Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the property of Baire for every $F$?

Question

Let $F\subset\Bbb{R} $ intersect every uncountable $\mathcal{F}_{\sigma}$ set.

$B\subset \Bbb{R}$ is said to have the property of Baire if $B=U\triangle M$ where $U$ is open and $M$ is meager.

Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the property of Baire for every $F$ defined above ?

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Edit $1$ : $F$ intersects every uncountable $F_{\sigma}$ set iff F intersects every uncountable closed set.

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Edit $2$: If $f:\Bbb{R}\to \Bbb{R}$ is additive i.e $f(x+y) =f(x) +f(y) $ then $f$ is $\Bbb{Q}$-linear.In other words $f$ is a linear function if we consider $\Bbb{R}$ as a vector space of $\Bbb{Q}$.

1. $f(x+y) =f(x) +f(y) $

2. $f(qx) =qf(x) $

$\forall x, y\in\Bbb{R} $ and $\forall q\in\Bbb{Q}$

$\underline{ \text{Case } 1}$ : $( f \text{ is } \Bbb{R} \text{ linear}) $

Then $f(x) =ax$ for some $a\in \Bbb{R}$.

Hence clearly $f$ is additive onto map $($ and moreover $f$ is a linear homeomorphism / topological isomorphism $)$.

But $f$ fails to hold the third property mentioned above. For an example ,if we take $F=\mathcal{B}( \text{Bernstein set})$ then $f(F) =a\mathcal{B}$ doesn't have the property of Baire.

$\boxed{\text{ Required function can't be $\Bbb{R}$ -linear}}$

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$\underline{\text{Case }2} $: $f$ is $\mathbb{Q}$-linear but not $\Bbb{R}$-linear.

Points to be considered:

1. If $g:\Bbb{R}\to \Bbb{R}$ defined by $g(x) =qx , q\in\Bbb{Q}$ is continuous then $g=f$ on $\Bbb{Q}$ implies $g=f $ on $\Bbb{R} $ i.e any continuous $\Bbb{Q}$-linear map extends linearly on $\Bbb{R}$.

$\boxed{ \text{ So $g$ can't be a continuous on $\Bbb{Q}$}}$

2. A linear map is completely determined by it's action on the basis. So our task is reduced to construct a discontinuous linear map from the vector space $\Bbb{R}_{\Bbb{Q}}$ to $\Bbb{R}$.

3. A non-continuous solution of an additive function ( called ugly function) is non-measurable.

Conjecture $1$: $f:\Bbb{R}\to \Bbb{R} $ is $\Bbb{Q}$-linear and non-measurable function. Then $\exists F\subset \Bbb{R}$ closed uncountable set such that $f(F) $ doesn't have the property of Baire.

4. There exists a discontinuous additive function $f:\Bbb{R}\to\Bbb{R}$ satisfying Darboux property ( It can be shown by defining $f$ linealy on a hamel basis of $\Bbb{R}_{\Bbb{Q}}$ and then extending additively on $\Bbb{R}$ )

Conjecture $2$ : The Darboux property of $f$ ( discontinuous $\Bbb{Q}$-linear function) is sufficient to conclude that $f(F) $ have property of Baire for every closed uncountable set $F$.

5. Any second category (non meager) subset of $\Bbb{R}$ contains a set that fails to have the property of Baire. $[$ Suppose $ A\subset \Bbb{R}$ non meager. Then $A\cap \mathcal{B} $ or $A\cap \mathcal{B}^c$ atleast one of the set doesn't have the B.P otherwise both would be meager and eventually $A$ would be meager $]$

6. $f:\Bbb{R}\to \Bbb{R}$ is an additive onto function such that $f$ is not injective then $f^{-1}(y)$ is dense in $\Bbb{R}$ for all $y\in\Bbb{R}$.

Question: Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the property of Baire for every $F$ defined above ?

Here is the MO post of this question.