Boundary-compressible incompressible surfaces at torus boundary are parallel annuli

First apply [F9] to the given surface, fixed pointwise on its boundary. The resulting surface is neat. Incompressibility passes to it, and a boundary-compressing witness passes in either direction; absence of boundary compressions therefore passes as well. Work with this neat representative for the following smooth collar constructions. The boundary-fixing topological ambient extension pulls any resulting boundary-parallel product back to the actual original surface. Thus this preparation does not add a neatness hypothesis to the Statement. If $S$ is boundary-incompressible, the first conclusion holds. Otherwise choose a boundary-compressing disk $D$, with $\mathrm{\partial }D=\alpha \cup \beta$ and $\alpha$ an essential proper arc of $S$. A boundary disk on a torus whose interior misses $\mathrm{\partial }S$ can be pushed inward to compress $S$, unless $S$ is itself a disk: a boundary circle bounding a disk on the connected surface forces the surface to be that disk. An innermost choice among the finitely many inessential boundary circles provides such a boundary disk. But a disk has no essential proper arc. Therefore every boundary circle of $S$ on the torus $T$ containing $\beta$ is essential. By [F2] its complementary regions are annulus strips. The arc $\beta$ lies in one of them. If both endpoints lie on the same boundary copy of that strip, it cuts off a disk there. Gluing this disk to $D$ gives a disk with boundary $\alpha$ and an interval of $\mathrm{\partial }S$; incompressibility of $S$ forces that circle to bound a disk on $S$, so $\alpha$ would cut off a disk on $S$, contrary to essentiality. Thus $\beta$ spans its annulus strip $A_0$.

The two endpoints of $\alpha$ lie on distinct actual boundary circles of $S$. The only alternative after step 1.1 would be that the two ends of $A_0$ are the two copies of one circle $c$, with $c$ the only boundary circle of $S$ on $T$. Two-sidedness rules this out. Along the connected arc $\alpha$, the interior of $D$ lies on one fixed normal side of $S$. Along $c$, that same side restricts to one fixed side of $c$ in $T$, since the proper surface is two-sided. Hence $\beta$ must approach the same side of $c$ at both endpoints; a spanning arc of $T$ cut along $c$ approaches opposite sides. This contradiction proves the assertion. Let $P$ be a small regular neighborhood in $S$ of $\alpha$ and its two incident boundary circles. It is a pair of pants. Its third circle $d$ bounds a disk just off $S$: boundary-compress the annulus strip $A_0$ along $D$ and push the resulting disk a little inward. Its interior misses $S$, because the interior of $A_0$ misses $\mathrm{\partial }S$ and the interior of $D$ misses $S$. Incompressibility makes $d$ bound a disk on $S$. This disk is on the side opposite $P$: if it contained $P$, the two boundary circles of $S$ in $P$ would lie in the interior of that disk, contrary to the surface half-space charts. Gluing the disk to $P$ fills a whole connected component of $S$, so $S$ is an annulus. The original $D$ is its boundary compression. Apply [F3] to get the second conclusion. Every neighborhood and disk surgery used only finitely many explicit surface and quadrant charts; [F2] and [F3] introduce no choice assumption.