Suppose $f:X \to Y$ is a map of sets and $F$ a filter on $X$ such that its image filter is contained in an ultrafilter $G$ on $Y$. Can I find an ultrafilter $H$ on $X$ whose image is $G$?
If this question is too elementary, I apologize. I have not worked much with ultrafilters, so sometimes basic properties escape me.


EDIT: I messed up the formulation of the question earlier, sorry!


(The previous formulation said that $F$ was on $Y$ etc., to make sense of the responses already posted).



Ultrafilters containing the image of a filter
Note that the image of a filter on $X$ will be $G$ if and only if it contains the filter consisting of all preimages of ''big'' sets in $Y$. Note also that we can combine two filters $F$ and $F'$ (that is, find a filter containing both of them) if and only if the intersection of any pair of sets $S\in F$, $S'\in F'$, is nonempty. Simply take the filter consisting of all sets which contain such an intersection.


So it suffices to show that if $S\in F$ and $T\in G$, then $S\cap f^{-1}(T)$ is nonempty. Suppose it were empty. Then $f(S)$ lies in the complement of $T$, hence is not in $G$. But $S\subset f^{-1}(f(S))$ implies that $f^{-1}(f(S))$ is in $F$, which contradicts that the image filter of $F$ was contained in $G$.


