$f(\mathcal F):=\{f(F):F \in \mathcal F\}$ is filter-basis and $G = \{G \subset N: f^{-1}(G) \in \mathcal F\}$ Let $\mathcal F$ be a filter on $M$ and $f:N \rightarrow N$.
I want to show that $f(\mathcal F):=\{f(F):F \in \mathcal F\}$ is a filter-basis of a filter $\mathcal G$ on N and that $\mathcal G = \{G \subset N: f^{-1}(G) \in \mathcal F\}$
I got the first part now, the only thing left to show is $$\mathcal G = \{G \subset N: f^{-1}(G) \in \mathcal F\}$$
 A: For the 2nd part assume empty set in f(F),
conclude empty set in F.  
For the third part, assume f(K), f(L) in f(F).
Thus K $\cap$ L in F;  f(K $\cap$ L) subset f(K) $\cap$ f(L).
Equality is incorrect.  Subset is all that is needed.
A: $f[\mathcal{F}]$ is a filter base (it's non-empty, as it always contains $f[X]$):
(i) As all $F \in \mathcal{F}$ are non-empty, all $f[F]$ are non-empty too.
(ii) If $f[F_1], f[F_2] \in f[\mathcal{F}]$, for some $F_1, F_2 \in \mathcal{F}$
then $F_1 \cap F_2 \in \mathcal{F}$ as we have a filter and $f[\mathcal{F}] \ni f[F_1 \cap F_2] \subseteq f[F_1] \cap f[F_2]$, so that any two elements are refined by a third.
This finishes the filter base check.
This generates the filter $f[\mathcal{F}]^{\uparrow} = \{A \subseteq Y: \exists f[F] \in f[\mathcal{F}]: f[F] \subseteq A\}$ and the claim is that this equals $\mathcal{G} = \{A \subseteq Y: f^{-1}[A] \in \mathcal{F}\}$.
To see that show two inclusions. So take $A \in f[\mathcal{F}]^{\uparrow}$, and $F \in \mathcal{F}$ such that $f[F] \subseteq A$. Then $F \subseteq f^{-1}[A]$ (as $x \in F$ implies $f(x) \in f[F]$ so $f(x) \in A$, so $x \in f^{-1}[A]$) and so as $F \in \mathcal{F}$ is a filter, $f^{-1}[A] \in  \mathcal{F}$ and so $A \in \mathcal{G}$.
Take $A \in \mathcal{G}$, then $f^{-1}[A] \in \mathcal{F}$ and $f[\mathcal{F}] \ni f[f^{-1}[A]] \subseteq A$ so that $A \in f[\mathcal{F}]^\uparrow$ and we have the other inclusion.
