I was thinking about perfect matchings in the graph of the unit-cube of dimension $n$: $Q_n = [0,1]^n$. ($0$-$1$-strings of length n are vertices. Two of such are connected by an edge iff. they differ by only 1 bit (i.e. Hamming distance is equal to 1)).
Suppose you're given an arbitrary perfect matching $M$ in $Q_n$. Is it possible to find a facet $F$ of $Q_n$ such that all matching edges $e = (v,w) \in M$ that cover a vertex in $F$ lie in $F$? Stated differently, if $v \in F$ and $(v,w) \in M$ for some vertex $w$, does that imply that $w \in F$? This statement should be equivalent to finding a separating hyperplane, that separates $Q_n$ into two $(n-1)$-cubes, such that no matching edge crosses the hyperplane.
What do you think, is this true or false? Obviously its false for $n=1$ but I think it should hold for all $n \geq 2$, but can't prove it!
Any advice on how to prove it or a counterexample is greatly appreciated! Thanks!
Here is a perfect matching in $Q_4$ that does not have this property.