= Solution
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 (mathematics)> sequence give
$$
H_2(V_r,\partial V_r;\mathbb F_2)=0,
\qquad
H_1(\partial V_r;\mathbb F_2)\hookrightarrow H_1(V_r;\mathbb F_2)=0.
$$
The <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 are
$$
u=\alpha\gamma,\qquad v=\alpha\beta^{-1}\gamma\delta^{-1},\qquad
W=u\delta^{-1}v^{-1}u\delta.
$$
In the other configuration they are
$$
u=\alpha\gamma^{-1},\qquad v=\alpha\delta\gamma\beta,\qquad
W=u(\gamma\delta)^{-1}u^{-1}v(\gamma\delta).
$$
Changing 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 proved
$$
\boxed{\exists\ e:(D^2,\partial D^2)\hookrightarrow(M,\partial M)
\quad\text{with }e(\partial D^2)\text{ not bounding a disk in }\partial M.}
$$
The 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 \b[$K$ is the unknot].
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