Question

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 b₁≤b₂ then f(b₁)≤f(b₂). (previously there was requirement that ∧ should be preserved, but actually I don't need it)

2) f(b₁∨b₂)=f(b₁)∨f(b₂)

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 = {periodic sequences of non-negative reals with integer period}⊆ 2^{r≥0}

Q = {periodic sequences of non-negative integers}⊆ 2^{k≥0}

f(α)={i : [i,i+1)∩α≠∅}

Answer 1

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 = { a_n | n in ω } and the nonzero elements of Q as { q_n | n in ω}. We will associate a_n with q_n. In order to define f, suppose that b is any element of B. Let A_b = { q_n | b ∧ a_n not = 0 } be the associated set in Q. Define the function f:B to Q by f(b) = ∨ A_b, if A_b is finite, and otherwise f(b)=1.

We now make several observations about this function f. First, the function is clearly onto, since f(a_n)=q_n. 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 A_{b ∨ c} = A_b union A_c. 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 a_n) = 1 for every n. And f definitely does not respect meet, since any two elements of A have meet 0, but the corresponding q_n 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 q_n's), then your f and my construction are the same.

Answer 2

Edit: This is the answer to the original question. See my other answer for the answer to the revised question.

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The answer is that every such f is an isomorphism. Thus, there is such an f only when B is isomorphic to Q.

The main reason is that your conditions imply that f must preserve negation. That is, for any b in B, then

b ^ -b = 0, so you have 0 = f(0) = f(b ^ -b) = f(b) ^ f(-b),

and

b v -b = 1, so you have 1 = f(1) = f(b v -b) = f(b) v f(-b).

So f(b) and f(-b) are complements in Q, and thus f(-b) = -f(b) in Q.

Thus, f preserves all the Boolean algebra structure. Next, since you have f(b)=0 IFF b=0, it means that f must be an isomorphism.

In particular, there is no such f for your example B and Q, since B is uncountable.

If you relax that IFF, then it just means that Q is a quotient of B by the filter F=f^inverse(1).