Image of a $z-$ultrafilter is contained in a unique $z-$ultrafilter

Question

I understand that for a continuous map $f:X\longrightarrow Y$, if $\mathcal{F}$ is a $z-$ultrafilter on $X$, then $f(\mathcal{F})=\{A\in Z(Y):f^{-1}(A)\}$ is a prime $z-$filter on $Y$ and may not be a $z-$ultrafilter.

Stephan Willard said that $f(\mathcal{F})$ is contained in a unique $z-$ultrafilter. How so?

Answer 1

A basic fact about the zero-sets $Z(X)$ on a space $X$:

If $A,B \in Z(X)$ are such that $A \cap B = \emptyset$, then there are $C,D \in Z(X)$ such that $A \cap C= B \cap D = \emptyset$ and $C \cup D = X$. (see Willard 12E, 8.)

Note that the following statements are equivalent for a $z$-filter $\mathcal{F}$:

1. $\mathcal{F}$ is a $z$-ultrafilter (i.e. it's maximal by inclusion among $z$-filters).
2. For all $A \in Z(X)$, if $\forall F \in \mathcal{F}: F \cap A \neq \emptyset$, then $A \in \mathcal{F}$.
3. For all $A \in Z(X)$: either $A \in \mathcal{F}$ or there exists $F_1 \in Z(X)$ such that $A \cap F_1 = \emptyset$ and $F_1 \in \mathcal{F}$.

This is quite easy to see.

Now, if $\mathcal{F}$ is a prime $z$-filter and $\mathcal{U}_1$ and $\mathcal{U}_2$ are two different ultrafilters extending $\mathcal{F}$, then the previous equivalence tells us we can find zerosets $A \in \mathcal{U}_1$ and $B \in \mathcal{U}_2$ such that $A \cap B = \emptyset$. The first fact then gives us $C,D$ zerosets such that $A \cap C = B \cap D = \emptyset$ and $C \cup D =X \in \mathcal{F}$, so $\mathcal{F}$ being prime gives us that, say, $C \in \mathcal{F}$. But then $C \in \mathcal{U}_1$ and this is a contradiction as $A \in\mathcal{U}_1$ too and these sets are disjoint. A similar contradiction is found when $D \in \mathcal{F}$, of course.

So all prime $z$-filters have a maximal larger $z$-ultrafilter (Zorn), and by the previous, this is a unique one.

It is indeed easy to see from the definition that $f(\mathcal{F})$ is a prime $z$-filter, so the previous fact applies and it is contained in a unique prime $z$-ultrafilter.