I am trying to prove that a countable dense subset $Q\subset 2^\omega$ (where $2^\omega$ denotes the Cantor space) is not Wadge reducible to its complement. I am trying to prove this because I want to claim that $Q$ is Wadge incomparable to its complement. (For $A,B\subset 2^\omega$, we say $A$ is Wadge reducible to $B$ if there exists continuous $f:2^\omega\to 2^\omega$ such that $f^{-1}(B)=A$).
I tried proving this by contradiction. If I assume that there is a continuous $f:2^\omega\to 2^\omega$ such that $f^{-1}(Q)=Q^c$ and $f^{-1}(Q^c)=Q$, then I see that the image of $f$, $f(2^\omega)$, is a countable subset of $2^{\omega}$. Also, I noticed that there exists $x\in 2^{\omega}$ such that $f^{-1}(x)\subset Q^c$ is uncountable.
I couldn't notice any further things, and couldn't draw a contradiction.
Any help is appreciated.
Update: Here is one possible approach that just came up in my mind, though not sure whether this is correct:
$Q$ is an $F_\sigma$ subset that is not $G_\delta$. For if such a map $f$ existed, then $f^{-1}(Q)=Q^c$ is also an $F_\sigma$ set. But then $Q=(Q^c)^c$ is a $G_\delta$ set which is a contradiction.