If ${\scr U}$ and ${\scr V}$ are ultrafilters on nonempty 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 nonprincipal 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 offtopic by Joseph Van Name, Dan Petersen, Dima Pasechnik, András Bátkai, Piotr Hajlasz Mar 26 at 0:03
This question appears to be offtopic. The users who voted to close gave this specific reason:
 "This question does not appear to be about research level mathematics within the scope defined in the help center." – Joseph Van Name, Dan Petersen, Dima Pasechnik, Piotr Hajlasz

$\begingroup$ No. . . . . . . $\endgroup$ – Joseph Van Name Mar 25 at 11:55

1$\begingroup$ 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 $\mathcal U \otimes \mathcal U$? $\endgroup$ – Will Brian Mar 25 at 13:15

1$\begingroup$ @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. $\endgroup$ – Andreas Blass Mar 25 at 14:18

1$\begingroup$ @AndreasBlass: Thanks  I was reading that in the wrong direction. $\endgroup$ – Will Brian Mar 25 at 14:27

$\begingroup$ 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. $\endgroup$ – 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 onetoone on any set in $\mathcal U\otimes\mathcal U$.