ at these finitely many distinct parameters; interpolate the positive $q$ values by disjoint nonnegative smooth bumps against a positive constant. Distinct crossing parameters are disjoint because there are no triples. Make $q$ constant near the endpoints. Put $\mathrm{\Omega }=\{(s,t):0\le s\le 1,t^2\le s(1-s)q(s)\}$. The map $(s,v)\mapsto (s,v\sqrt{q(s)})$ is a smooth diffeomorphism from the round disk $v^2\le s(1-s)$ onto $\mathrm{\Omega }$, including its two endpoint tips. Thus $\mathrm{\Omega }$ is a smooth disk.

Define $f:\mathrm{\Omega }\to {\mathbb{R}}^3\subset S^3$ by $f(s,t)=(\gamma (s),t)$. Since ${\gamma }^{\mathrm{\prime }}(s)\mathrm{\ne }0$, its two derivative vectors $({\gamma }^{\mathrm{\prime }}(s),0)$ and $(0,0,1)$ are independent, so $f$ is an immersion, including in the smooth disk charts at the tips. The only possible multiple points occur at the crossing pairs of $\gamma$. For a pair $(u,v)$, assume $h(u)<h(v)$. The image double arc is $\{(\gamma (u),t):\mathrm{\mid }t\mathrm{\mid }\le h(u)\}$, with preimages $\{u\}\times [-h(u),h(u)]$ and $\{v\}\times [-h(u),h(u)]$. The first is properly embedded in $\mathrm{\Omega }$, whereas the second lies entirely in its interior because $h(u)<h(v)$. The image sheets are transverse: their planar tangent directions are independent, and both contain the vertical tangent. Double arcs from different crossings are disjoint because their planar crossing points are distinct. There are no triple points. No two boundary points have the same image, since boundary heights at a crossing are $\pm h(u)$ and $\pm h(v)$, all distinct. Thus $f$ satisfies exactly [F4].

The plane $t=0$, completed by the point at infinity, is an equatorial sphere of $S^3$. It cuts the boundary of $f(\mathrm{\Omega })$ at exactly the two tips. The upper boundary arc is $(\gamma (s),h(s))$ and the lower is its reflection $(\gamma (s),-h(s))$ with opposite boundary orientation. Both are smooth proper ball-arc tangles: near a tip the equation $t^2=s(1-s)q(s)$ solves smoothly for $s$ as a function of $t^2$, so the arc is transverse to the equatorial plane. Close the upper arc by the deleted trivial arc in the equatorial sphere and round its two corners in crossing-free endpoint neighborhoods. This closure represents $K^{\mathrm{\prime }}$: it has exactly the projected crossings of the diagram in step 1.1 and their original height ordering. Explicitly, after choosing smooth endpoint roundings, interpolate its heights to those of the regular diagram realization of $K^{\mathrm{\prime }}$ while keeping the planar immersion fixed. At every crossing, each endpoint height difference is positive in its marked ordering, so their convex interpolation stays positive; the planar immersion ensures immersion of every slice. This gives a smooth isotopy of embedded circles, and [F2] makes it ambient. The endpoint rounding takes place in disjoint balls without other strands and changes no knot type. Consequently [F3] identifies the oriented boundary of $f(\mathrm{\Omega })$ with $K^{\mathrm{\prime }}\mathrm{\#}{K}^{\mathrm{\prime }†}$, which represents $K\mathrm{\#}{K}^{†}$ by step 1.1. Together with step 3.1, this is the required ribbon immersion.