Surgery on a product of spheres produces a sphere in the standard framing

The disk chart at $y_0$ exhibits the embedding $\phi :S^p\times D^q\hookrightarrow S^p\times S^q$, $\phi (x,y)=(x,y)$ in the chart, with image in the interior; its restriction to the disk factor is the product trivialization, so $\phi$ is a framed embedded surgery sphere with underlying sphere $S^p\times \{y_0\}$, framed by the $S^q$-factor.

Use the hemisphere parameterizations given by the map of [F3]. The removed neighbourhood in the second factor is one labelled closed hemisphere and its closed complement is the other, with matching boundary coordinate $S^{q-1}$. Thus removing the open tube leaves exactly $S^p\times D^q$ with the boundary identification specified by the product framing; no complement assertion for an arbitrary disk chart is needed.

Glue $D^{p+1}\times S^{q-1}$ to the complement of step 1.2 by the product boundary identification. By [F2] this is the rounded boundary of $D^{p+1}\times D^q$. Choose a convex rounding of the product corners: its boundary is transverse to each ray from the origin, so it is $\{\rho (u)u:u\in S^m\}$ for a positive smooth $\rho$. The radial map $u\mapsto \rho (u)u$ has smooth inverse $z\mapsto z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$, proving that this boundary is diffeomorphic to $S^m$. Rounding independence gives the same diffeomorphism type for other compatible roundings. This proves the first computation, including $p=0$ and $q=1$.

For the dual reading, regard $S^m=\mathrm{\partial }(D^{p+1}\times D^q)$ with the decomposition of [F2]; the standard framed $S^p=S^p\times \{0\}$ lies in the solid piece $S^p\times D^q$ and has tubular neighbourhood $S^p\times D^q\subseteq S^m$, framed by the $D^q$-factor. The $p$-surgery on $S^m$ along this sphere removes $S^p\times \mathrm{int}{D}^q$ and glues in $D^{p+1}\times S^{q-1}$ along $S^p\times S^{q-1}$, leaving two copies of $D^{p+1}\times S^{q-1}$ glued along their common boundary; that double is the product $S^{p+1}\times S^{q-1}$ of the double of $D^{p+1}$, which is $S^{p+1}$ by [F3], with the closed factor $S^{q-1}$, the gluing being the product identification. Hence the surgery on $S^m$ along the standard framed $S^p$ produces $S^{p+1}\times S^{q-1}$.

For $p\ge 1$, the starting product is connected, so [F5] shows that the inverse of the first computation uses the belt sphere of dimension $q-1$ in $S^m$ and returns $S^p\times S^q$. For $p=0$, compute that inverse directly: in $S^m=\mathrm{\partial }(D^1\times D^q)$, remove the interior of the belt tube $D^1\times S^{q-1}$ and insert $S^0\times D^q$. The complement is another $S^0\times D^q$, with the product boundary identification; their union is $S^0\times (D^q{\cup }_{S^{q-1}}{D}^q)\cong S^0\times S^q$ by [F3]. The second computation operates on the other sphere, of dimension $p$, in $S^m$, and produces $S^{p+1}\times S^{q-1}$; its inverse returns $S^m$. Thus the two $p$-surgeries are not identified with each other's dual operations.