Image of Ultrafilter is Ultrafilter

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

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

Proof
From Image Filter is a Filter, we have 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.