A problem with an edge labeling on the boolean lattices 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:  


*

*For every maximal chain, exactly one edge is labeled by $\alpha$.  

*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: 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 $B_n$.
A: John Machacek's elegant answer solves the question, but we can also answer the question by characterizing all the labelings. For each labeling $\lambda$, consider a labeling $\pi$ of elements in $B_n$ such that $\pi(a)$ is the number of edges with label $\alpha$ on each chain from $\hat 0$ to $a$, for every $a\in B_n$. $\pi$ is well-defined as each chain from $\hat 0$ to $a$ must have the same number of edges with label $\alpha$. And the domain of $\pi$ is $\{0,1\}$ and $\pi$ is monotone, that for every $a,b\in B_n$, $a\geq b$ implies $\pi(a)\geq \pi(b)$. Let $S\subset B_n$ be the set of elements that $\pi(a)=1$. By monotony, $a\geq b$ and $b\in S$ implies $a\in S$. If $a$ and $b$ has the same cardinal and differ by exactly one element, in other words, there is element $c$ and distinct atoms $x$ and $y$ that $a=c\cup\{x\}$ and $b=c\cup\{x\}$, then $a\in S$ and $b\in S$ implies $c\in S$ by the second assumption in the problem. So by induction, $a\in S$ and $b\in S$ implies $(a\cap b)\in S$. So $S$ must be an interval $[c, \hat 1]$ for some $c\in B_n$. And evidently $[\hat 0, \hat1]$ does not correspond to any labeling $\lambda$ while any other $S=[c, \hat 1]$ corresponds to a valid labeling $\lambda$. So we can find an atom $x\not \in c$, and then the edge $[x^c, \hat 1]$ is labeled with $\alpha$.
A: If we add the following third condition:  


*For every element $b \in B_n$, $b \neq \hat{0}$, then at most one edge $[x,b]$ is labeled by $\alpha$.  


Then we can prove:    
Proposition: There exists an atom $a$ such that an edge $[x,y]$ is labeled by $\alpha$ iff $y=x \vee a$.    
Proof: Let $\lambda$ be the labeling map. We start with an induction as for John's answer. If $n=1$ it is ok. We assume that it is true at rank $<n$ and we will prove for rank $n>1$. There exists an atom $s$ such that $\lambda([\hat{0},s]) = \beta$, then we can apply the induction on $[s,\hat{1}]$, and it follows the existence of an atom $a$ of $[\hat{0}, s^c]$ such that an edge $[x,y]$ of $[s,\hat{1}]$  is labeled by $\alpha$ iff $y=x \vee a$. So $\lambda([a^c,\hat{1}]) = \alpha$, then $\lambda([s^c,\hat{1}])=\beta$. We apply the induction again on $[\hat{0},s^c]$, it follows the existence of an atom $a'$ of $[\hat{0}, s^c]$ such that an edge $[x,y]$ of $[\hat{0}, s^c]$ is labeled by $\alpha$ iff $y=x \vee a'$. We just need to prove that $a'=a$. If $a \neq a'$ then there is a maximal chain starting by $[\hat{0},a']$ and finishing by $[a^c,\hat{1}]$, but both are labeled by $\alpha$, contradiction. $\square$
