Suppose $Q$ is an atomless countable boolean algebra, and $B$ is an arbitrary atomless boolean algebra. $Q$ is unique modulo isomorphisms. There is a subalgebra in $B$ that is isomorphic to $Q$. There is probably a mapping from $B$ to $Q$ that preserves all boolean operations, but I need something different. Let $f$ be an epimorphism (cover) from $B$ to $Q$ with the following properties:


1. If $b\_1\le b\_2$ then $f(b\_1)\le f(b\_2)$. (previously there was requirement that $\wedge$ should be preserved, but actually I don't need it)
2. $f(b\_1\vee b\_2)=f(b\_1)\vee f(b\_2)$
3. $f(b)=0$ iff $b=0$
4. $f(1)=1$


The negation operation does not need to be preserved.


Does $f$ exist for any $B$?


Example:


$B = \{\textrm{periodic sequences of non-negative reals with integer period}\}\subseteq 2^{\{r\ge 0\}}$


$Q = \{\textrm{periodic sequences of non-negative integers}\}\subseteq 2^{\{k\ge 0\}}$


$f(\alpha)=\{i : [i,i+1)\cap\alpha\ne\varnothing\}$



Countable atomless boolean algebra covered by a larger boolean algebra
The answer to the revised version of the question is **Yes**. In fact, there is no need to assume that B is atomless, but rather, only that it is infinite. 


Suppose that B is any infinite Boolean algebra. It follows that there is a countable maximal antichain A subset B. The idea of the proof is to map A arbitrarily into your countable atomless Boolean algebra Q, and then extend to B in a way I will describe. Enumerate the maximal antichain A = { an | n in ω } and the nonzero elements of Q as { qn | n in ω}. We will associate an with qn. In order to define f, suppose that b is any element of B. Let Ab = { qn | b ∧ an not = 0 } be the associated set in Q. Define the function f:B to Q by f(b) = ∨ Ab, if Ab is finite, and otherwise f(b)=1.


We now make several observations about this function f. First, the function is clearly onto, since f(an)=qn. Also, f(1)=1, since 1 meets every element of A, and f(0)=0 since 0 meets no elements of A. Moreover, f(b)=0 iff b=0, since no nonzero element of B has zero meet with every element of A, as A was a *maximal* antichain. 


Because (b ∨ c) ∧ a = (b ∧ a) ∨ (c ∧ a), it follows that Ab ∨ c = Ab union Ac. From this, it follows that f(b ∨ c) = f(b) ∨ f(c), since if either set is infinite, then the answer is 1, and if they are finite, we are taking the join of two finite joins. Thus, f is join-preserving.


It follows that f is an order-homomorphism, since b <= c implies b ∨ c = c implies f(b) ∨ f(c) = f(b ∨ c) = f(c) implies f(b) <= f(c). 


So f has all the desired properties. 


Note that f definitely does not respect negation, since f(neg an) = 1 for every n. And f definitely does not respect meet, since any two elements of A have meet 0, but the corresponding qn must sometimes be nonzero.


This construction has some affinity with your example. Namely, if you take the various half-open unit characteristic functions as the elements of the maximal antichain (and use the corresponding qn's), then your f and my construction are the same.


