We say that a Boolean algebra $B$ is projective if for all Boolean algebras $C$ and $D$, if $f:C\to D$ is an onto homomorphism, and $g:B\to D$ is any homomorphism, there is some homomorphism $h:B\to C$ making the obvious triangle commute. As an example, any free Boolean algebra is projective, and in fact, a Boolean algebra is projective if and only if it is a retract of a free one. There are some nice characterizations of projectivity in the Handbook of Boolean Algebras (Chapter 20), in terms of the existence of sequences of relatively complete subalgebras.

Let us say that a Boolean algebra $B$ is weakly projective if $$ \text{ for all Boolean algebras $C$, whenever $B$ is a surjective homomorphic image of $C$, then it is a subalgebra of $C$. }$$ (there is another notion of weak projectivity in the literature on Boolean algebras which seems to be unrelated to this one). Weak projectivity (of other kinds of algebras, but defined the same way) is useful in abstract algebraic logic and universal algebra to characterize varieties such that all of their subquasivarieties are already varieties. It is not hard to see that projectivity implies weak projectivity, however, I am having difficulty with showing the concepts are independent. I suspect that such a counterexample should exist for Boolean algebras, but I would be interested to know whether this notion has been studied in this setting before. I suspect it might have been, but since the naming conventions are so murky I can't find reference to it.

Any references or ideas are very welcome!

  • $\begingroup$ This is interesting (+1). Do you know of any example of a weakly projective Boolean algebra which you don't know if it is projective? $\endgroup$
    – amrsa
    Jan 28 at 19:58

1 Answer 1


I would be interested to know whether this notion has been studied in this setting before.

Projective Boolean algebras have been studied in many papers, starting with Halmos's 1961 paper. I do not know whether anyone has tried to study weakly projective Boolean algebras. I tried to find something about this today on MathSciNet but I was unsuccessful. Nevertheless:

Theorem. There exist weakly projective Boolean algebras that are not projective.

To show this, I will describe a way to generate some weakly projective Boolean algebras, and then show that some of them are not projective.

Lemma 1. Let $P$ be a projective algebra in some variety $\mathcal V$. If $W$ is an algebra in $\mathcal V$ for which there exists an injective homomorphism $\alpha\colon W\to P$ and a surjective homomorphism $\beta\colon W\to P$, then $W$ is weakly projective.

Proof. If $A\in {\mathcal V}$ and $h\colon A\to W$ is surjective, then $\beta\circ h\colon A\to P$ is surjective. Since $P$ is projective in $\mathcal V$, the homomorphism $\beta\circ h$ has a section, say $\sigma\colon P\to A$. Sections of surjective homomorphisms are injective, so the composite map $\sigma\circ\alpha\colon W\to A$ is injective. This embeds $W$ into $A$, verifying the weak projectivity of $W$. \\\

Corollary 2. If $F$ is a free Boolean algebra of infinite rank and $W\leq F\times F$ is the graph of a congruence on $F$, then $W$ is a weakly projective Boolean algebra.

Proof. If $F$ is a free Boolean algebra of infinite rank, then $F\cong F\times F$. Since $W$ is the graph of a congruence on $F$, $W\subseteq (F\times F)\cong F$, and the composition of the implied maps yields an embedding $\alpha\colon W\to F$. On the other hand, $W$ is a subdirect subalgebra of $F\times F$ since it is the graph of a congruence, so the first coordinate projection is a surjective homomorphism $\beta\colon W\to F$. Now apply Lemma 1 to conclude that $W$ is weakly projective. \\\

Now the hard part. It is known that any countable Boolean algebra of more than one element is projective. Therefore, to exhibit a weakly projective Boolean algebra that is not projective I must work with uncountable Boolean algebras. For this part I will argue that if $F$ is a free Boolean algebra of uncountable rank and $\theta$ is a congruence on $F$ of index two, then the graph $W$ of $\theta$ is not a projective algebra. (By Corollary 2, this $W$ will be weakly projective.)

Lemma 3. Let $F$ be a free Boolean algebra of uncountable rank $\kappa$ and let $\chi\colon F\to 2$ be a character/homomorphism. Let $I=\chi^{-1}(0)$ be the prime ideal of $F$ associated to $\chi$ and let $U=\chi^{-1}(1)$ be the ultrafilter of $F$ associated to $\chi$. The graph of $\ker(\chi)$ (which is $W=(I\times I)\cup (U\times U)\leq (F\times F)$) is not a projective Boolean algebra.

Proof. [It is a fact that any prime ideal or ultrafilter of a free Boolean algebra of rank $\kappa$ requires $\kappa$ generators. In particular, neither $I$ nor $U$ is principal.]

To prove that $W$ is not projective I will refer to the Bockstein Separation Property (BSP), which is a property that is satisfied by any projective Boolean algebra. The BSP asserts that if $J$ and $K$ are ideals satisfying $J\cap K = \{0\}$, then there exists countably generated ideals $J'\supseteq J, K'\supseteq K$ such that $J'\cap K'=\{0\}$. We will see that this property fails in $W$.

Let $J = I\times \{0\}$ and let $K=\{0\}\times I$. $J$ and $K$ are disjoint ideals of $W$. If $W$ were projective, then by the BSP there must be disjoint, countably generated ideals $J'\supseteq J, K'\supseteq K$. We will complete the argument by showing that (i) the disjointness of $J'$ and $K'$ forces $J'=J, K'=K$, while (ii) neither $J$ not $K$ is countably generated. Together, (i) and (ii) refute the BSP for $W$, so $W$ is not projective.

For Goal (i), suppose that $J'$ properly contains $J=I\times \{0\}$. If $(a,b)\in J'-J$, then since $(a,b)\in W=(I\times I)\cup (U\times U)$, we must have $a, b\in I$ or $a, b\in U$. In either case, $b\neq 0$. (If $b\in U$, then $b\neq 0$ since $0\notin U$; if $b\in I$, then $a\in I$ and since $(a,b)\notin J$ we have $b\neq 0$.) In either case we may choose $c<b$ such that $a, b, c\in I$ or $a, b, c\in U$. I am using that $U$ is not principal in the latter case. Thus, $(a,c) < (a,b)\in J'$, so $(a, c)\in J'$. We have $(a,c) \vee (0,b - c) = (a,b)\in J'$, $(0,b-c)\in J'\cap K \subseteq J'\cap K' = \{(0,0)\}$, which contradicts $c<b$. This contradiction establishes Goal (i).

For Goal (ii), assume that $J'=J=I\times \{0\}$ is a countably generated ideal in $W$. Apply the first projection map to this set of pairs we obtain that $I$ is countably generated in $F$. But, as noted in the first paragraph, $I$ requires $\kappa$ generators, and $\kappa$ is uncountable. This completes the proof of Goal (ii). \\\

  • $\begingroup$ This is excellent! It is great to see such an example. I would also be interested to know if this has been studied in these settings before, though I suspect it has not. My current research interest is in Heyting algebras and int. logic, and there this notion of weak projectivity is mostly considered for finite algebras; but also there small counterexamples seem unlikely to me (no finite ones can exist, and I suspect no countable ones either, but the structure of projective Heyting algebras is much less understood). Thank you for your excellent answer! $\endgroup$ Jan 29 at 13:17

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