Local resolution of ribbon singularities

Identify $B^4$ with the unit ball in ${\mathbb{R}}^4$ and use the explicit boundary collar $c:S^3\times [0,\epsilon )\to B^4$, $c(y,s)=(1-s)y$, with $0<\epsilon <1$. On the unit disk set $r(x)=1-\parallel x{\parallel }^2$, so $r=0$ on $\mathrm{\partial }{D}^2$, $r>0$ in the interior, and $dr$ is nonzero in the inward normal direction at the boundary. By [F1], the finitely many preimage arcs form disjoint compact subsets of $D^2$. Choose pairwise disjoint neighborhoods of the interior arcs ${\beta }_i$, all disjoint from every proper arc ${\alpha }_j$ and from $\mathrm{\partial }{D}^2$, and choose smooth bump functions ${\varphi }_i$ supported in those neighborhoods with ${\varphi }_i=1$ on ${\beta }_i$ and $0\le {\varphi }_i\le 1$. This is a finite choice of smooth functions on a disk.

Let $M=\max \{0,{\max }_{i,s}(r({\alpha }_i(s))-r({\beta }_i(s)))\}$ when there is at least one double arc, and let $M=0$ otherwise. Choose $\delta >0$ so that $\delta (2M+3)<\epsilon$, set $a=2\delta (M+1)$, and define $h=\delta r+a{\sum }_i{\varphi }_i$. The supports are disjoint, so $0<h<\epsilon$ on $\mathrm{Int}{D}^2$ and $h=0$ exactly on $\mathrm{\partial }{D}^2$. For every double arc and every parameter $s$, $h({\beta }_i(s))-h({\alpha }_i(s))=a+\delta (r({\beta }_i(s))-r({\alpha }_i(s)))\ge \delta (M+2)>0$. Near $\mathrm{\partial }{D}^2$ all bumps vanish, so $h=\delta r$ there and has positive inward derivative.

Define $\tilde{f}(x)=c(f(x),h(x))$. If $\tilde{f}(x)=\tilde{f}(y)$, injectivity of the collar coordinates gives $f(x)=f(y)$ and $h(x)=h(y)$. By [F1], either $x=y$ or $x,y$ are the two preimages of one of the double arcs; the strict height inequality in step 2.1 excludes the latter. Thus $\tilde{f}$ is injective.

The map $f$ is an immersion, so its derivative is injective; adjoining the height coordinate preserves injectivity of the derivative, and the collar map $c$ is a local diffeomorphism. Hence $\tilde{f}$ is an immersion. A continuous injection from compact $D^2$ to the Hausdorff space $B^4$ is a homeomorphism onto its image, so this injective immersion is a smooth embedding. Since $h^{-1}(0)=\mathrm{\partial }{D}^2$, its preimage of $S^3=\mathrm{\partial }{B}^4$ is exactly $\mathrm{\partial }{D}^2$; the nonzero inward derivative established in step 2.1 gives the proper boundary collar condition in [F2]. The disk orientation transports through the embedding and its boundary remains the original oriented knot $K$.