Let $P \subseteq 2^{\mathbb{N}}$ be a $\Pi_1^1$-complete set. Show that $$R = \{(A,B) \in 2^{\mathbb{N}} \times 2^{\mathbb{N}} : B \subseteq A \, \& \, B \in P\}.$$ is also $\Pi_1^1$-complete.
Definition: A set $P ⊂ X$ is $\Pi_1^1$-complete if $P \in \Pi_1^1(X)$ and for every Polish space $Y$ and every $Q \in \Pi_1^1(Y)$, there is a Borel function $f : Y \rightarrow X$ such that $Q = f^{-1}(P)$.
To prove this, I want to define a Borel function $g: 2^{\mathbb{N}} \rightarrow 2^{\mathbb{N}} \times 2^{\mathbb{N}}$ such that $g^{-1}(R)=P.$
There are a few trivial ways to embed some $X$ into $X\times X$. The most canonical $g$ simply works:
I leave the answer behind the curtains in case you want to give it another try.
This $g$ is Borel and we have $$ g̣^{-1}(R) = \{B: g(B)_2\subseteq g(B)_1 \mathrel{\&} g(B)_2\in P\} = \{B: B \in P\} = B, $$
where $(A_1,A_2)_i = A_i$ for $i=1,2$.