Net of a polyhedron

Remarks

Example: let the unit cube have top face $ABCD$ and bottom face $EFGH$, with $A$ above $E$, $B$ above $F$, $C$ above $G$ and $D$ above $H$. Let $T$ consist of the four vertical edges $AE$, $BF$, $CG$, $DH$ and the three top edges $BC$, $CD$, $DA$. These $7$ edges join all $8$ vertices without a cycle, so $T$ is a spanning tree. The unfolding along $T$ puts the bottom square in the middle, the four side squares around it, and the top square attached to the side $ABFE$ along $AB$. This is the familiar cross of six squares, and the squares do not overlap, so it is a net. Up to symmetry the cube has $11$ distinct nets, and none of its edge unfoldings overlaps.

Non-example: cutting the cube only along the four edges of its top square does not use a spanning tree, since these edges form a cycle and miss the bottom vertices. The top face falls off, so the surface is not unfolded as one piece.

For a convex polytope, the cut edges of any unfolding of the surface into one flat piece along edges form a spanning tree, so the definition loses no unfoldings. Some edge unfoldings of some convex polytopes overlap, which is why a net requires the extra disjointness condition.