# Relation between ultrafilters ${\scr U}$ and ${\scr U} \otimes {\scr U}$ [closed]

If $${\scr U}$$ and $${\scr V}$$ are ultrafilters on non-empty sets $$A$$ and $$B$$ respectively, then the tensor product $${\scr U}\otimes{\scr V}$$ is the following ultrafilter on $$A\times B$$: $$\big\{X\subseteq A\times B: \{a\in A:\{b\in B: (a,b)\in X\}\in {\scr V}\}\in {\scr U}\big\}.$$ Is there a non-principal ultrafilter $${\scr U}$$ on $$\omega$$ and a function $$f:\omega \to (\omega\times\omega)$$ such that $$f({\scr U}) = {\scr U}\otimes {\scr U}$$?

## closed as off-topic by Joseph Van Name, Dan Petersen, Dima Pasechnik, András Bátkai, Piotr HajlaszMar 26 at 0:03

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• No. . . . . . . – Joseph Van Name Mar 25 at 11:55
• Actually, now that I think about it, shouldn't the answer be "yes" because you can just use a projection mapping $\omega \times \omega \rightarrow \omega$ to recover $\mathcal U$ from $\mathcal U \otimes \mathcal U$? – Will Brian Mar 25 at 13:15
• @WillBrian Projections give you $f(\mathcal U\otimes\mathcal U)=\mathcal U$, but the OP is asking for the other direction, $f(\mathcal U)=\mathcal U\otimes\mathcal U$, and that is indeed impossible. – Andreas Blass Mar 25 at 14:18
• @AndreasBlass: Thanks -- I was reading that in the wrong direction. – Will Brian Mar 25 at 14:27
• The answer on your question is negative as Andreas Blass mentioned. But Katetov proved the existence of filter with property $f(\mathcal{F})=\mathcal{F}\otimes\mathcal{F}$. See details here. – ar.grig Mar 25 at 18:21

As Will Brian noted in a comment, either projection $$p:\omega^2\to\omega$$ satisfies $$p(\mathcal U\otimes\mathcal U)=\mathcal U$$. If you also had a function $$f$$ as in the question, with $$f(\mathcal U)=\mathcal U\otimes\mathcal U$$, then the composite function $$f\circ p$$ would send $$\mathcal U\otimes\mathcal U$$ to itself. By a result that I quoted at Selective ultrafilter and bijective mapping, and for which Ali Enayat supplied a reference, it would follow that $$f\circ p$$ becomes the identity function when restricted to a suitable set in $$\mathcal U\otimes\mathcal U$$. That is absurd, because $$p$$ is not one-to-one on any set in $$\mathcal U\otimes\mathcal U$$.