Ultrafilters containing the image of a filter

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).

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 do you mean a filter $F$ on $X$? or do you mean inverse image filter (how is this defined)? – Martin Brandenburg May 10 2010 at 22:50 I totally made a big typo, I'm sorry! It's fixed now! – David Carchedi May 10 2010 at 23:08

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 $G$.

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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.

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 This is my fault, I made a typo. – David Carchedi May 10 2010 at 23:09

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 $f[U]$.

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 Yes, there is a typo, it's fixed now, sorry about that. – David Carchedi May 10 2010 at 23:09 Maybe I'm missing something, but, how does the first paragraph you typed answer the question posed in the last? If $f:X \to Y$ and $F$ is a filter on $X$ whose image=$(A \subset Y| f^{-1}(A) \in F)$ is contained in an ultrafilter $G$ (on $Y$), what do I take as an ultrafilter $U$ on X so that its image is $G$? – David Carchedi May 11 2010 at 0:53 I have edited to explain, and Kevin has explained it also. – Joel David Hamkins May 11 2010 at 1:57 The assignment of $X$ its set of ultrafilters $\beta X$ turns into a functor on SET (it is in fact a monad). By the image of an ultrafilter, I mean its image under the induced map $\beta(f):\beta X \to \beta Y$ which sends $U$ to $(A \subset Y | f^{-1}(A)\in U)=:\beta(f)(U)$. Clearly $G \subset \beta(f)(U)$, by construction BUT, since they are both ultrafilters, this must be an equality. So f shouldn't need to be onto, right? – David Carchedi May 11 2010 at 16:12 Yes, that is fine, and this is what the very last part of my answer amounts to. There is another sense of image, meaning that you just look at pointwise images of the subsets of X, and this is attainable, if the map is onto, but not if it isn't. – Joel David Hamkins May 11 2010 at 16:20
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