= Free group action on a tree
Suppose a <group> $G$ acts on a <tree> without <edge inversions>, and only the identity fixes a vertex. Then $G$ is a <free group>. Equivalently, the induced action on the <geometric realization> is free on every point: no nonidentity element fixes a vertex or an edge midpoint.
The quotient is a <graph>, and the map from the tree is a <covering graph>. Indeed, two incident edge germs with the same image would be related by a vertex stabilizer, which is trivial. Since the tree is <simply connected>, it is the <universal cover> of the quotient. Choosing lifts of a maximal subtree identifies $G$ with the quotient graph's <fundamental group>, freely generated by the edges outside that subtree.
The no-inversion hypothesis matters. The order-two group acting on a single edge by swapping its endpoints fixes no vertex, yet fixes the midpoint. That group is not a <free group>, and its quotient map folds the edge instead of giving a <covering graph>. <Barycentric subdivision> exposes the midpoint as a fixed vertex, removing this ambiguity.
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