Interior one-handle trades and the fully framed circle-surgery pair

Let $A$ be the lower stage and join the feet by the boundary arc required in the Statement. A small regular neighborhood $Q$ of its union with the two foot disks in the attaching face is a $(d-1)$-ball: in the arc chart $I\times D^{d-2}$ its two end half-balls close the tube to a ball. Thicken $Q$ inward by its fixed collar and round its inner edge. This gives a $d$-ball $B\subset A$ meeting the closure of $A\setminus B$ along one $(d-1)$-disk $D$, with both feet on the complementary outer boundary of $B$. Thus $B$ with the one-handle attached is the ball-plus-tube neighborhood $S^1\times D^{d-1}$, with $D$ as its designated lower-stage interface disk. Trivialize the normal bundle of the core closed by a small inward push of the boundary arc over a cut interval. Its endpoint transition has positive determinant by the stated orientability condition, so a path in $G{L}^{+}(d-1,\mathbb{R})$ closes the frame. The foot and arc product coordinates therefore give the displayed product thickening. Its core and a smaller tube lie in the interior, off the protected boundary. This boundary-ball construction is the reason a joining arc in the attaching face is required; an arbitrary interior arc would not provide the disk $D$.

For the four-dimensional bridge choose coordinates $S^1\times D^3$ about the level circle, with $D_{\mathrm{e}\mathrm{q}}^2$ the level equator and lower/upper half-balls $D_{-}^3,D_{+}^3$. Remove this tube and insert $D_a^2\times S^2$, using its given full rank-three framing. Split $S^2=S_{-}^2{\cup }_{S_{\mathrm{e}\mathrm{q}}^1}{S}_{+}^2$. The lower product with $S^1\times D_{-}^3$ removed is a product collar with an indented outgoing face: in the normal coordinates the half-ball removal moves the old upper face to an inward rounded graph. A smooth increasing interval map on each normal line, fixed at the lower end, carries that face to the graph; these maps are identity outside the tube and the seam rounding is in a smaller collar. The upper half has the same product description. Thus the surgery is a product, the two inserted pieces, and an absorbable outgoing product.

Interior surgery replaces a smaller $S^1\times D^{d-1}$ by $D^2\times S^{d-2}$. The remaining shell of the original product is $S^1\times S^{d-2}\times I$, so it is a retained boundary collar. Split the sphere into its two hemispheres. The first $D^2\times D_{+}^{d-2}$ is a ball. The second, read with core factor $D_{-}^{d-2}$, is a $(d-2)$-handle with attaching region $D^2\times S^{d-3}$ and outgoing region $S^1\times D_{-}^{d-2}$. Choose the product coordinates so that the interface disk $D$ of step 1.1 lies in $S^1\times \mathrm{int}{D}_{+}^{d-2}$, which is on the first ball's boundary. The first ball is attached to the unchanged complement of $B$ along $D$, just as $B$ was. Absorbing either ball by [F4] identifies that first stage with $A$; carry the outer boundary markings through this same collar comparison. The remaining handle has index $d-2$. The outer collar is unchanged point by point; every later attachment is carried by that boundary identification. This is an interior modification, not a cancellation of the original ambient manifold. For $d=4$ it trades index one for two, and for $d=5$ one for three; reversing dimension five gives index four to two.

The first piece $D_a^2\times S_{-}^2$ is a two-handle with core $D_a^2$, normal disk $S_{-}^2$ and attaching region $S_a^1\times S_{-}^2$. On the old level this is the knot $K$ with its prescribed level framing. Its outgoing region is $D_a^2\times S_{\mathrm{e}\mathrm{q}}^1$. The second piece $D_a^2\times S_{+}^2$ is also a two-handle, but its disk factors are interchanged: core $S_{+}^2$, normal disk $D_a^2$, attaching region $D_a^2\times S_{\mathrm{e}\mathrm{q}}^1$. This is exactly the outgoing region of the first handle; using the same core factor for both would give a different attaching region.

The second attaching circle is $\{p\}\times S_{\mathrm{e}\mathrm{q}}^1$ for an interior point $p\in D_a^2$. Move $p$ along a radial arc toward $S_a^1$, first shrinking its normal thickening uniformly. This is a compact isotopy of the actual attaching product embedding in the boundary after the first handle. At the seam its product corner rotation carries it to the old level, where it is a small meridian of $K$. Its tangential $S_a^1$ direction is a constant push direction around this meridian and its other normal direction is the complementary radial direction in $D_{\mathrm{e}\mathrm{q}}^2$. Thus its framing has no turn relative to the meridian disk, so is zero. Its other boundary piece $S_a^1\times S_{+}^2$ is where the indented upper product attaches. Absorbing that product retains the original upper boundary map. By [F1] these product isotopies compare the full handle attachments, not just their cores.

For an arbitrary prescribed rank-three frame on a circle in a two-handle trace, [F2] first moves the circle to the base and realizes its framing by an integral level frame. Its framing homotopy is realized by an isotopy of small tubular embeddings, so step 1.2, step 2.2, step 3.1 apply to the prescribed surgery relative to the whole ambient boundary. By [F3] the resulting pair has the O3 boundary-relative identification on its small solid torus; old attaching links outside that torus and external base collars are fixed. This proves the bridge for every framing, without identifying the two rank-three parity classes, as well as the existence trade in step 1.1, step 2.1.