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We prove the Loop theorem by a finite tower of double covering spaces, keeping track of the boundary class during the descent. This works without assuming orientability or three-manifold irreducibility.

Let $F$ be the boundary component containing the basepoint. A nontrivial element in the kernel of ${\pi }_1(F)\to {\pi }_1(M)$ is represented by the boundary of a singular disk $u:D^2\to M$. Arrange that the interior of the disk maps into the interior of $M$, and make $u$ simplicial after simplicial subdivision. Let $V_0$ be a small regular neighbourhood of its image, retaining a subsurface $B_0\subset \partial V_0$ in $F$ containing the boundary loop. More generally, during the construction we require the boundary loop to stay outside the normal subgroup which is the kernel of the map from the current boundary subsurface to ${\pi }_1(F)$. This records exactly the information needed at the end.

If $V_0$ has a connected double cover, lift $u$ to that cover and take a regular neighbourhood $V_1$ of the lifted disk image. Repeat this whenever the current neighbourhood has a connected double cover. This procedure is finite. Indeed, let $C_i$ be the simplicial image of the disk at stage $i$. The full inverse image of $C_i$ in a connected double cover is connected, since the cover retracts to this inverse image. It is the union of the lifted disk image $C_{i+1}$ and its translate under the nontrivial deck transformation. These two images must meet. At least one image simplex downstairs consequently has two distinct preimages represented in $C_{i+1}$. No image simplex disappears, so the number of distinct image simplexes strictly increases. This number is at most the number of simplexes of the fixed domain disk. Thus the tower proof of the loop theorem terminates.

At the top, $V_r$ has no connected double cover. Equivalently, $H^1(V_r;{\mathbb{F}}_2)=0$. Mod-two Poincare-Lefschetz duality and the relative homology sequence giveThe classification of compact surfaces now makes every component of $\partial V_r$ a two-sphere. The lifted boundary subsurface $B_r$ is planar. Its fundamental group is normally generated by its boundary circles. Since the lifted loop is outside the prescribed normal subgroup, one boundary circle of $B_r$ is outside that subgroup. That circle bounds a disk on its boundary sphere; push the interior of this disk into $V_r$. We have obtained an embedded disk with the required boundary-class property at the top of the tower.

We next descend one double cover. Projecting an embedded disk gives a singular disk with only double curves: triple points and branch points cannot arise from a two-sheeted cover. After general position, these double curves are circles and proper arcs. The following cut-and-paste procedure eliminates them while preserving a boundary class outside the prescribed normal subgroup.

First eliminate the circular double curves. Take a double circle innermost on one of its preimage sheets. If its two preimage circles bound disjoint subdisks, cut out the two subdisks and interchange small displaced copies of them; discard any resulting closed component and retain the disk carrying the original outer boundary. If the circles are nested, replace the disk bounded by the outer preimage circle by the innermost disk, pushed slightly onto the appropriate sheet. Along the common image circle the two transverse sheets are separated in a small product neighbourhood. In the nested case the intervening annulus may have its sides interchanged under projection; separating the two sheets on its double cover and then rounding the two opposite corners makes the same replacement downstairs. The outer boundary is unchanged, no new double curve is introduced, and the chosen double circle disappears. Repeating leaves only double arcs.

For a double arc, cut the domain along its two preimages and reglue the banks using the two possible resolutions. There are two disk choices. At least one has boundary outside the prescribed normal subgroup. Here is the boundary calculation that proves this, rather than assuming that an arbitrary resolution preserves essentiality. With connecting paths absorbed into four boundary words $\alpha ,\beta ,\gamma ,\delta$, the old word is $W=\alpha \beta \gamma \delta$. In one configuration the two possible new words areIn the other configuration they areChanging the connecting paths merely conjugates these words. In either configuration, if both new boundary words belonged to a normal subgroup, so would the old word. Choose a disk whose word stays outside the subgroup. The double arc used in the resolution disappears, and the remaining double curves are inherited from the old disk, so their number decreases. Remove any circular double curves again, and repeat. The resulting disk is embedded.

Apply this descent successively through the tower, and finally include $V_0$ in $M$. Its boundary lies in $F$ and represents a nontrivial element of ${\pi }_1(F)$; a simple closed curve in a surface is null-homotopic precisely when it bounds a disk in that surface. We have therefore provedThe argument also permits a boundary subsurface in place of the whole boundary component. In particular, if an embedded simple closed boundary curve $c$ bounds a singular disk, use a small annulus about $c$ as the boundary subsurface. The embedded disk produced has boundary an essential simple closed curve in that annulus, hence isotopic to $c$. Extend that isotopy through a boundary collar. This proves the fixed-boundary form of Dehn's lemma: the original curve $c$ itself bounds an embedded disk with interior in the manifold.

Now consider the knot map. Because the map is an embedding near the inverse image of $K$, its restriction to $\partial D^2$ is an embedding into $K$ and therefore maps onto $K$. No interior point can map to $K$, since it would duplicate a boundary image. Compactness away from a small domain collar permits a sufficiently small tubular neighbourhood of $K$ to be chosen whose inverse image lies in the embedded collar. Its intersection with this collar cuts off an annulus. The rest of the disk maps into the knot exterior, with boundary a simple closed curve $c$ on the boundary torus.

The annulus shows that $c$ runs once along $K$. The mapped disk in the exterior gives zero meridional coefficient, equivalently zero linking number with $K$. Thus $c$ is a longitude. Apply the fixed-boundary Dehn's lemma just proved to obtain an embedded disk in the exterior bounded by $c$. Attach the embedded collar annulus between $c$ and $K$. This gives an embedded spanning disk for $K$. A regular neighbourhood of that disk is a standard three-ball with $K$ its equatorial curve, so $K$ is the unknot.

A piecewise linear ball is a simplicial complex admitting a homeomorphism to an $n$-simplex that is linear on the simplexes of suitable finite simplicial subdivisions. Its boundary is a piecewise linear sphere.

In the illustrated simplicial subdivision, inserting the three edge midpoints turns one filled triangle into four. The counts change from $(V,E,F)=(3,3,1)$ to $(6,9,4)$, but $6-9+4=1$. Splitting a single edge adds one vertex and one edge, so its net contribution is $1-1=0$. Adding a diagonal inside a polygonal face adds one edge and one face, giving $-1+1=0$.

These local accounts explain common refinements. For arbitrary finite simplicial subdivisions, including barycentric subdivision in any dimension, the underlying geometric realization is unchanged. The topological invariance of Euler characteristic therefore guarantees the same alternating count, even when the individual counts change substantially.