Weakened Boolean algebra homomorphism I think this is almost trivial, but I am having some trouble visualizing why. Let $\mathcal{A}$ and $\mathcal{B}$ be Boolean algebras and $f$ a function from $\mathcal{A}$ to $\mathcal{B}$ satisfying, for all $a$ and $b$ in $\mathcal{A}$,

*

*$f(a\vee b)=f(a)\vee f(b)$;

*$f(a\wedge b)=f(a)\wedge f(b)$;

*$f(a\rightarrow b)=f(a)\rightarrow f(b)$.

It seems to me that $f$ does not need to be a Boolean algebra (BA) homomorphism,: taking the two-valued BA with $\{0,1\}$ as elements, and the four-valued BA with $\{0, a, b, 1\}$ as elements ($\neg a=b$ and vice-versa), $f$ defined by $f(0)=a$ and $f(1)=1$ seems to satisfy all requirements, and it is clearly not a homomorphism.
Furthermore, it also seems to me that the class of Boolean algebras, equipped with these functions as morphisms, should be a category. However, I have never seem it defined anywhere: am I correct that this is indeed a category, different from the usual category of Boolean algebras? If so, does it have a standard name?
 A: I don't know a name for your precise definition but it is almost the same as the category of non-unital Boolean algebras.  What is a non-unital Boolean algebra?  Well, roughly speaking, it is a Boolean algebra without the element $1$, where the complement operation $\neg$ is replaced by a relative complement operation $a\setminus b$ (since there is no "universe" in which to be taking the absolute complement).  There are several equivalent ways of making this precise.  One is to say that a non-unital Boolean algebra is (up to isomorphism) a collection of sets containing $\emptyset$ and closed under the operations $\cup,\cap,$ and $\setminus$.  Another is to say that a non-unital Boolean algebra is an unbounded distributive lattice with a least element and with relative complements, i.e. for each $a$ and $b$ there exists an element $a\setminus b$ such that $(a\setminus b)\vee (a\wedge b)=a$ and $(a\setminus b)\wedge(a\wedge b)=0$.  (This means that for each $a$, the set of elements below $a$ forms an ordinary Boolean algebra with $a$ as the greatest element.)  Homomorphisms of non-unital Boolean algebras are just functions that preserve $0,\wedge,\vee,$ and $\setminus$ (though preserving $0$ is automatic since it is $a\setminus a$ for any element $a$).  If $\mathcal{A}$ and $\mathcal{B}$ are ordinary Boolean algebras, then a non-unital homomorphism $\mathcal{A}\to\mathcal{B}$ is the same as an ordinary Boolean homomorphism $\mathcal{A}\to(b)$ where $(b)$ is the ideal in $\mathcal{B}$ generated by some element $b\in\mathcal{B}$ (which is itself a Boolean algebra with $b$ as its greatest element, and a sub-non-unital Boolean algebra of $\mathcal{B}$).  Another equivalent definition is that a non-unital Boolean algebra is a Boolean rng, i.e. a non-unital ring in which every element is idempotent, with the homomorphisms being the usual homomorphisms of non-unital rings.
What does all this have to do with your definition?  Well, given a function $f:\mathcal{A}\to\mathcal{B}$ between Boolean algebras, you can equivalently consider the "dual" map $g(a)=\neg f(\neg a)$.  Then $g$ preserves joins iff $f$ preserves meets and $g$ preserves meets iff $f$ preserves joins.  Moreover, since $\neg(\neg a\to\neg b)=\neg a\wedge b=b\setminus a$, $f$ preserves $\to$ iff $g$ preserves relative complements.  So your homomorphisms are just the duals of the non-unital Boolean algebra homomorphisms.  So your category is isomorphic to the full subcategory of non-unital Boolean algebras consisting of the algebras that happen to have a greatest element (but that greatest element does not have to be preserved by the morphisms).  Concretely, a map $f:\mathcal{A}\to\mathcal{B}$ is a homomorphism in your sense iff it is a Boolean homomorphism when considered as a map $\mathcal{A}\to\mathcal{C}$ where $\mathcal{C}=\{b\in \mathcal{B}:b\geq f(0)\}$ (which is a Boolean algebra with $f(0)$ as its least element).
