There is an arithmetical set $A\subseteq 2^{\omega}\times \omega^{<\omega}$ so that for any $x\in 2^{\omega}$, $A(x)=\{\sigma\mid \exists n(x|n,\sigma)\in A\}$ is an $x$-recursive tree which has an infinite path but no infinite path hyperarithmetic in $x$.

Now let $B$ be an arithmetical set so that $(x,y)\in B$ if and only if $y$ is an infinite path through $A(x)$. $B$ has the following property:

(1). For any $x$, there is some $y$ so that $(x, y)\in B$;

(2) No analytic function uniformizing $B$. Suppose not, then there is a real $z$ and $\Sigma^1_1(z)$-function $f$ uniformizing $B$. Let $x\geq_h z$, then $f(x)\leq_h x\oplus z\leq_h x$, a contradiction.

There is an arithmetical set $A\subseteq 2^{<\omega}\times \omega^{<\omega}$ so that for any $x\in 2^{\omega}$, $A(x)=\{\sigma\mid \exists n(x|n,\sigma)\in A\}$ is an $x$-recursive tree which has an infinite path but no infinite path hyperarithmetic in $x$.

Now let $B$ be an arithmetical set so that $(x,y)\in B$ if and only if $y$ is an infinite path through $A(x)$. $B$ has the following property:

(1). For any $x$, there is some $y$ so that $(x, y)\in B$;

(2) No analytic function uniformizing $B$. Suppose not, then there is a real $z$ and $\Sigma^1_1(z)$-function $f$ uniformizing $B$. Let $x\geq_h z$, then $f(x)\leq_h x\oplus z\leq_h x$, a contradiction.