Image Filter is Filter

Theorem
Let $$X, Y$$ be sets.

Let $$\mathcal P \left({X}\right)$$ and $$\mathcal P \left({Y}\right)$$ be the power sets of $$X$$ and $$Y$$ respectively.

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

Let $$\mathcal F \subset \mathcal P \left({X}\right)$$ be a filter on $$X$$.

Then the image filter of $$\mathcal F$$ with respect to $$f$$:
 * $$f \left({\mathcal F}\right) := \left\{{U \subseteq Y: f^{-1} \left({U}\right) \in \mathcal F}\right\}$$

is a filter on $$Y$$.

Proof
From the definition of a filter we have to prove four things:
 * 1) $$f \left({\mathcal F}\right) \subset \mathcal P \left({Y}\right)$$
 * 2) $$Y \in f \left({\mathcal F}\right), \varnothing \notin f \left({\mathcal F}\right)$$
 * 3) $$U, V \in f \left({\mathcal F}\right) \implies U \cap V \in f \left({\mathcal F}\right)$$
 * 4) $$U \in f \left({\mathcal F}\right), U \subseteq V \subseteq Y \implies V \in f \left({\mathcal F}\right)$$

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

Since $$f^{-1} \left({\varnothing}\right) = \varnothing \notin \mathcal F$$ we know that $$\varnothing \notin f \left({\mathcal F}\right)$$.

Therefore, $$f \left({\mathcal F}\right) \ne \mathcal P \left({Y}\right)$$, which implies (1).

Because $$f^{-1} \left({Y}\right) = X \in \mathcal F$$, we have $$Y \in f \left({\mathcal F}\right)$$.

Since we've already shown $$\varnothing \notin f \left({\mathcal F}\right)$$, this implies (2).

Let $$U, V \in f \left({\mathcal F}\right)$$.

From Preimage of Intersection $$f^{-1} \left({U \cap V}\right) = f^{-1} \left({U}\right) \cap f^{-1} \left({V}\right) \in \mathcal F$$ (since $$\mathcal F$$ is a filter).

Thus $$U \cap V \in f \left({\mathcal F}\right)$$, and so (3) holds.

Finally, let $$U \in f \left({\mathcal F}\right)$$ and $$V \subseteq Y$$ such that $$U \subseteq V$$.

Then from Subset of Image $$f^{-1} \left({U}\right) \subseteq f^{-1} \left({V}\right)$$.

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