On page 187 of Jech's Set Theory, there is a proof sketch of the fact that $\alpha \mapsto L_{\alpha}$ is $\Delta_{1}$. As far as I can tell, Jech's argument only shows that this operation is $\Sigma_{1}$. Can anyone explain why it is also $\Pi_{1}$?
More generally, I am wondering how to show that $\alpha \mapsto L_{\alpha}[A]$ is $\Delta_{1}$.
I think it would be enough to show that $Y=$def$_{A}(X)$ is absolute between transitive models of ZF, since then
"$s$ is a function on $\alpha \wedge$
$s(0)=0 \wedge$
$s(\beta + 1)=$def$_{A}(s(\beta)) \wedge$
$s(\beta)=\bigcup_{\gamma < \beta}s(\gamma)$ if $\beta$ is a limit"
is also absolute between transitive models. So, if $\phi(\alpha, s)$ expresses the above relation, then $X= L_{\alpha}[A]$ can be expressed either as
$\exists s [\phi(\alpha +1, s) \wedge s(\alpha)=X]$ or as
$\forall s [\phi(\alpha +1, s) \rightarrow s(\alpha)=X]$ and therefore $\alpha \mapsto L_{\alpha}[A]$ is absolute between transitive models.
This is a consequence of it being a function. Let us say that $"y=f(x)"$ is a function and is defined by a $\Sigma_1$ formula $\varphi(x,y)$ then it is also $\Delta_1$ since it has the following $\Pi_1$ definition,
$$\forall z (\varphi(x, z)\rightarrow z=y)$$ which of course is equivalent to
$$\forall z(\neg \varphi(x,z) \vee z=y)$$ this is clearly $\Pi_1$.