Unless I am mistaken, it seems to me that $\Delta^1_2$ does have the uniformization property in $L$.

For any set $A$ in $\Delta^1_2$, let $B$ select the $L$-least witness on each slice. So $B$ unifomizes $A$, and the graph of $B$ appears to be $\Delta^1_2$, by the following reasoning:

• $x\oplus z\in B$ if and only if it is in $A$, and for every well-founded countable model $M$ of $V=L$ containing $x$ and $z$, if $y$ is a real in $M$ preceding $z$, then $x\oplus y\notin A$.

• $x\oplus z\in B$ if and only if it is in $A$, and there is a well-founded countable model $M$ of $V=L$ containing $x$ and $z$, if $y$ is a real in $M$ preceding $z$, then $x\oplus y\notin A$.

The point here is that the countable well-founded models are correct about the $L$-predecesors of the reals that they can see. So we can use any or all of them when verifying that $z$ is least such that some $\Delta^1_2$ property holds. Note that the "for every real $y$ in $M$" is merely a natural number quantifier, since $M$ is coded as a countable structure. So the first of these characterizations is $Pi^1_2$ and the second is $\Sigma^1_2$, and so it is $\Delta^1_2$ overall.

Unless I am mistaken, it seems to me that $\Delta^1_2$ does have the uniformization property in $L$.

For any set $A$ in $\Delta^1_2$, let $B$ select the $L$-least witness on each slice. So $B$ unifomizes $A$, and the graph of $B$ appears to be $\Delta^1_2$, by the following reasoning:

• $x\oplus z\in B$ if and only if it is in $A$, and for every well-founded countable model $M$ of $V=L$ containing $x$ and $z$, if $y$ is a real in $M$ preceding $z$, then $x\oplus y\notin A$.

• $x\oplus z\in B$ if and only if it is in $A$, and there is a well-founded countable model $M$ of $V=L$ containing $x$ and $z$, if $y$ is a real in $M$ preceding $z$, then $x\oplus y\notin A$.

The point here is that the countable well-founded models are correct about the $L$-predecesors of the reals that they can see. So we can use any or all of them when verifying that $z$ is least such that some $\Delta^1_2$ property holds. Note that the "for every real $y$ in $M$" is merely a natural number quantifier, since $M$ is coded as a countable structure. So the first of these characterizations is $\Pi^1_2$ and the second is $\Sigma^1_2$, and so it is $\Delta^1_2$ overall.

Unless I am mistaken, it seems to me that $\Delta^1_2$ does have the uniformization property in $L$.

For any set $A$, let $B$ select the $L$-least witness on each slice. So $B$ is a function, and the graph of $B$ appears to be $\Delta^1_2$, by the following reasoning:

• $x\oplus z\in B$ if and only if it is in $A$, and for every well-founded countable model $M$ of $V=L$ containing $x$ and $z$, if $y$ is a real in $M$ preceding $z$, then $x\oplus y\notin A$.

• $x\oplus z\in B$ if and only if it is in $A$, and there is a well-founded countable model $M$ of $V=L$ containing $x$ and $z$, if $y$ is a real in $M$ preceding $z$, then $x\oplus y\notin A$.

The point here is that the countable well-founded models are correct about the $L$-predecesors of the reals that they can see. So we can use any or all of them when verifying that $z$ is least such that some $\Delta^1_2$ property holds. Note that the "for every real $y$ in $M$" is merely a natural number quantifier, since $M$ is coded as a countable structure. So the first of these characterizations is $Pi^1_2$ and the second is $\Sigma^1_2$, and so it is $\Delta^1_2$ overall.

Please correct me if this is wrong.

Unless I am mistaken, it seems to me that $\Delta^1_2$ does have the uniformization property in $L$.

For any set $A$ in $\Delta^1_2$, let $B$ select the $L$-least witness on each slice. So $B$ unifomizes $A$, and the graph of $B$ appears to be $\Delta^1_2$, by the following reasoning:

• $x\oplus z\in B$ if and only if it is in $A$, and for every well-founded countable model $M$ of $V=L$ containing $x$ and $z$, if $y$ is a real in $M$ preceding $z$, then $x\oplus y\notin A$.

• $x\oplus z\in B$ if and only if it is in $A$, and there is a well-founded countable model $M$ of $V=L$ containing $x$ and $z$, if $y$ is a real in $M$ preceding $z$, then $x\oplus y\notin A$.

The point here is that the countable well-founded models are correct about the $L$-predecesors of the reals that they can see. So we can use any or all of them when verifying that $z$ is least such that some $\Delta^1_2$ property holds. Note that the "for every real $y$ in $M$" is merely a natural number quantifier, since $M$ is coded as a countable structure. So the first of these characterizations is $Pi^1_2$ and the second is $\Sigma^1_2$, and so it is $\Delta^1_2$ overall.