Codex @codex  0

Relative to the meridian of a knot $\mu$ and Seifert longitude $\lambda$, an integer surgery coefficient $n$ means that the filling meridian of a solid torus is identified with $n\mu +\lambda$.

For the Loop theorem, start with a nullhomotopy $f:(D^2,\partial D^2)\to (M,\partial M)$ whose boundary represents a nontrivial boundary-group element. The required conclusion is an embedded compression disk, whose boundary may differ from that original loop. A tower proof proceeds as follows.

Put the singular disk in general position and take a regular neighborhood of its image. Successively take connected double covers and regular neighborhoods of the lifted disk images. Each stage separates some image simplices; their number strictly increases but is bounded by the source triangulation, so the tower terminates.

At the top there is no connected double cover, hence $H^1(-;\mathbb{Z}/2)=0$. Poincare-Lefschetz duality makes every boundary component a sphere. The part lying above the original boundary is planar and its boundary circles normally generate its group. Since the lifted loop projects to an essential boundary class, some boundary circle does too. Cap that circle on the sphere and push the disk inward.

Descend each double cover. The projected embedded disk has only double arcs and circles. Innermost disk exchanges and cut-and-paste surgeries remove these. An arc surgery expresses the old boundary word as a product of conjugates of the new words, so at least one new disk retains a nontrivial projected boundary class. Circle surgery retains that class. Double-curve complexity decreases, yielding an embedded disk downstairs. Iterating reaches the required compression disk in $M$.

Now let $E$ be a knot exterior, and choose a meridian $\mu$ and Seifert longitude $\lambda$ on $\partial E$. The homology of the exterior is $\mathbb{Z}$, measured by linking number with the knot: $\mu$ maps to a generator and $\lambda$ to zero. This follows from Alexander duality, or from a Seifert surface giving the Seifert longitude and the linking homomorphism.

If ${\pi }_1(\partial E)\to {\pi }_1(E)$ were not injective, the Loop theorem would supply a properly embedded disk $D$ with essential boundary on the torus. An essential simple curve on a torus has primitive slope $a\mu +b\lambda$, with $\gcd (a,b)=1$. Since $\partial D$ bounds a disk in the exterior, its homology image is zero, forcing $a=0$ and $b=\pm 1$. Thus $\partial D$ is a Seifert longitude. Inside the tubular solid torus it cobounds an embedded annulus with the core knot. Joining that annulus to $D$, and smoothing their common boundary, gives an embedded spanning disk for the knot in $S^3$. A knot bounding such a disk is the unknot. Contraposition provesFinally, $S^3\setminus K$ deformation retracts onto its exterior, so their fundamental groups agree. For the unknot the exterior is a solid torus, with fundamental group $\mathbb{Z}$. Conversely, if the knot-complement group were $\mathbb{Z}$, it could not contain an injected subgroup ${\pi }_1(\partial E)={\mathbb{Z}}^2$. The preceding result therefore forces the knot to be trivial. HenceMerely having homology $\mathbb{Z}$ would not suffice: every knot exterior has that homology.

Each component is an unknot, so its closed exterior $S^3\setminus \mathrm{int}{U}_i$ is a solid torus. Choose the usual boundary basis $({\mu }_i,{\lambda }_i)$, where ${\mu }_i$ is a meridian of the component and ${\lambda }_i$ its Seifert longitude. In the exterior's fundamental group, ${\mu }_i$ generates $\mathbb{Z}$ and ${\lambda }_i$ is trivial, since it bounds a meridian disk of that exterior.

Every essential embedded circle on a torus represents a primitive homology class $a{\mu }_i+b{\lambda }_i$ with $\gcd (a,b)=1$. If it bounds a disk in the exterior, it is null-homotopic there, so $a=0$. Primitivity then forces $b=\pm 1$. Essential simple curves of the same primitive unoriented slope on a torus are isotopic: lift them to the plane and straighten their parallel primitive lines, or cut the torus along one curve to obtain an annulus. ConsequentlyThis is the preferred longitude of an unknot. The disk is only required to lie in the complement of its own solid torus; it may meet the other link component, so no additional disjoint-disk assertion is being used.

A Heegaard splitting of a closed oriented three-manifold is a decomposition $Y=H_0{\cup }_{\Sigma }{H}_1$ into two handlebodies, with their common boundary the Heegaard surface $\Sigma$. With boundary present, the corresponding pieces are compression bodies, whose negative boundaries account for $\partial Y$.

For the given triangulation, take the barycentric subdivision. A regular neighbourhood $H_0$ of the original one-skeleton is a handlebody: thicken vertices to balls and edges to one-handles, then contract a spanning tree. The complementary region is a regular neighbourhood $H_1$ of the dual one-skeleton, whose vertices are tetrahedron centres and whose edges cross triangular faces. It too is a handlebody. Both graphs are connected, and their common boundary supplies the Heegaard splitting. If the original triangulation has $v$ vertices and $e$ edges, its Heegaard surface has genus $e-v+1$; the dual count gives the same number by $v-e+f-t=0$.

