The upper boundary of the surgery trace is the surgered manifold

The trace is obtained from $N=M\times [0,1]$ by attaching the standard $(p+1)$-handle along the embedding $\phi \times \{1\}$ in the upper face $M\times \{1\}$, with a handle of index $p+1$ in an ambient manifold of dimension $m+1$. The boundary trade of [F2] applies with $k=p+1$ and $n=m+1$, so $\mathrm{\partial }{W}_{\phi }=(\mathrm{\partial }N\setminus (\phi \times \{1\})(S^p\times \mathrm{int}{D}^q))\cup (D^{p+1}\times S^{q-1}),$ the two parts meeting along $S^p\times S^{q-1}$ with the identification induced by $\phi$. Since $\mathrm{\partial }N=M\times \{0\}\bigsqcup M\times \{1\}$ and the removed piece lies in the upper face, the first summand is $M\times \{0\}\bigsqcup (M\times \{1\}\setminus \phi (S^p\times \mathrm{int}{D}^q))$.

Put $C=M\times I$ and $k=p+1$. By [F8], $(W_{\phi },C)$ is equivalent, relative to $C$, to $E=C{\cup }_{{\phi }_0\times \{1\}}{D}^k$; hence this equivalence also fixes $M\times \{0\}$. Let $Y=M{\cup }_{{\phi }_0}{D}^k$. Collapse the cylinder coordinate to define $q:E\to Y$, leaving the cell coordinates unchanged. An inverse $j:Y\to E$ is the identity on the lower face and sends $u=rx$ in the cell to $2u$ in the upper cell for $r\le 1\mathrm{/}2$, and to $({\phi }_0(x),2(1-r))$ in the cylinder for $r\ge 1\mathrm{/}2$. The formulas agree at $r=1\mathrm{/}2$ and at the attaching boundary. The composite $qj$ radially expands $r$ to $\min (2r,1)$, homotopic to the identity relative to the cell boundary. For $jq$, a homotopy on $E$ sends $(a,t)$ in the cylinder to $(a,(1-s)t)$ and sends $u=rx$ in the cell to $(1+s)u$ in that cell when $r\le 1\mathrm{/}(1+s)$, and to $({\phi }_0(x),2-(1+s)r)$ in the cylinder otherwise. These prescriptions agree on all seams, start at the identity, end at $jq$, and fix the lower face. This proves the first equivalence in (ii), without gluing incompatible retractions.

The second boundary component just computed, namely $(M\setminus \phi (S^p\times \mathrm{int}{D}^q))\cup (D^{p+1}\times S^{q-1})$ with the identification induced by the framing on the overlap, is exactly the surgered manifold $M_{\phi }$ of [F3]: the removed sets agree, the glued pieces agree, and the gluing identification is the same framing datum. Hence the outgoing face is diffeomorphic to $M_{\phi }$ by the identity on $M\setminus \phi (S^p\times \mathrm{int}{D}^q)$, and this diffeomorphism carries the belt sphere $\{0\}\times S^{q-1}$ of the handle to the belt sphere of the surgery. The incoming face $M\times \{0\}$ is a union of boundary components untouched by the attachment, and it is disjoint from the outgoing face. This proves (i).

Reverse the local handle presentation: the product $D^{p+1}\times D^q$ becomes $D^q\times D^{p+1}$, with attaching region $S^{q-1}\times D^{p+1}$, the former outgoing region. To see the reversed collar presentation, use the local handle height $-\mathrm{\mid }u{\mathrm{\mid }}^2+\mathrm{\mid }v{\mathrm{\mid }}^2$; changing its sign exchanges $u$ and $v$, its lower and upper faces, and core and cocore, while outside the handle the collars are read backwards. Its attaching sphere is therefore the belt sphere $\beta$ in the outgoing face. Applying [F8] to this $q$-handle and then the explicit cylinder-cell equivalence of step 1.2, now with $M_{\phi }$ and $k=q$, gives $(W_{\phi },M_{\phi })\simeq (M_{\phi }{\cup }_{\beta }{D}^q,M_{\phi })$. Both indices lie between $1$ and $m$; only these interior indices of [F8] are used. This proves (ii).

The trace is a compact smooth $(m+1)$-manifold whose boundary is the disjoint union of the incoming face $M\times \{0\}$, identified with $M$ by the product structure, and the outgoing face, identified with $M_{\phi }$ by step 2.1; the collars of the two faces are those of the cylinder and of the handle attachment, and the triad data are those of [F7]. By [F6] the incoming face is the level $M\times \{0\}$ of the product presentation, so the trace is a bordism from $M$ to $M_{\phi }$. This proves (iii).