---

Source. Sections 4.3–4.4, printed pp. 454–455 (published PDF).

Statement. In a planted tree for $M$, the path from its root $r$ to every outer vertex $v$ is a stem. Any alternating path inside the tree which ends at $v$ with its matching edge is an initial segment of the backward path from $v$ to $r$. An edge from $v$ to an exposed vertex $w$ outside the tree therefore gives an augmenting path from $r$ to $w$.

Proof. For $v=r$ the root path has length zero. At any other outer vertex, start backwards with its unique matching edge. This reaches an inner vertex. That vertex has degree two in the tree, so its other, nonmatching edge is the only possible next edge. At the next outer vertex the matching edge is again unique unless it is the exposed root.

These forced steps cannot repeat a vertex, since the tree has no circuit. Finiteness forces termination. No inner vertex can stop the tracing, and the only outer vertex without a matching edge is $r$, by Section 4.2. The traced path is thus the unique tree path to $r$. It alternates, and its forward final edge at $v$ is a matching edge, making it a stem. The same forced-step argument gives the stated containment for every shorter alternating path inside the tree.

The new edge $vw$ is outside $M$, as $w$ is exposed. Appending it to the stem gives a simple alternating path with distinct exposed endpoints $r,w$. The augmenting-path criterion applies. $\square$