Image of Ultrafilter is Ultrafilter

Theorem

Let $X, Y$ be sets.

Let $f: X \to Y$ be a mapping.

Let $\FF$ be an ultrafilter on $X$.

Then the image filter $f \sqbrk \FF$ is an ultrafilter on $Y$.

Proof 1

From Image Filter is Filter, we have that $\FF$ is a filter on $Y$.

Let $\GG$ be a filter on $Y$ such that $f \sqbrk \FF \subseteq \GG$.

We have to show that $f \sqbrk \FF = \GG$.

Aiming for a contradiction, suppose there exists $U \in \GG$ such that $U \notin f \sqbrk \FF$.

By the definition of $f \sqbrk \FF$ this implies that $f^{-1} \sqbrk U \not \in \FF$.

By definition of ultrafilter:

$V := X \setminus f^{-1} \sqbrk U \in \FF$

Because $V \subseteq f^{-1} \sqbrk {f \sqbrk V}$ it follows that:

$f^{-1} \sqbrk {f \sqbrk V} \in \FF$

Thus also:

$f \sqbrk V \in f \sqbrk \FF$

By assumption we have:

$f \sqbrk \FF \subseteq \GG$

thus:

$f \sqbrk V \in \GG$

But:

$f \sqbrk V \cap U = f \sqbrk {X \setminus f^{-1} \sqbrk U} \cap U = \O \notin \GG$

Thus $\GG$ is not a filter, which is a contradiction to our assumptions.

Thus there can be no $U \in \GG$ such that $U \notin \FF$.

Therefore:

$f \sqbrk \FF = \GG$

Hence $f \sqbrk \FF$ is an ultrafilter.

$\blacksquare$

Proof 2

From Image Filter is Filter, we have that $\FF$ is a filter on $Y$.

In view of definition of ultrafilter, we need to show that:

for every $A \subseteq Y$, either $A \in f \sqbrk \FF$ or $\relcomp Y A \in f \sqbrk \FF$

Let $A \subseteq Y$ be arbitrary.

Suppose $A \not \in f \sqbrk \FF$.

Then, by definition of image filter:

$f^{-1} \sqbrk A \not \in \FF$

By definition of ultrafilter:

$\relcomp X {f^{-1} \sqbrk A} \in \FF$

As Complement of Preimage equals Preimage of Complement:

$\relcomp X {f^{-1} \sqbrk A} = f^{-1} \sqbrk {\relcomp Y A}$

Thus:

$f^{-1} \sqbrk {\relcomp Y A} \in \FF$

That is, by definition of image filter:

$\relcomp Y A \in f \sqbrk \FF$

$\blacksquare$