necessity of $f(0)=0$ and $f(1)=1$ in homomorphisms of boolean algebras Let $A,B$ be boolean algebras and let $f \colon A \rightarrow B$. $f$ is a homomorphism of boolean algebras if $f$ is a homomorphism of the corresponding lattices and $f(0)=0$ and $f(1)=1$.
Why is it necessary to specify that $f(0)=0$ and $f(1)=1$? 
Suppose $f$ is a homomorphism of lattices. We know that $a \vee 0 =a \Longleftrightarrow a \wedge  0 = 0$ for all $a \in A$, so $f(0) = f(0 \wedge a) = f(0) \wedge f(a)$ for all $a \in A$. That means $f(0)$ acts as the zero element for every element in the image of $f$, but it doesn't mean it's a zero for all the elements in $B$. 
Is this why the condition is necessary? Or am I not seeing something else? (if $f$ was an isomorphism, would the condition be redundant?)
 A: Usually, when we consider a class of algebras, we want the homomorphisms to preserve all operations used in the algebra.  
The nullary operations $0,1$ are parts of the Boolean algebra structure just as the $\vee$ or $\wedge$ are. So we want them to be preserved by homomorphisms. We want to preserve the complement $x\mapsto \neg x$ as well; however in the case of Boolean algebras, we get the preservation of the complement for free. So it is not necessary to include the preservation of the complement in the definition of the homomophism of Boolean algebras. But we cannot do this with $0$ and $1$.
Another way how to look at these things:
Fact 1 The category of Boolean algebras is a full subcategory of the category of bounded distributive lattices.
As you (basically) observed yourself, we cannot drop "bounded" from this Fact. Nor we can drop "distributive". 
There is exactly the same relationship between groups and semigroups: we do not need to include preservation of the inverses and of the unit element in the definition of a group homomorphism.
