Why is this set $\mathbf{\Pi}^0_\xi$? $\newcommand{\C}{\mathcal C}$
Let $\C$ denote the Cantor space and let $U\subseteq \C\times\C$ be $\C$-universal for $\mathbf{\Sigma}^0_\xi(\C)$, for some $1\leq\xi<\omega_1$, meaning that $U\in\mathbf{\Sigma}^0_\xi(\C\times \C)$ and $\Sigma^0_\xi(\C)=\{U_y\mid y\in \C\}$, where for $y\in\C$, $U_y=\{x\in\C\mid (y,x)\in U\}$.
Let $A=\{y\in\C\mid (y,y)\not\in U\}$. Why is $A\in\mathbf{\Pi}^0_\xi(\C)$? I see that $A$ cannot be in $\mathbf{\Sigma}^0_\xi(\C)$, but it's not clear to me why it must be in the complementary class. This is a step in a proof in Kechris's book.
 A: The key point is the following closure property of the Borel classes (and well-behaved pointclasses in general):

Each Borel class is closed under continuous preimages.

More precisely, given Polish spaces $\mathcal{A},\mathcal{B}$ and a continuous map $f:\mathcal{A}\rightarrow\mathcal{B}$, taking $f$-preimages does not increase topological complexity: if $X\subseteq\mathcal{B}$ is in ${\bf\Sigma^0_\xi}(\mathcal{B})$ then $f^{-1}(X)$ is in ${\bf\Sigma^0_\xi}(\mathcal{A})$.
This is a good exercise; the key point is to think about how taking preimages interacts with intersections, unions, and complements. Note that images are much more poorly behaved than preimages in this sense (and this leads to the difference between Borel and analytic sets).

In this case the continuous map we're interested in is the diagonal embedding $$d:\mathcal{C}\rightarrow\mathcal{C}\times\mathcal{C}: c\mapsto (c,c).$$ This is indeed a continuous map from $\mathcal{C}$ to $\mathcal{C}\times\mathcal{C}$, so we can apply the closure result above. In particular we have $d^{-1}(U)\in{\bf\Sigma^0_\xi}(\mathcal{C})$. Now consider $\mathcal{C}\setminus d^{-1}(U)$.
