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If a boundary surface inclusion has noninjective fundamental group map, there is a compression disk whose boundary represents a nontrivial element of that kernel. More generally the boundary can be required to avoid a prescribed normal subgroup if a singular disk with that property exists.

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.

The Loop theorem converts a nonzero kernel of ${\pi }_1(F)\to {\pi }_1(M)$, for a boundary surface $F$, into a properly embedded compression disk. It is stronger than simply producing a singular nullhomotopy, and does not assert that every original self-intersecting loop is the boundary of the final disk.

The proof outline uses a strengthened version: keep the boundary class outside a prescribed normal subgroup $N◃{\pi }_1(F)$. Begin with a singular disk in general position. Local cut-and-paste removes branch points; when a disk splits, at least one boundary stays outside $N$, since otherwise their product would lie in $N$. Take a regular neighborhood of its image and lift the disk through connected double covers of successive neighborhoods. Each lift decreases a finite lexicographic singularity complexity, so the tower terminates. At the top no double cover remains; mod-two duality forces its boundary components to be spheres. A boundary curve of the disk neighborhood lying outside the pulled-back $N$ then bounds an embedded disk on such a sphere, pushed inward. Project down one stage at a time. Exchange disk sectors along double curves, retaining a boundary outside $N$, to remove the projection's double intersections. Repeating yields an embedded disk downstairs. Taking $N=1$ proves the Loop theorem; controlling a prescribed simple boundary gives Dehn's lemma. The key points are termination of the covering tower and preservation of an essential boundary during surgery, not an unjustified removal of all intersections by general position alone.

Two classification applications are particularly useful. First, a two-sided embedded surface is incompressible precisely when its inclusion is fundamental-group injective, with the usual exclusion of inessential sphere components. A compression disk immediately supplies a nontrivial kernel element. In the other direction, cut along the surface and use the Seifert-van Kampen theorem: if both boundary-copy maps were injective, the amalgamated-product or HNN normal-form theorem would make the original inclusion injective. Hence a nontrivial kernel appears on one cut side, and the Loop theorem produces a compression disk. This is what makes algebraically detected essential surfaces usable in three-manifold splitting and incompressible hierarchies.

Second, consider a compact orientable irreducible three-manifold with one torus boundary. If that boundary is not fundamental-group injective, the Loop theorem supplies a compression disk. An essential simple curve on a torus is nonseparating, so compression changes the torus into a sphere. Pushing this sphere into the manifold, irreducibility gives a ball on the side away from the original torus boundary; the region between the sphere and the torus is the compression neighborhood. Undoing compression therefore attaches one one-handle to a ball. This proves the solid-torus recognition by a compression disk criterion:As a concrete consequence, a knot with knot group $\mathbb{Z}$ is the unknot. Its exterior is irreducible: a sphere disjoint from a connected knot bounds a ball on the side not containing the knot. The map from its boundary group ${\mathbb{Z}}^2$ to $\mathbb{Z}$ cannot be injective, so its exterior is a solid torus. The boundary of a meridian disk of this exterior lies in the kernel of the linking-number map; it is therefore the primitive preferred longitude. Joining that longitude to the knot through its tubular neighborhood makes an embedded spanning disk for the knot. This proves the cyclic-knot-group unknot criterion, rather than merely identifying a familiar consequence by name.

A compact orientable irreducible three-manifold with connected torus boundary which is compressible is a solid torus. Compressing along a compression disk gives an embedded sphere. The compressed component has one sphere boundary and irreducibility makes it a ball; undoing compression attaches one one-handle, giving $D^2\times S^1$.