Image of Ultrafilter is Ultrafilter/Proof 2

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

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:

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

$\blacksquare$