Rooted spanning trees and Schreier systems correspond

Let $T$ be a rooted spanning tree. For each vertex $v$, let $p_v$ be the unique simple path in $T$ from $H$ to $v$, and let $\tau (v)$ be its label. A simple path in a tree never backtracks, so $\tau (v)$ is reduced. Because $T$ is spanning, every coset has some label $\tau (v)$; because $T$ is a tree, the path $p_v$ is unique, so no two distinct words $\tau (v)$ label the same vertex. Any initial segment of $\tau (v)$ labels an initial subpath of $p_v$, hence labels the tree path to an earlier vertex. Therefore $\{\tau (v)\}$ is a Schreier system.

Conversely, let $\mathcal{T}$ be a Schreier system. For each non-base representative $t=x_1\cdots x_n\in \mathcal{T}$, join the vertex $Ht$ to the vertex $H{x}_1\cdots x_{n-1}$ by the final labeled edge used to read $t$. Because prefixes stay in $\mathcal{T}$, every non-base vertex acquires exactly one parent; because the parent has smaller word length, repeatedly following parent edges must terminate at $H$. Thus every vertex is connected to $H$, and a cycle cannot occur because along a cycle one could not keep decreasing length and return to the starting vertex. So these parent edges form a rooted spanning tree.

The two constructions are inverse. Reading the tree path from $H$ recovers each representative in the Schreier system, and taking parent edges from the prefixes of those representatives reconstructs the original rooted tree.