Filters on $\omega$ I am currently reading the book "Set theory on the real line" by Bartoszynski and Judah and I do have problems to proof the following statement:
Suppose $\mathcal{F}$ is a filter on $\omega$ including the Fréchet-Filter $\mathcal{G}$ ($\mathcal{G}\subset\mathcal{F}$).
Then the following is equivalent:
(i) For every partition of $\omega$ into finite sets $\{I_n:n\in\omega\}$, there exists $X\in\mathcal{F}$ such that $X\cap I_n=\emptyset$ for infinitely many $n\in\omega$.
(ii)For every function $f\in\omega^\omega$ which is finite to one, $f(\mathcal{F})=\{X\subset\omega:f^{-1}(X)\in\mathcal{F}\}$ is not the Fréchet-Filter.
[A function is finite to one if each point in its range space is the image of only finitely many points in the domain]
The book says, $(i)\Leftrightarrow (ii)$ is obvious. But I can't proof it. 
 A: A finite-to-one function $f : A \to B$ partitions $A$ into finite sets, namely $f^{-1}(\{b\})$ for each $b \in B$.
In the other direction, given a partition $I_n$ of $A$, you may define $f : A \to \omega$ as follows: each $a \in A$ is in exactly one $I_n$ (since this is what it means to be a partition), so you can unambiguously define $f(a) = n$. Then the preimage of each $n$ is $I_n$, so if the $I_n$ are finite, then $f$ is finite-to-one.
This correspondence is the key to the question: prove that an $I_n$ and $X$ satisfying the properties of (i) correspond to an $f$ satisfying the properties of (ii) (recall that $f(\mathcal F)$ is not the Fréchet filter precisely if it has an element with infinite complement, so you just need to find such an element) and likewise if (ii) holds there is an element of $f(\mathcal F)$ with infinite complement, so you need to demonstrate that it satisfies the conclusions of (i).
A: I employ the set-theorists' notation: $\{f(a):a\in A\}=f''A$ (read $f$-double-tick-$A$) when $A\subset$ dom $f$. Also $f^{-1}B=\{x: f(x)\in B\}.$
For $(i)\implies (ii)$:
First, if $\omega$  \  $f''\omega$ is infinite, then $f^{-1}f''\omega=\omega\in \mathbb F$ but $f''\omega$ is not in the Frechet filter.
Second, if $\omega$  \  $f''\omega$  is finite, let $\quad J=\{f^{-1}\{n\}:n\in \omega\}$  \  $\{\phi\}.$  $$\text {Let } Y\in \mathbb F \text { such that } K=\{j\in J:Y\cap j=\phi\} \text {is infinite.}$$ $$\text {Let } Z= \cup \{j\in J: Y\cap j\ne \phi\}=\cup (J  \backslash  K).$$ Then $Z\supset Y\in \mathbb F$ so $Z\in \mathbb F.$ Let $X=f''Z.$ Then $f^{-1}X=Z\in \mathbb F,$ but $\omega$  \   $X$ is infinite (and hence $X$ is not in the Frechet filter): Because $L=\omega$  \   $Z=\cup K$ is infinite, so $\omega$  \  $X=f''L$ is infinite . 
For $(ii)\implies (i)$:
Let $\{I_n\}$ be a partition of $\omega$ into (non-empty) finite sets . Let $f''I_n=\{n\}$ for each $n\in \omega.$ Let $X\subset  \omega$ such that $\omega$      \  $X$ is infinite and such that $Y=f^{-1}X\in \mathbb F.$ $$\text {Then }\quad   \{n\in \omega: \phi =Y\cap I_n\}=\omega   \backslash   X \quad  \text {which is infinite.}$$ Remark: The notation $f''A$ is not merely convenient, but avoids the ambiguity of $f(A) $ when $A$ is both a member and a subset of dom$(f)$.
