About the existence of a particular kind of "splitting" function on atomless complete Boolean algebras Let $\mathbb{B} = \langle B, \wedge, \vee, \leq, \neg, 0, 1 \rangle$ be an atomless complete Boolean algebra.
We call $f$ a splitting function on $\mathbb{B}$ iff


*

*$f : B-\{1\} \longrightarrow B \times B \ \ \ (b \mapsto (b_0, b_1))$,

*$b_i \leq b$ for all $b \in B-\{1\}$ and $i \in \{0, 1\}$,

*$b_0 \wedge b_1 = 0$ for all $b \in B-\{1\}$.


A splitting function on $\mathbb{B}$ is monotone iff 


*$b \leq b'$ implies $b_i \leq b_{i}'$, for all $\{b, b'\} \subset B-\{1\}$ and $i \in \{0, 1\}$.


Monotone splitting functions trivially exist: consider eg. the map $b \mapsto (0, b)$. However, it is not so clear to me how non-trivial these can be. 
In particular, is there always a monotone splitting function on $\mathbb{B}$ satisfying 


*$\sup\{b_{i}' : b \leq b'\} = \neg b_{1-i}$ for all $b \in B-\{1\}$ and $i \in \{0, 1\}$?


If not, what additional conditions on $\mathbb{B}$ can guarantee its existence, and in what cases does it fail to exist?
 A: I think you are asking too much.
Assume we have such a function and let $a$ be nonzero such that both $a_0$ and  $a_1$ are nonzero. Then $b\le a_0$ implies $b_1=0$ and $b\le a_1$ implies $b_0=0$. 
If $a$ is nonzero such that $a_0=0$ and $0<a_1<a$ then $b\le a\setminus a_1$ implies both $b_0=0$ and $b_1=0$, so the function is trivial below $a\setminus a_1$.
This shows that the union of the three sets $\{a:a_0=a_1=0\}$, $\{a:a_0=0\land a_1=a\}$ and $\{a:a_0=a\land a_1=0\}$ is open and dense in the algebra.
A: Claim: Let $a \mapsto (a_0, a_1)$ be any map (for $a < 1_B$) satisfying (1)-(4). Call a condition $b < 1_B$ left-trivial (resp. right-trivial) if $(\forall a \leq b)(a_0 = 0_B)$ (resp. $(\forall a \leq b)(a_1 = 0_B)$). Then we can partition $1_B$ into two trivial conditions.
Proof: Let $D$ be the family of all trivial members of $B \setminus \{0_B, 1_B\}$. Note that if $0_B < x < 1_B$ and $x$ is nontrivial then, then for some $y \leq x$, both $y_0, y_1 > 0_B$. It follows that $y_0$ is right-trivial and $y_1$ is left-trivial. So $D$ is dense in $B$. Let $A$ be a maximal antichain of members of $D$. Let $x$ (resp. y) be the union of all left-trivial (resp. right-trivial) members of $A$. Then $\{x, y\}$ is such a partition.
It follows that for any $z < 1_B$, $z_0 = z \cap y$ and $z_1 = z \cap x$. This determines all such maps.
