This question is a particular take on the following theme. Suppose $A$ and $B$ are two notions of "large subset of $\omega^\omega$;" when is there a uniform method for turning an element of $A$ into an element of $B$?


We work in ZF (although results under strengthenings of ZF are also interesting).


For $U, V\subseteq\mathcal{P}(\omega^\omega)$, say $U$ *spreads onto* $V$ if there is some $F:\omega^\omega\rightarrow\omega^\omega$ such that $F(X)\in V$ for every $X\in U$, where $F(X)=\{F(x): x\in X\}$.


*The original motivation for this was in the context of working with cardinal characteristics of the continuum without choice - about which I have [asked](https://mathoverflow.net/questions/191545/cardinal-characteristics-without-choice) some [questions](https://mathoverflow.net/questions/243461/comparing-the-sizes-of-uncountable-sets-of-reals-under-ad?noredirect=1&lq=1) before. This also was a natural, and hopefully less trivial, outgrowth of [this question](https://mathoverflow.net/questions/243837/spreading-sets-especially-without-choice).*


*However, especially in light of question 2 below, it is unclear whether there's actually any relation. So now it's just a pure curiosity question.*




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Here's a trivial example. If $A=\{$dominating families$\}$ and $B=\{$escaping families$\}$, then $A$ spreads onto $B$ via the identity map.




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Here's a less trivial - maybe even interesting! :P - example. Let $C=\{$pointed perfect sets$\}$ (where a pointed perfect set is a nonempty closed subset $K$ of $\omega^\omega$ with no isolated points, such that every element of $K$ computes the tree representation of $K$ - that is, the set of strings $\sigma\in\omega^{<\omega}$ such that $\exists f\in K(\sigma\prec f)$). Then $C$ spreads onto $A$ (where $A$ is the set of dominating families as above). This argument hinges on two observations:


* There is a definable way to assign to each perfect $K\subseteq\omega^\omega$ a surjection $s\_K: K\rightarrow \omega^\omega$. (Look at the tree representations.)
* Every real computes only countably many tree representations of perfect sets.


We combine these facts as follows. Given a real $f\in\omega^\omega$, let $T\_i$ ($i\in\omega$) be the perfect sets whose tree representation is computable in $f$ (these come ordered by index of Turing reduction), and let $g\_i=s\_{T\_i}(f)$. Then let $F$ be defined as $$F(f)(n)=1+\sum\_{i=0}^ns\_{T\_i}(f)(n).$$ That is, $F$ takes in $f$, and spits out a real dominating each of the countably many reals which $f$ could correspond to in any of the perfect sets $f$ computes.


Now suppose $T$ is any pointed perfect set; I want to argue that $F(T)$ is a dominating family. Since $s\_T$ is surjective, it's enough to show that for each $f\in T$, there is some $g\in F(T)$ such that $g$ dominates $s\_T(f)$. But since $T$ is perfect, $f$ computes the tree representation of $T$; so we may take $g=F(f)$, and problem solved.




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However, each of the two results above has a natural follow-up question:



> 
> **Question 1.** Does $B$ (escaping) spread onto $A$ (dominating)?
> 
> 
> 


Of course the answer is consistently "no," since it is consistent that the dominating number is strictly greater than the bounding number. However, I would like a ZF answer. Less valuable, but still very interesting, would be answers in either ZFC or ZF+AD+etc (which don't decide whether $\mathfrak{b}<\mathfrak{d}$ and which prove a weak CH, respectively - so the question is not trivial in either case).



> 
> **Question 2.** Can we drop "pointed" in the second example? That is, does the family of *merely perfect* sets spread onto $A$?
> 
> 
> 


I don't see how to do this even in ZFC; we seem to need some way to assign to each real $f$ a family of "few" (specifically, smaller than the escaping number) reals which it needs to dominate. Via pointedness, we were able to cheat - every real is in only countably many pointed perfect sets. However, without this I don't see how to get the argument to go through. (Note that there are perfect sets whose tree representations aren't computed by *any* of their elements!)


*Actually, I'm beginning to think that the right approach for this type of question is to restrict attention to "pointed subsets of $\omega^\omega$" - that is, dominating/escaping/perfect/whatever sets, every element of which computes "the right representation" of the set in question. However, I still don't know quite what I mean by that. Any thoughts on the matter, while not the main focus of this question, would be welcome!*



Reducing largeness notions, uniformly
I hope that I didn't misunderstand the question. If escaping family means a family of reals which is unbounded in $\leq^\*$ then an Erdős-Sierpinski-type proof gives that in ZFC+CH we have that $B$ spreads onto $A$ as follows:


From an enumeration of the Borel non-dominating sets $\{B'\_\alpha:\alpha<\mathfrak{c}\}$ taking unions of the initial segments (and regrouping if necessary) we can get an increasing sequence of non-dominating Borel sets $\{B\_\alpha:\alpha<\mathfrak{c}\}$ such that for every $\alpha$ the ring $B\_{\alpha+1} \setminus B\_\alpha$ has cardinality $\mathfrak{c}$. Do the same with non-escaping Borel sets and obtain the enumeration $\{C'\_\alpha:\alpha<\mathfrak{c}\}$ and the increasing sequence $\{C\_\alpha:\alpha<\mathfrak{c}\}$ with $|C\_{\alpha+1} \setminus C\_\alpha|=\mathfrak{c}$. (We can also assume that $|C\_0|=|B\_0|=\mathfrak{c}$). 


Fix bijections $(f\_\alpha)\_{\alpha<\mathfrak{c}}$ between the sets $B\_0$ and $C\_0$, and $B\_{\alpha+1} \setminus B\_\alpha$ and $C\_{\alpha+1} \setminus C\_\alpha$, respectively. Let $F:\omega^\omega \to \omega^\omega$ be the function which consists of these bijections, i. e., $F(x)=f\_\alpha(x)$ iff $x \in B\_{\alpha+1} \setminus B\_\alpha$.


We claim that a set $S$ is dominating if and only if $F(S)$ is escaping. So suppose that $S$ is dominating but $F(S)$ is not escaping. Then $F(S)$ is covered by a Borel non-escaping set, which must be enumerated in $\{C'\_\alpha:\alpha<\mathfrak{c}\}$ therefore $F(S)$ is covered by $C\_\alpha$ for some $\alpha<\mathfrak{c}$. But $F^{-1}(C\_\alpha)=B\_\alpha$ by definition of $F$, so $S \subset B\_\alpha$, contradicting the assumption that $S$ was dominating.
The other direction is the same.


This proof works in a more general context, namely for $\sigma$-ideals which have cofinality $\mathfrak{c}$.


