Let $X$ and $Y$ be uncountable Polish spaces and $f:X\to Y$ a Borel bijection (which automatically has Borel inverse); such a map $f$ exists by Kuratowski's theorem.
Given countable ordinals $1\leq \alpha,\beta < \omega_1$, let's say $f$ has type $(\alpha,\beta)$ when:
- the preimage of open sets in $Y$ are $\boldsymbol{\Sigma}^0_\alpha$ in $X$,
- the image of open sets in $X$ are $\boldsymbol{\Sigma}^0_\beta$ in $Y$.
For example:
- $f:X\to Y$ has type $(1,1)$ iff $f$ is a homeomorphism.
- If $f:X\to Y$ has type $(1,\beta)$, then $f$ is continuous; if $f$ has type $(\alpha,1)$ then $f$ is an open map.
- Second-countability of $X, Y$ yields that every Borel bijection $f:X\to Y$ has type $(\alpha,\beta)$ for some $\alpha,\beta < \omega_1$. Indeed, take countable bases $U_n$ and $V_n$ for the topologies of $X$ and $Y$ respectively. Then the rank of the preimage (resp. image) of any open set is bounded by the supremum of the ranks of $f^{-1}[V_n]$ (resp. $f[U_n]$). In particular, $f$ is continuous iff there is $1\leq \alpha<\omega_1$ such that $f$ has type $(\alpha,1)$; a similar result characterizes openness.
Let's call $$\mathfrak{b}(f)=\min\{\max\{\alpha,\beta\}:1\leq \alpha,\beta < \omega_1,\,f\text{ has type }(\alpha,\beta)\}$$ the Borel birank of $f$; intuitively, if $\mathfrak{b}(f)$ is high then $f$ has to be complicated in some direction. Set $$\mathfrak {a}(X,Y) = \min\{\mathfrak b(f) : f:X\to Y \text{ is a Borel bijection}\}.$$
I feel like this has to have been studied before, but I haven't been able to find anything on it. For concreteness, let $X$ be an uncountable Polish space and $f:X\to 2^\omega$ be a Borel bijection; I have no idea how I'd begin to compute $\mathfrak b(f)$, especially since in general $f$ is going to be challenging to write down.
Here's an example of the kind of thing I'm interested in: what is $\mathfrak{a}(\mathbb R,\mathbb R^2)$? The two spaces aren't homeomorphic, so $\mathfrak{a}(\mathbb R,\mathbb R^2) > 1$. But computing an upper bound is much more annoying. For example, since Borel sets are $\omega$-Lusin we can continuously biject $\mathbb R\setminus \mathbb Q$ with a cocountable subset $E\subseteq \mathbb R^2$. Do whatever you want to Borel biject $\mathbb Q\to \mathbb R^2\setminus E$. Preimages of open sets in $\mathbb R^2$ are in fact $\boldsymbol{\Delta}^0_2$ in $\mathbb R$, but I'm not sure if images of open sets are so nice that we can conclude $\mathfrak{a}(\mathbb R,\mathbb R^2) = 2.$ Any ideas would be much appreciated - thanks!
