It will be useful to have other ways of showing that a given function between boolean rings is a morphism. The definition, of course, is that and , since and here denote our addition and multiplication, respectively.
It’s also sufficient to show that and . Indeed, we can build both and from and :
So if and are preserved, then so are and .
A little more subtle is the fact that for surjections we can preserve order in both directions. That is, if is equivalent to for a surjection from one underlying set onto the other, then is a homomorphism of Boolean rings. We first show that preserves by using its characterization as a least upper bound: , , and if and then .
So, since preserves order, we know that , and that . We can conclude from this that , and we are left to short that . But f(E)\cup f(F)=f(G)$ for some , since is surjective. Since , we must have , and similarly . But then , and so , as we wanted to show.
We thus know that preserves the operation , and we can similarly show that preserves , using the dual universal property. In order to build and from and , we need complements. But complements satisfy a universal property of their own: , and if then .
First, I want to show that by showing it is below all elements of . Indeed, by the surjectivity of , every element of is the image of some element . Thus we want to show that . But this is true because for all , and so
Now given a set and its complement , we know that . Since preserves , we must have . Let’s say is another set so that . By the surjectivity of , we have for some . Thus , and thus . This tells us that , and thus . And so .
Therefore, since preserves , , and complements, it preserves and , and thus is a morphism of Boolean rings.