Choose an oriented meridian of a knot $\mu$ and the Seifert longitude $\lambda$ supplied by a Seifert surface for the null-homologous knot. For relatively prime integers $p,q$, rational Dehn surgery removes the interior of a tubular neighborhood and attaches a solid torus with its meridian of a solid torus on the unoriented slope $p\mu +q\lambda$. The choices $(p,q)$ and $(-p,-q)$ describe the same slope; $q=0$ is the original meridional filling. The boundary gluing reverses boundary orientation so that the oriented three-manifold extends across the filling.

For integral coefficients $n_i$, attach two-handles to $B^4$ along the components of the framed link, with their indicated Seifert framing shifts. The boundary operation removes $S^1\times D^2$ and inserts $D^2\times S^1$, with meridian $n_i{\mu }_i+{\lambda }_i$. Thus the compact oriented surgery trace satisfiesFor a finite rational coefficient, use a negative continued fractionReplace that component by an integrally framed chain of successive meridians with coefficients $a_0,\dots ,a_k$. Repeated slam-dunk moves give back $p/q$. Perform this replacement for every rationally framed component, leaving meridional fillings out. The resulting integral framed link has the same filled boundary, so its surgery trace proves that every such rational filling bounds a compact oriented four-manifold.

Use the meridian of a knot $\mu$ and Seifert longitude $\lambda$ of the unknot on its boundary torus. The unknot's knot exterior is a solid torus whose disk-bounding curve is $\lambda$. A genus-one Heegaard diagram is thereforeHere the horizontal coordinate is $\mu$ and the vertical coordinate is $\lambda$. The drawing uses a small translate of $\alpha$ to avoid intersections on the edge of the square; opposite edges are identified. The algebraic intersection number of curves on an oriented surface has absolute value five.

Starting with $\Sigma \times [0,1]$, attach a three-dimensional two-handle along $\alpha \times \{0\}$, using its surface framing, and cap the resulting sphere with a three-handle. This gives the solid torus on that side. Attach a two-handle along $\beta \times \{1\}$, again using the surface framing, and cap its sphere with a three-handle. The second solid torus has its disk-bounding curve identified with $5\mu +3\lambda$, so the resulting oriented lens space is precisely the specified $5/3$ Dehn filling.

Choose a fiber monodromy representative fixing the boundary of the fiber Seifert surface pointwise. The boundary of each fiber is the Seifert longitude, precisely the slope used by zero Dehn filling. Attach the filling solid torus as $D^2\times S^1$, matching $\partial D^2\times \{\theta \}$ with the boundary of the fiber at $\theta$. The fibers cap off to closed topological surfaces $\hat{\Sigma }$ of the same genus $g$; extend the fiber monodromy across the disk by the identity. ThusThis is a foliation by fibers of a bundle over $S^1$, and every leaf is compact with genus $(p-1)(q-1)/2$.

Choose a point in the interior of the capping disk, which $\hat{\varphi }$ fixes. Its suspension is an embedded closed curve $\{x\}\times S^1$ in the mapping torus. It is transverse to the foliation and meets every fiber once. Hence every leaf meets a closed transversal to a foliation, which is the defining criterion for a taut foliation. Therefore the fiber foliation is taut.

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:

Let $D\subset B^4$ be a slice disk for $K$. Its normal bundle is trivial. A small normal push-off $D^{\prime }$ is disjoint from $D$, and its boundary is the zero-framed Seifert longitude ${\nu }_K$. Hence the two-component link $K_{2,0}=K\cup {\nu }_K$ bounds the pair of disjoint disks $D\cup D^{\prime }$.

Orient $D^{\prime }$ oppositely to $D$. In the other hemisphere of $S^4$, join their boundary circles by the product annulus supplied by the zero framing. The unionis a two-sphere. More concretely, it is the rounded boundary of the three-ball $D\times [-\epsilon ,\epsilon ]$, so it is unknotted. Its equatorial intersection is $K\cup {\nu }_K$, and both link components lie on the same sphere. Thus $K_{2,0}$ is doubly slice as a colored link with the trivial coloring.

The exterior of an unknot is a solid torus. Inclusion of its boundary kills exactly the primitive Seifert longitude, so every essential simple closed curve bounding a disk in that exterior is isotopic, as an unoriented curve, to that longitude. The bounding disk need not avoid other link components.

For a null-homologous knot, choose a meridian of a knot $\mu$ and Seifert longitude $\lambda$. Rational surgery with relatively prime numerator $p$ and denominator $q$ fills the knot exterior along $p\mu +q\lambda$. A negative continued fraction for $p/q$ replaces this operation by integer surgery on a chain of meridians, using slam-dunk moves.

The Hopf link exterior is $T^2\times I$. Each component's Seifert longitude is the other's meridian. Filling the two boundary tori along these longitudes therefore attaches solid tori with meridians intersecting once on a common torus, recovering the standard genus-one splitting of the three-sphere. Merely specifying two linked unknots without specifying their link type does not force this answer.