A knot summed with its reverse mirror bounds a ribbon disk

Cut a short trivial arc from $K$ inside a small ball. Identify its complementary ball with the unit ball $B^3\subset {\mathbb{R}}^3$ so that the remaining arc is a smooth embedding $a:[0,1]\to B^3$, its endpoints lie on $S^2$, and its interior lies in the open ball. Straighten its endpoint collars to be radial and transverse to $S^2$. A small perturbation supported away from the endpoints makes $a$ avoid the origin; compactness and the fact that a one-dimensional embedded arc has empty interior allow the perturbation to be chosen arbitrarily small in $C^1$, so it remains embedded and fixed on the endpoint collars. Put $f(s)=\parallel a(s){\parallel }^2$. Then $f>0$ on $[0,1]$, $f(0)=f(1)=1$, $f<1$ on $(0,1)$, and transversality gives $f^{\mathrm{\prime }}(0)<0<f^{\mathrm{\prime }}(1)$.

The zeros of $f^{\mathrm{\prime }}$ lie in a compact subinterval of $(0,1)$ because $f^{\mathrm{\prime }}$ is nonzero on the endpoint collars. Cover the zero set by finitely many intervals carrying smooth functions ${\varphi }_i$ supported in $(0,1)$ whose derivatives equal $1$ on smaller intervals still covering the zero set. For $u=(u_i)$ define $f_u=f+{\sum }_i{u}_i{\varphi }_i$. On a neighborhood of every zero of $f^{\mathrm{\prime }}$ the total family $\mathcal{F}(s,u)=f_u^{\mathrm{\prime }}(s)$ is a submersion in the parameter direction, since some $\mathrm{\partial }\mathcal{F}\mathrm{/}\mathrm{\partial }{u}_i={\varphi }_i^{\mathrm{\prime }}(s)=1$. On the compact complement of that neighborhood, $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }$ has a positive lower bound, so for all sufficiently small $u$ no new zeros occur there. Thus the total map is transverse to $0$ for parameters in a sufficiently small ball. By [F3], choose a parameter in that ball for which $0$ is a regular value of $f_u^{\mathrm{\prime }}$. The critical points of $f_u$ are then nondegenerate and finite. Shrink the parameter ball so $0<f_u<1$ on the perturbation support. The radial arcs $a_{\lambda }(s)=\sqrt{f_{\lambda u}(s)\mathrm{/}f(s)}a(s)$, $0\le \lambda \le 1$, form a smooth isotopy of embeddings fixed near the endpoints: the formula is smooth since $f>0$, and sufficiently small $C^1$ perturbations preserve immersion and injectivity on the compact source. By [F4], extend this isotopy in an ambient isotopy supported away from the removed trivial arc, so replacing $a$ by $a_u$ preserves the knot type of its closure. Rename $a_u$ as $a$ and $f_u$ as $f$.

In the unit ball $B^4=\{(x,t)\in {\mathbb{R}}^3\times \mathbb{R}:\parallel x{\parallel }^2+t^2\le 1\}$ define $\mathrm{\Omega }=\{(s,t):0\le s\le 1,t^2\le 1-f(s)\}$ and $D=\{(a(s),t):(s,t)\in \mathrm{\Omega }\}$. The map $(s,t)\mapsto (a(s),t)$ is an injective immersion, hence an embedding because $\mathrm{\Omega }$ is compact. The function $q(s)=(1-f(s))\mathrm{/}(s(1-s))$ extends smoothly and positively to $s=0,1$, with $q(0)=-f^{\mathrm{\prime }}(0)>0$ and $q(1)=f^{\mathrm{\prime }}(1)>0$. Consequently $(s,v)\mapsto (s,v\sqrt{q(s)})$ is a smooth diffeomorphism from the round disk $v^2\le s(1-s)$ onto $\mathrm{\Omega }$. The boundary of $\mathrm{\Omega }$ is regular: at an interior boundary point $t\mathrm{\ne }0$, and at its two endpoints $f^{\mathrm{\prime }}(s)\mathrm{\ne }0$, so the differential of $f(s)+t^2$ is nonzero. Thus $D$ is a smooth disk, its boundary lies in $S^3$, and its interior lies in the open four-ball. By [F6], its boundary is exactly the boundary knot specified by properness.

The two boundary arcs of $D$ are the lifts $(a(s),\pm \sqrt{1-f(s)})$ and meet the equatorial $S^2=\{t=0\}\cap S^3$ only at their endpoints. Each hemisphere of $S^3$ is a standard three-ball. Away from a small collar of its equator, projection $(x,t)\mapsto x$ is a smooth diffeomorphism onto the corresponding truncated ball and identifies the lifted arc with $a$. At an endpoint, use the equator collar coordinate $t\ge 0$ (or $-t\ge 0$ in the lower hemisphere). Since $f^{\mathrm{\prime }}(0)\mathrm{\ne }0$ and $f^{\mathrm{\prime }}(1)\mathrm{\ne }0$, the implicit function theorem writes the lifted arc near each endpoint as $s=\sigma (t^2)$, with $\sigma$ smooth; in a chart $y$ on the equator its tangential coordinates are $y(t)=y_0+O(t^2)$, while $t$ is the inward collar coordinate. Choose a cutoff $\chi (y,t)$ supported in this chart and equal to $1$ near the arc, and use $(y,t)\mapsto (y-\chi (y,t)(y(t)-y_0),t)$. Its derivative is the identity plus $O(t)$, so after shrinking the collar it is a diffeomorphism and straightens the arc to $\{y_0\}\times [0,ϵ)$. This is the same standard ball-arc collar as the radial endpoint collar of $a$. Thus the lifted upper arc is the original ball-arc tangle and the lower one is its image under $(x,t)\mapsto (x,-t)$, up to collar-supported isotopy. The boundary orientation traverses one lifted arc in the direction of $a$ and the other in the reverse direction. By [F2] and [F5], closing these two arcs with the fixed connected-sum gluing gives precisely the oriented type $K\mathrm{\#}{K}^{†}$. Therefore $\mathrm{\partial }D$ has the required oriented knot type.

On $D$, the squared radial function is $\rho (s,t)=f(s)+t^2$. Its interior critical points are exactly $(s,0)$ where $f^{\mathrm{\prime }}(s)=0$. At each such point the Hessian is $\mathrm{diag}(f^{\mathrm{\prime }\mathrm{\prime }}(s),2)$, which is nondegenerate and has index $0$ when $f^{\mathrm{\prime }\mathrm{\prime }}(s)>0$ and index $1$ when $f^{\mathrm{\prime }\mathrm{\prime }}(s)<0$ by [F7]. Thus there are no index-$2$ critical points, and step 4.1 identifies the boundary knot as $K\mathrm{\#}{K}^{†}$. This proves the statement.