# Is there a 'local' version of Near Coherence of Filters?

The axiom Near Coherence of Filters (NCF) is known to be independent of ZFC.

Axiom (NCF I): For any two free ultrafilters $\mathcal D$ and $\mathcal E$ on $\mathbb N$, there exist finite-to-one functions $f,g: \mathbb N \to \mathbb N$ such that $f(\mathcal D) = g(\mathcal E)$.

Where we define $f(\mathcal D) = \{A \subset \omega: f^{-1}(A) \in \mathcal D\}$. Equivalently $f(\mathcal D)$ is the unique ultrafilter generated by $\{f(D): D \in \mathcal D\}$

There is an equivalent version of the axiom.

Axiom (NCF II): For any two free ultrafilters $\mathcal D$ and $\mathcal E$ on $\mathbb N$, there exists a finite-to-one monotone function $f :\mathbb N \to \mathbb N$ such that $f(\mathcal D) = f(\mathcal E)$.

One can ask if the same property holds over an upper-subset of $\mathbb N^*$: First define the preorder $\le$ on $\mathbb N^*$ where $\mathcal F \le \mathcal D$ means that $\mathcal F = f(\mathcal D)$ for some finite-to-one monotone function $f :\mathbb N \to \mathbb N$. Now fix some $\mathcal F \in \mathbb N^*$ and define $(\mathcal F \uparrow) = \{ \mathcal D \in \mathbb N^*: \mathcal F < \mathcal D \}$

Is anything known about the following proposition?

Axiom?: For any two free ultrafilters $\mathcal D, \mathcal E \in (\mathcal F \uparrow)$, there exists a finite-to-one monotone function $f :\mathbb N \to \mathbb N$ such that $f(\mathcal D) = f(\mathcal E) \in (\mathcal F \uparrow)$.

In particular:

1. Is the proposition consistent?

3. Does it hold for any known $\mathcal F$ under ZFC?
• Maybe this is standard notation, but what do you mean by $f(\mathcal{D})$? Do you mean the set of inverse images of elements of $\mathcal{D}$ under $f$, or the set of forward images? – Paul McKenney Feb 14 '18 at 21:38
• The collection of sets whose inverse image is in $\mathcal D$. Equivalently the ultrafilter generated by all the the forward images. – Daron Feb 14 '18 at 23:19
• If the filter $\mathcal F$ is meager, then your Axiom seems to be equivalent to NCF. If $\mathcal F$ is not meager, then everything depends on the choice of $\mathcal F$. For example your Axiom trivially holds if $\mathcal F$ is an ultrafilter. So, maybe you want to ask if NCF is equivalent to your Axiom for any non-meager filter $\mathcal F$? – Taras Banakh Feb 18 '18 at 11:10
• Sorry, there was a mistake in the definition of the preorder. The two elements should have been the other way around. I see how this makes the axiom trivial for $\mathcal F$ an ultrafilter (we assume everything is an ultrafilter). I have corrected the mistake now. Is the axiom still trivial? – Daron Feb 18 '18 at 13:13