Image of Ultrafilter is Ultrafilter

Theorem
Let $$X, Y$$ be two sets, $$f : X \rightarrow Y$$ a mapping and $$\mathcal{F}$$ an ultrafilter on $$X$$.

Then the image filter $$f(\mathcal{F})$$ is an ultrafilter on $$Y$$.

Proof
We already know that $$\mathcal{F}$$ is a filter on $$Y$$.

Let $$\mathcal{G}$$ be a filter on $$Y$$ such that $$f(\mathcal{F}) \subseteq \mathcal{G}$$. We have to show that $$f(\mathcal{F}) = \mathcal{G}$$.

Let $$U \in \mathcal{G}$$.

Assume that $$U \not \in f(\mathcal{F})$$.

By the definition of $$f(\mathcal{F})$$ this implies that $$f^{-1}(U) \not \in \mathcal{F}$$.

By the equivalent definitions of ultrafilters we thus know that $$V := X \setminus f^{-1}(U) \in \mathcal{F}$$.

Because $$V \subseteq f^{-1}(f(V))$$ it follows that $$f^{-1}(f(V)) \in \mathcal{F}$$ and thus also $$f(V) \in f(\mathcal{F})$$.

By assumption we have $$f(\mathcal{F}) \subseteq \mathcal{G}$$, thus $$f(V) \in \mathcal{G}$$.

But $$f(V) \cap U = f(X \setminus f^{-1}(U)) \cap U = \emptyset \not \in \mathcal{G}$$.

Thus $$\mathcal{G}$$ is not a filter, a contradiction to our assumptions.

Thus $$U \in f(\mathcal{F})$$ and therefore $$f(\mathcal{F}) = \mathcal{G}$$.

Hence $$f(\mathcal{F})$$ is an ultrafilter.