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).
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 $G$.
If $f:X\to Y$ and you have an ultrafilter $G$ on $X$, then it induces an ultrafilter $U$ on $Y$ by defining $A\in U\iff f^{-1}A\in G$. The ultrafilter $U$ is said to be Rudin-Kiesler below $G$, and this ordering on ultrafilters is intensely studied in large cardinal set theory. It follows that $f[G]\subset U$, and so this may be the answer you seek.
But in general, I am not sure what you mean by the image filter, when your maps go in that direction. Perhaps there is a typo?
Perhaps you meant to ask the following question: you have a map $f:X\to Y$ and a filter $F$ on $X$ (not $Y$), and the image $f[F]\subset G$ for some ultrafilter $G$ on $Y$. The question is whether there is an ultrafilter $U$ on $X$ with $f[U]\subset G$.
In this case, let $U$ be any ultrafilter containing $F$ and all $f^{-1}A$ for $A\in G$. There is such an ultrafilter (as Kevin also explains in his answer) since if $B\in F$ and $A\in G$, then $f[B]\cap A\in G$ and so $B\cap f^{-1}A$ is nonempty (and so the collection forms a filter base). For any such $U$, we have $f[U]\subset G$. If $f$ is onto $Y$, then we get actually $f[U]=G$.
Note, if $f$ is not onto $Y$, then there is no possiblity that $f[U]=G$, if this should mean $G$ consists precisely of $f[A]$ for $A\in U$, since all such $f[A]$ are subsets of the range of $f$. (So in this technical sense, the answer to your question is negative.) But on the positive side, $G$ is the image of $U$ in the sense that it is the filter gnerated by the sets in $f[U]$.
Yes you certainly can: let the ultrafilter H be the set of all sets V in Y such that the inverse image of the V's is inside the ultrafilter G on X. Then the inverse image of H is G. It should be that unless I misunderstood your question.
