Minimal van Kampen area agrees with minimal algebraic relator area

Let $D$ be any van Kampen diagram for $w$ with $m$ faces. We show by induction on $m$ that $[w]$ is a product of at most $m$ conjugates of defining relators or their inverses in the free group. For $m=0$, $D$ is a tree by the graph argument in [L2]'s cited proof; its outer walk freely reduces to the empty word, including the one-vertex case.

Conversely, let $[w]={\prod }_{k=1}^m[u_k{r}_k^{{\epsilon }_k}{u}_k^{-1}]$ be an algebraic expression with $m$ minimal as in [F1]. The converse construction in [L1] produces a diagram whose literal boundary word is $w$ and whose face count is at most $m$. Therefore the minimal diagram area is at most the algebraic relator area.

Suppose $m>0$. A generic ray from an interior point of the finite union of face closures to infinity crosses edge interiors and avoids vertices. At its last exit from that union it crosses an edge $e$ with one face on its inner side and the unbounded complementary region on its other side, so $e$ occurs once in the outer walk. Collapse that face across $e$ as in [L2], obtaining a planar simply connected diagram $D^{\mathrm{\prime }}$ with $m-1$ faces. Write the old outer word from the fixed basepoint as $aeb$, with $e$ denoting its oriented label, and let $p$ be the complementary face-boundary walk joining the same edge endpoints. The new outer walk reads $apb$ (including any bridge traversals). Consequently $[aeb][apb{]}^{-1}=[ae{p}^{-1}{a}^{-1}]$, a conjugate of the cyclic face label, hence of a defining relator or its inverse. By induction $[apb]$ needs at most $m-1$ factors, so $[w]$ needs at most $m$. Thus the algebraic area is at most every diagram area.

Steps 1.1 and 2.1 give one inequality between the two minima and step 1.2 gives the reverse inequality. Therefore the two minimal areas are equal.