Every embedded torus in a ball is compressible

The interior of the parametrized PL ball is a contractible boundaryless topological three-manifold. Its second mod-two homology is zero by [F4]. Hence [F1] and [F4] show that its complement of $T$ cannot be connected. The same is true in $B$: push its outer boundary collar slightly inward, fixing the compact interior torus; if $B\setminus T$ were connected, its connected pushed image and the collar homotopy would make the interior complement connected. There are exactly two components. Indeed the two open product-collar sides of connected $T$ are connected, and every complementary component must meet one: otherwise its closure misses $T$ and it is open and closed in connected $B$. More explicitly, at any point of $T$ in the closure of a component, a local flat chart puts some collar-side patch in that component; the connected full side collar then belongs to it. Thus there are at most two components. Their closures $W_1,W_2$ are compact orientable topological three-manifolds, with the corresponding copies of $T$ in their boundaries.

If both maps ${\pi }_1(T)\to {\pi }_1(W_i)$ were injective, choose their coset transversals by [A1]. The given topological product collar gives the connected-overlap van Kampen cover, and [F3] expresses ${\pi }_1(B)$ as their amalgam with ${\pi }_1(T)$ embedded. By [F5] this would embed ${\mathbb{Z}}^2$ into the trivial group of the ball, a contradiction. Thus one side boundary map has nontrivial kernel. This conclusion is topological and has required no triangulation of the cut pieces.

Apply [F2]'s ACTUAL TOP triangulation theorem to that compact orientable side $W_i$, using [A1]. Let $q:\mathrm{\mid }L\mathrm{\mid }\to W_i$ be its resulting finite manifold triangulation. The preimage $q^{-1}(T)$ of the whole boundary component is a connected PL boundary surface subcomplex; no pre-existing compatibility of the original embedded torus with the original ball triangulation was needed. Conjugating by $q$ identifies its inclusion kernel with the nontrivial kernel from step 2.1. Apply finite PL Loop with that entire boundary component and normal subgroup $1$. It supplies a proper embedded PL disk in $\mathrm{\mid }L\mathrm{\mid }$ with boundary representing a nonidentity element of its torus group. Transport it by $q$ and then by the cut-side inclusion into $B$. These maps are homeomorphisms on the disk: the disk meets only this single boundary face and its interior misses all boundary faces. Hence its image $D$ is a TOP embedded disk in $\mathrm{int}B$, with $D\cap T=\mathrm{\partial }D$ and essential boundary on $T$. It is proper in the cut side, as the Statement requires. This completes the unqualified TOP two-sided torus case with the original full AC hypothesis.