Can boolean homomorphisms of boolean algebras correspond to ultrafilters?

691 Views Asked by At

I am trying to solve 5th problem in Exercises 2.9 in Awodey's book on page 55:

Show that for any boolean algebra $B$, boolean homomorphisms $h : B \to 2$ correspond exactly to ultrafilters in $B$.

I do not understand what the question is asking me to prove because:

a) Boolean homomorphisms in a category $\mathsf{Bool}$ are functors mapping objects of a boolean algebra to objects of some other boolean algebra and then mapping the arrows in a similar fashion.

b) an ultrafilter is a kind of a filter of a boolean algebra - it is a subset of the objects of a boolean algebra.

Therefore how can a function correspond to a set of objects? It cannot be equivalence since the types of the mathematical objects (function, set) are different. What correspondence does Awodey mean here?

1

There are 1 best solutions below

1
On BEST ANSWER

By "correspondence" we mean "bijection" in this case. That is, there is a (natural) bijection between the set of ultrafilters of $B$ and the set of Boolean homomorphisms $B\to\textbf{2}$.

Note that a Boolean homomorphism $h:B\to\textbf{2}$ is determined precisely by its kernel in $B$. (Why?) A kernel of a Boolean homomorphism is always an ideal of its domain, but in the case of Boolean homomorphisms into $\textbf{2}$, the kernel is a prime ideal of its domain, and so its complement is an ultrafilter. In this fashion, we find a natural bijection as described.