A continuous map relating co-constructible reals My question is the following:

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*Given $x,y \in \omega^\omega$ such that $x\equiv_c y$ is there an $L$-definable continuous map $\varphi: \omega^\omega\rightarrow  \omega^\omega$ such that $\varphi(x) = y$?

By $x\equiv_c y$ I mean that they are in the same constructibility degree, i.e. $L(x) = L(y)$ and by $\varphi$ being $L$-definable I mean that the map $\varphi$ is defined by a formula with constructible parameters. Also I can assume (if it is of any help) $V=L(x)$.
At first look I would say that it is unlikely, but I cannot say why.
Do you have any idea or suggestion?
Thanks!
 A: By "formula" I will mean "formula of set theory with ordinal parameters." Note that "ordinal parameters" is equivalent to "constructible parameters" for our purposes, since $L$ carries a definable bijection with $\mathsf{Ord}$.
Also, since the answer to your question is trivially true if $V=L$, I'm interpreting your question as asking whether we always have the situation you describe. Under this interpretation we get a highly robust negative answer: every $M\models\mathsf{ZFC}$ has a forcing extension in which the principle you ask about fails. So your intuition is correct in a very strong way.

We start with two simple observations:

We always have $x\equiv_c x'$.

Proof: since $x'\ge_Tx$ we have $x'\ge_cx$, and conversely since $x'$ is $\Sigma^0_1(x)$ we have $x\ge_cx'$. $\quad\Box$

Given countably many continuous functions $f_i:\omega^\omega\rightarrow\omega^\omega$ ($i\in\omega$), there is a real $a$ such that whenever $x\ge_Ta$ we have $f_i(x)\le_Tx$ for each $i\in\omega$ - and consequently $f_i(x)\not=x'$ for any $i\in\omega$.

Proof: basically, have $a$ "code" all the $f_i$s in some appropriate way. $\quad\Box$

Now given a (transitive, for simplicity) model $M$ of $\mathsf{ZFC}$, let $\mathsf{ODC}(M)$ ("ordinal-definable continuous") be the set of continuous functions $f:\omega^\omega\rightarrow\omega^\omega$ which are definable over $M$ by a formula with ordinal parameters.

If $\mathsf{ODC}(M)$ is countable in $M$, then $M\models$ "There is a counterexample to your question."

Proof: apply the two simple observations above, with $y=x'$. $\quad\Box$
Of course, on the face of it $\mathsf{ODC}(M)$ could be quite large (e.g. size continuum$^M$). Fortunately, we can control it relatively easily:

If $\mathbb{P}\in M$ is a homogeneous forcing and $G$ is $\mathbb{P}$-generic over $M$, then "$\mathsf{ODC}(M)=\mathsf{ODC}(M[G])$" - or, more precisely, if $f\in\mathsf{ODC}(M[G])$ then $f\upharpoonright M\in\mathsf{ODC}(M)$.

Proof: this is the usual homogeneity argument: the formulas defining elements of $\mathsf{ODC}$ only involve parameters from the ground model, so their behavior on ground model inputs is independent of the choice of generic. Now use definability of forcing. $\quad\Box$
The point is that if $M,M[G]$ are as above, then $$M[G]\models\vert\mathsf{ODC}(M[G])\vert=\vert\mathsf{ODC}(M)\vert.$$ And now to wrap up we just note that $Col(\omega,\kappa)^M$ is homogeneous for any $\kappa$, including $\kappa=\vert\mathsf{ODC}(M)\vert^M$.
