I was thinking about Duhrer's problem the other day: "Can every convex polyhedron be unfolded into a non-intersecting net?" The cuts that you make to form the net must be a spanning tree of the edges of the polyhedron being cut.
Since a net is also a tree (whose vertices are instead faces of the original polyhedron), it occurred to me that "every tree of cuts corresponds to a net of the dual of the polyhedron." When it first occurred to me it was so intuitive that I nearly accepted that it was true, but I'm not sure how to prove it. Perhaps by induction on the number of faces: that would perhaps allow us to enumerate all the possible cases for how many vertices are added (for each addition of a face) and therefore how many faces are added to the dual, which could perhaps allow us to determine how the net and the spanning tree have to change.
This sounds like a hefty task, so I thought I'd instead ask here whether it has been proven or disproven, and whether or not the proof is trivial.
The essential point in your statement is the convexity of the original polyhedron. This ensures that the initial tree of cuts has to reach indeed every vertex of that polyhedron.
The remainder of the proof then is just to observe that both the cuts and the obtained net (of the original polyhedron) form a tree. One being built from vertices and connecting edges, the other being built from faces and connecting edges. And for sure the polyhedral duality itself, which interchanges vertices of one and faces of its dual, resp. pairs of orthogonal edges of either one.
Thus replacing every vertex of the tree of cuts by the corresponding face of the dual polyhedron as such already ensures that you will have encountered all its faces exactly once. Further the original tree does provide a clue on how those vertices are to be connected (line segments of that tree). This is being transfered into a connection advice of those dual faces, that is, this other tree indeed represents a net of the dual polyhedron.
--- rk