Image Filter is Filter

Theorem
Let $$X, Y$$ be sets, $$f : X \rightarrow Y$$ a mapping and $$\mathcal{F} \subset \mathcal{P}(X)$$ a filter on $$X$$. Then the image filter of $$\mathcal{F}$$ with respect to $$f$$, $$f(\mathcal{F}) := \{ U \subseteq Y: f^{-1}(U) \in \mathcal{F} \}$$, is a filter on $$Y$$.

Proof
From the definition of a filter we have to prove four things:
 * 1) $$f(\mathcal{F}) \subset \mathcal{P}(Y)$$
 * 2) $$Y \in f(\mathcal{F})$$ and $$\emptyset \not \in f(\mathcal{F})$$
 * 3) If $$U, V \in f(\mathcal{F})$$ then $$U \cap V \in f(\mathcal{F})$$
 * 4) If $$U \in f(\mathcal{F})$$ and $$U \subseteq V \subseteq Y$$ then $$V \in f(\mathcal{F})$$.

By construction we have $$f(\mathcal{F}) \subseteq \mathcal{P}(Y)$$.

Since $$f^{-1}(\emptyset) = \emptyset \not \in \mathcal{F}$$ we know that $$\emptyset \not \in f(\mathcal{F})$$ and therefore $$f(\mathcal{F}) \ne \mathcal{P}(Y)$$, which implies (1).

Due to $$f^{-1}(Y) = X \in \mathcal{F}$$ we have $$Y \in f(\mathcal{F})$$ and since we've already shown $$\emptyset \not \in f(\mathcal{F})$$ this implies (2).

To show (3) let $$U, V \in f(\mathcal{F})$$. Then $$f^{-1}(U \cap V) = f^{-1}(U) \cap f^{-1}(V) \in \mathcal{F}$$ (since $$\mathcal{F}$$ is a filter) and thus $$U \cap V \in f(\mathcal{F})$$.

Finally, if $$U \in f(\mathcal{F})$$ and $$V \subseteq Y$$ such that $$U \subseteq V$$ then $$f^{-1}(U) \subseteq f^{-1}(V)$$.

Since $$f^{-1}(U) \in \mathcal{F}$$ and $$\mathcal{F}$$ is a filter it follows that $$f^{-1}(V) \in \mathcal{F}$$, which implies $$V \in f(\mathcal{F})$$ and thus (4).