# The Unapologetic Mathematician

## Specifying Morphisms Between Boolean Rings

It will be useful to have other ways of showing that a given function $f:\mathcal{S}\to\mathcal{T}$ between boolean rings is a morphism. The definition, of course, is that $f(E\Delta F)=f(E)\Delta f(F)$ and $f(E\cap F)=f(E)\cap f(F)$, since $\Delta$ and $\cap$ here denote our addition and multiplication, respectively.

It’s also sufficient to show that $f(E\setminus F)=f(E)\setminus f(F)$ and $f(E\cup F)=f(E)\cup f(F)$. Indeed, we can build both $\Delta$ and $\cap$ from $\setminus$ and $\cup$:

\displaystyle\begin{aligned}E\Delta F&=(E\setminus F)\cup(F\setminus E)\\E\cap F&=E\setminus(E\setminus F)=F\setminus(E\setminus F)\end{aligned}

So if $\setminus$ and $\cup$ are preserved, then so are $\Delta$ and $\cap$.

A little more subtle is the fact that for surjections we can preserve order in both directions. That is, if $E\subseteq F$ is equivalent to $f(E)\subseteq f(F)$ for a surjection $f$ from one underlying set onto the other, then $f$ is a homomorphism of Boolean rings. We first show that $f$ preserves $\cup$ by using its characterization as a least upper bound: $E\subseteq E\cup F$, $F\subseteq E\cup F$, and if $E\subseteq G$ and $F\subseteq G$ then $E\cup F\subseteq G$.

So, since $f$ preserves order, we know that $f(E)\subseteq f(E\cup F)$, and that $f(F)\subseteq f(E\cup F)$. We can conclude from this that $f(E)\cup f(F)\subseteq f(E\cup F)$, and we are left to short that $f(E\cup F)\subseteq f(E)\cup f(F)$. But f(E)\cup f(F)=f(G)\$ for some $G\in\mathcal{S}$, since $f$ is surjective. Since $f(E)\subseteq f(G)$, we must have $E\subseteq G$, and similarly $F\subseteq G$. But then $E\cup F\subseteq G$, and so $f(E\cup F)\subseteq f(G)=f(E)\cup f(F)$, as we wanted to show.

We thus know that $f$ preserves the operation $\cup$, and we can similarly show that $f$ preserves $\cap$, using the dual universal property. In order to build $\setminus$ and $\Delta$ from $\cup$ and $\cap$, we need complements. But complements satisfy a universal property of their own: $E\cap E^c=\emptyset$, and if $E\cap F=\emptyset$ then $F\subseteq E^c$.

First, I want to show that $f(\emptyset)=\emptyset$ by showing it is below all elements of $\mathcal{T}$. Indeed, by the surjectivity of $f$, every element of $\mathcal{T}$ is the image of some element $E\in\mathcal{S}$. Thus we want to show that $f(\emptyset)\subseteq f(E)$. But this is true because $\emptyset\subseteq E$ for all $E\in\mathcal{S}$, and so $f(\emptyset)=\emptyset$

Now given a set $E$ and its complement $E^c=X\setminus E$, we know that $E\cap E^c=\emptyset$. Since $f$ preserves $\cap$, we must have $f(E)\cap f(E^c)=f(\emptyset)=\emptyset$. Let’s say $G\in\mathcal{T}$ is another set so that $f(E)\cap G=\emptyset$. By the surjectivity of $f$, we have $G=f(F)$ for some $F\in\mathcal{S}$. Thus $f(E\cap F)=f(E)\cap f(F)=\emptyset=f(\emptyset)$, and thus $E\cap F=\emptyset$. This tells us that $F\subseteq E^c$, and thus $G=f(F)\subseteq f(E^c)$. And so $f(E^c)=f(E)^c$.

Therefore, since $f$ preserves $\cup$, $\cap$, and complements, it preserves $\cup$ and $\setminus$, and thus is a morphism of Boolean rings.

August 11, 2010 - Posted by | Algebra, Ring theory