Weakening of projectivity for Boolean algebras independent from projectivity 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!
 A: 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).
\\\
