Self-Dual Pointclass with a Universal Set? I'm a little bit embarrassed to be asking this, but I can't find my mistake!
Assume AD. I am going to construct a self-dual pointclass with a universal set, swhich is impossible.
Let A $\equiv_w A^c$ and consider the pointclass $\Delta = \{ B | B\leq_w A\} $. By Wadge, $\Delta$ is a self-dual pointclass. Next, define $U(x,y) \leftrightarrow x(y)\in A$ where we view reals x as coding continuous functions in some reasonable way. The function $(x,y) \mapsto x(y)$ is continuous so $U \in \Delta$ (we have written U as a continuous preimage of A). Also U is clearly universal for $\Delta$, a contradiction.
Can you please help me find my mistake? 
Thank you,
Cody
 A: The only way you can define $U$ the way you define it (so that it can be universal in the end), is for $A$ to be $\Delta$-complete set, for some non-selfdual pointclass $\Delta$. (Sorry for using $\Delta$ for non-seldual pointclasses!). So the mistake is in assuming at the outset that $A$ is a $\Delta$-complete set, where $\Delta$ is $\textit{selfdual}$. But since $A\leq_W A^c$ (and in any case $\Delta$ is selfdual in your example), this can't be.
Notice that $A^c$ is a $\Delta$ set, but if $U$ were universal and a section of $U$ realized $A^c$, then this would give, for some $x\in \mathbb{R}$, that if $x(y)\in A^c\leftrightarrow x(y)\notin A \leftrightarrow \neg U(x,y)$, but then $U(x,y)\leftrightarrow x(y)\in A^c$, so $U(x,y)\leftrightarrow \neg U(x,y)$. (but I guess you saw that already...)
So you can't just conclude $U$ is universal unless you secured first that $A$ is a $\Lambda$-complete set for some nonselfdual pointclass $\Lambda$, otherwise the definition is not valid.
A: After a few conversations I figured it out. Henning's suspicion was right. The point is that not all reals code continuous functions (it is true that if x does code a continuous function $(x,y) \mapsto x(y)$ is continuous). So the computation $U(x,y) \leftrightarrow x(y) \in A$ is misleading, U is actually $\exists^\mathbb{R} \Delta$ 
