# Can we find $f\in \Bbb{R}^{[0, 1]}$ with the property $\mathcal{M}$ which doesn't satisfy the property $\mathcal{B}$?

$$f:[0, 1]\to \Bbb{R}$$ be a function.

1. $$f$$ satisfy the property $$\mathcal{M}$$ of $$f(A)$$ is meagre for every $$A\subset [0, 1]$$ meagre.

2. $$f$$ satisfy the property $$\mathcal{B}$$ if $$f(A)$$ is a set with the property of Baire for every $$A\subset [0, 1]$$ having the property of Baire.

I don't know whether the property $$\mathcal{M}, \mathcal{B}$$ has any name or not.

$$f\in \Bbb{R}^{[0, 1]}$$ with property $$\mathcal{B}$$ and we know every meager set satisfy the property of Baire this implies image of every meager set is a set with the baire property and this means it is a symmetric difference of an open set and meager set.

I believe we can find $$f\in \Bbb{R}^{[0, 1]}$$ with the property $$\mathcal{B}$$ which doesn't satisfy the property $$\mathcal{M}$$.

Question : Can you give me an explicit example of $$f\in \Bbb{R}^{[0, 1]}$$ which map every set with Baire property to a set with Baire property but doesn't map a meager set to a meager set?

Question : Can we find $$f\in \Bbb{R}^{[0, 1]}$$ with the property $$\mathcal{M}$$ which doesn't satisfy the property $$\mathcal{B}$$?

The sets with the Baire property forms a $$\sigma$$-algebra generated by open sets and meagre sets.

Suppose $$F\in \Bbb{R}^{[0, 1]}$$ satisfy the property $$\mathcal{M}$$.

$$F(M) \subset \Bbb{R}$$ is meagre for $$\forall M\subset [0, 1]$$ meagre. Hence $$F$$ maps the Cantor set $$\mathcal{C}$$ and all subsets of $$\mathcal{C}$$ to meagre set.

So $$F(\mathcal{C}) =\mathcal{C}$$.

And suppose $$F(U)$$ is not open for some $$U\subset [0, 1]\setminus\mathcal{C}$$ open.

Now again we have to map every meagre subset of $$U$$ to a meagre set. This is difficult and how to map rest of points.

Does this type of function exists? If yes how to construct?

Claim 1: Property $$\mathcal{B}$$ implies property $$\mathcal{M}$$.
Proof: Suppose $$f:[0, 1] \to \mathbb{R}$$ is a counterexample and fix a meager $$M \subseteq [0, 1]$$ such that $$f[M]$$ is non-meager and has BP (Baire property). Let $$W \subseteq f[M]$$ be non-meager and without BP (for example, $$W = B \cap f[M]$$ where $$B$$ is a Bernstein set). Let $$N = M \cap f^{-1}[W]$$. Then $$N$$ is meager and $$f[N] = W$$ does not have BP. Contradiction.
Claim 2: Property $$\mathcal{M}$$ does not imply property $$\mathcal{B}$$.
Proof: Let $$V$$ be a Vitali set. Define $$f:[0, 1] \to \mathbb{R}$$ by $$f(x) = y$$ iff $$y \in V$$ and $$x - y$$ is rational. Now for any meager $$M \subseteq [0, 1]$$, $$f[M] \subseteq M + \mathbb{Q}$$ is meager. So $$f$$ has property $$\mathcal{M}$$. As range of $$f$$ is $$V$$, $$f$$ doesn't send sets with BP to sets with BP. So it doesn't have property $$\mathcal{B}$$.