A Counterexample to a Corner-Peeling Conjecture for Subcomplexes of Z^3

Let G be a finite connected subgraph of the grid ℤ³, not necessarily induced, and let X(G) be the cube complex whose cells are the squares and 3-cubes of ℤ³ all of whose vertices and edges lie in G. A corner is a vertex that lies in exactly one maximal cell, and a corner peeling is an ordering v₁, …, vₙ of the vertices in which each vᵢ is a corner of the subgraph induced by v₁, …, vᵢ. In the problem session of the 2026 Oberwolfach workshop Median Geometry and Applications, V. Chepoi, from discussions with J. Chalopin and M. Kokkou, conjectured that G admits a corner peeling whenever X(G) is simply connected and every connected component of every section of X(G) by a coordinate plane is simply connected. We show that the conjecture is false. We give a subgraph G₆₂ of ℤ³ with 62 vertices in [0,3]³ and an induced subgraph H₇₃ with 73 vertices in [0,4]³ such that the cube complex and all its sections are collapsible, but there is no corner at all. Both examples are invariant under a cyclic group of order 6, and the disproof can be checked by hand; we also give explicit sequences of elementary collapses as machine-checkable certificates. A subdivision construction turns every finite subgraph into an induced subgraph whose pieces and corners are dilates of the original ones, so the conjecture for partial subgraphs and the conjecture for induced subgraphs are equivalent. Uncertified solver computations indicate that no counterexample fits into a 3 × 3 × 3 box of lattice points. We also record a smaller counterexample, with 49 vertices, found in an exploratory solver search during an independent verification of this note and checked by computer. This is an unrefereed note.