Let $B\_n$ be the boolean lattice of rank $n$. Let $\hat{0}$ and $\hat{1}$ be the minimum and the maximum, respectively.  

We identify the notion of edge with the notion of interval $[a,b]$ of cardinal $2$.  

We propose to label every edge with the symbols $\alpha$ or $\beta$, such that: 


1. For every maximal chain, exactly one edge is labeled by $\alpha$.
2. For every element $a \in B\_n$, $a \neq \hat{1}$, then at most one edge $[a,x]$ is labeled by $\alpha$.  

(note that $x$ is an atom of the interval $[a, \hat{1}]$)


*Question*: Is there a coatom $y$ of $B\_n$ such that the edge $[y,\hat{1}]$ is labeled by $\alpha$?



A problem with an edge labeling on the boolean lattices
Yes. For $n=1$, the result is clear as we must choose $\alpha$ for the one and only edge. For $n>1$, by the second condition there must be an atom $a \in B\_n$ such that $[\hat 0, a]$ is not labeled with $\alpha$. Now for the first condition to hold every maximal chain in $[a, \hat 1] \cong B\_{n-1}$ must have exactly one edge labeled with $\alpha$. Thus both conditions must hold for $[a, \hat 1] \cong B\_{n-1}$. By induction a edge from a coatom of $[a, \hat 1]$ must be labeled with $\alpha$, the result follows since coatoms of $[a, \hat 1]$ are coatoms of $B\_n$.


