Image Filter is a 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:

$(1) \quad f \left({\mathcal F}\right) \subset \mathcal P \left({Y}\right)$

$(2) \quad Y \in f \left({\mathcal F}\right), \varnothing \notin f \left({\mathcal F}\right)$

$(3) \quad U, V \in f \left({\mathcal F}\right) \implies U \cap V \in f \left({\mathcal F}\right)$

$(4) \quad U \in f \left({\mathcal F}\right), U \subseteq V \subseteq Y \implies V \in f \left({\mathcal F}\right)$

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 Mapping 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 $(4)$.

$\blacksquare$