This question is "take 2" of [this older one](https://mathoverflow.net/questions/420048/comparing-bornologies-for-domination-escaping), following a suggestion of Francois Dorais. Consider the following [bornologies](https://en.wikipedia.org/wiki/Bornology) $\mathbb{D},\mathbb{E}$ on the set $\mathcal{N}$ of all functions from $\mathbb{N}$ to $\mathbb{N}$:


* $\mathbb{D}=\{A: \exists f\in\mathcal{N}\forall g\in A\exists m\in\mathbb{N}\forall n>m(f(n)>g(n))\}$. (**Dominatable** sets)
* $\mathbb{E}=\{A: \exists f\in\mathcal{N}\forall g\in A\forall m\in\mathbb{N}\exists n>m(f(n)>g(n))\}$. (**Escapable** sets)


Clearly we must have $\mathfrak{b}=\mathfrak{d}$ in order for these to yield bornomorphic structures on $\mathcal{N}$, and Francois Dorais showed at the above-linked question that the converse holds as well. However, the "tame maps" version of the question is still open:



> 
> Suppose $\mathfrak{b}=\mathfrak{d}$. Must there be a **Borel** $i:\mathcal{N}\rightarrow\mathcal{N}$ such that the $i$-image of each set in $\mathbb{D}$ is in $\mathbb{E}$ and the $i$-preimage of each set in $\mathbb{E}$ is in $\mathbb{D}$?
> 
> 
> 


It's not even clear to me (per James Hanson's comment below) what the answer is assuming $\mathsf{V=L}$.



Comparing bornologies for cardinal characteristics via Borel maps
A bornomorphic map $i\colon\mathcal{N}\to\mathcal{N}$ for these bornologies cannot be Borel.


First, I shall use $f<^\infty g$ to mean that $(\forall m\in\mathbb{N})(\exists n>m)f(n)<g(n)$, and $f<^\*g$ to mean that $(\exists m\in\mathbb{N})(\forall n>m)f(n)<g(n)$.


For $f\in\mathcal{N}$, let $\mathbb{D}\_f=\{g\in\mathcal{N}\mid g<^\* f\}$, and let $\mathbb{E}\_f=\{g\in\mathcal{N}\mid g<^\infty f\}$. Since the $\mathbb{D}\_f$ and $\mathbb{E}\_f$ generate their respective ideals, $i$ is a bornomorphism if and only if for all $f\in\mathcal{N}$ there is $a(f)\in\mathcal{N}$ such that $i``\mathbb{D}\_f\subseteq\mathbb{E}\_{a(f)}$ and $u(f)\in\mathcal{N}$ such that $i^{-1}(\mathbb{E}\_f)\subseteq\mathbb{D}\_{u(f)}$.


We shall focus on the $u$ case. Another way of saying this is for all $g,f\in\mathcal{N}$, if $i(g)<^\infty f$ then $g<^\*u(f)$.


Adjoin to the universe a Cohen generic function $c\colon\mathbb{N}\to\mathbb{N}$, and let $d=\tilde{i}(c)$, where $\tilde{i}$ is the Borel function with the same code as $i$. Since ground model elements of $\mathcal{N}$ are a $<^\infty$-dominating family in the extension, there must be $f\in\mathcal{N}$ in the ground such that $d<^\infty f$. Let $h=u(f)$. Then, by our assumption, in the ground model we get


$$(\forall x\in\mathcal{N})(i(x)<^\infty f\implies x<^\*h)$$


This is a $\Pi^1\_1$-statement involving only $f$, $h$, and (a code for) $i$, so it remains true in our extension. In particular, generic cohen real $c$ is satisfies $c<^\*g$. However, Cohen generic functions $\mathbb{N}\to\mathbb{N}$ are never dominated by ground model reals.




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This proof is a small adjustment of Andreas Blass's Theorem 4.15 in *Combinatorial Cardinal Characteristics of the Continuum*.


*Blass, Andreas*, [**Combinatorial cardinal characteristics of the continuum**](https://doi.org/10.1007/978-1-4020-5764-9_7), Foreman, Matthew (ed.) et al., Handbook of set theory. In 3 volumes. Dordrecht: Springer (ISBN 978-1-4020-4843-2/hbk; 978-1-4020-5764-9/ebook). 395-489 (2010). [ZBL1198.03058](https://zbmath.org/?q=an:1198.03058).


