Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 141 / 3

Begin with the product cobordism $(X_0\bigsqcup X_1)\times [0,1]$. In its outgoing boundary choose one embedded $n$-ball in each component and attach an $(n+1)$-dimensional one-handle $D^1\times D^n$ along $S^0\times D^n$. On the outgoing boundary this deletes those two balls and joins their boundary spheres by $D^1\times S^{n-1}$, which is precisely the connected-sum construction. With the product orientation,This is the one-handle cobordism from a disjoint union to a connected sum.

Take a collar ${\partial }_{i}X\times [0,\epsilon ]$. In the quotient, the two annuli $A_1,A_2$ are folded together by $f$, while their common boundary curves ${\beta }_1,{\beta }_2$ become the boundary of a meridional disk. Because $f({\alpha }_1)={\alpha }_2$, the two halves of $\alpha$ join across this disk and bound it. The quotient of the collar is therefore a solid torus whose meridian is attached along $\alpha$; outside the collar nothing changes. Hence the annulus-quotient model of Dehn filling gives

Choose Seifert longitudes ${\lambda }_i$ disjoint from the small arcs removed in forming the connected sum of knots. Join the two resulting longitude arcs through the same connecting tube used for the knots. Their oriented band sum is a zero-linking parallel of $K_1\#K_2$, and therefore

Let $X_{\#}$ be the connected-sum exterior and fill it along its Seifert longitudeThe decomposing annulus in $X_{\#}$ has two parallel boundary circles, each meeting ${\lambda }_{\#}$ once. Applying part (b) cuts the filling back into $X_1$ and $X_2$ and identifies the peripheral curves byup to the signs required by the orientation-reversing torus gluing.

In $Y_1^{\prime }=X_1({\lambda }_1)$, the core $K_1^{\prime }$ has meridian ${\mu }_1^{\prime }={\lambda }_1$. By the orientation stipulated in the question its longitude is ${\lambda }_1^{\prime }={\mu }_1$. Thus the displayed gluing sends the meridian and longitude of $K_1^{\prime }$ to the longitude and meridian of $K_2$, respectively, which is exactly the splice of knots. Therefore the required integer surgery is zero surgery on the connected sum: