Linear isoperimetry implies uniformly thin geodesic bigons

Write $X_0$ for the simple realization and $X$ for the labelled realization of the same presented group and generating alphabet. In $X_0$, identity letters are constant paths and parallel labels share a single edge; in $X$ all prescribed labelled edges are retained. F1 is the identical word/relator convention for both. The inequalities in F2 therefore apply to $X_0$, with the same fixed $K,L$ for every presentation. They give a number ${\delta }_0={\delta }_0(K,L)\ge 0$ bounding all its chosen triangles. The common bound, rather than merely a separate bound for each presentation, is exactly the uniformity conclusion of F2 under F5.

By F4, $X_0$ satisfies the four-point condition with ${\kappa }_0=3{\delta }_0$. Since both spaces induce the same word metric on their common vertex set by F3, all vertex quadruples in $X$ satisfy that identical condition. Equivalently, each of their three opposite-pair sums is at most the maximum of the other two plus $2{\kappa }_0$; for a sum that is not largest this is automatic, and for a largest sum it is the four-point hypothesis.

For arbitrary points $a,b,c,d\in X$, choose a nearest endpoint vertex $\widehat{a},\widehat{b},\widehat{c},\widehat{d}$ on each of their unit edges, and use the point itself if it is already a vertex. Each distance to the selected vertex is at most $1\mathrm{/}2$, including loop edges; these are four finite choices. The triangle inequality yields $\mathrm{\mid }{d}_X(u,v)-d_X(\widehat{u},\widehat{v})\mathrm{\mid }\le 1$ for each pair. Consequently corresponding opposite-pair sums differ by at most $2$. Denote the three sums for the original points by $S_i$ and the corresponding vertex sums by ${\widehat{S}}_i$, $1\le i\le 3$. For each $i$, step 2.1 gives $S_i\le {\widehat{S}}_i+2\le {\max }_{j\mathrm{\ne }i}{\widehat{S}}_j+2{\kappa }_0+2\le {\max }_{j\mathrm{\ne }i}{S}_j+2{\kappa }_0+4.$ Thus the largest sum is at most the second-largest plus $2({\kappa }_0+2)$, including ties. This proves the four-point condition on all of $X$ with constant ${\kappa }_0+2$.

The geodesicity of $X$ in F3 and F4 now make every chosen triangle in $X$ $4({\kappa }_0+2)=12{\delta }_0+8$-slim. This depends only on $K,L$, even if many labels represent the same generator or the identity; rounding at a loop uses its shorter half-interval and needs no map collapsing that loop.

For two specified geodesics from $x$ to $y$, regard them as two sides of a triangle with vertices $x,y,x$ and with third side the constant segment at $x$. Step 4.1 places every point of either side within $12{\delta }_0+8$ of the other side together with $x$; since $x$ already lies on that other side, their Hausdorff distance is at most this same constant. This includes $x=y$, constant sides, an empty generating alphabet (a one-vertex group), empty relator sets, $K=0$ and $L=0$. All added realization-comparison choices were finite; the assumed AC is used only through F2's explicitly named cone and uniformity selections.