Take some stable theory $T$ with elimination of imaginaries, all sets appearing are small subsets of the monster model of $T$, elementary maps are restrictions of automorphism of the monster model of $T$.  

Is the following statement correct?


1. For all algebraic closed sets $A,B$, with additionaly that these sets are independent over $acl(\emptyset)$. And all elementary $f:acl(AB)\rightarrow acl(AB)$ fixing $A$ and $B$. And all $C,D$ two (closed) sets independent over $acl(\emptyset)$ with $A\subset C$ and $B\subset D$. We can extend $f$ to some automorphism fixing $C$ and $D$.
2. Or equivalently, there is no $c\in acl(A\cup B)-dcl(A\cup B)$, such that $Ac$ independent of $B$ over $acl( \emptyset)$ and $A$ not independent of $Bc$ over $acl(\emptyset)$ (or vice versa).



Extending elementary maps in stable theories.
Martin Ziegler provided this solution. 


We got that $C$ is independent of $B$ over $A$. Then $C$ is independent of $acl(A\cup B)$ over $A$. So by stationarity (since acl(A)=A) we can extend $f$ to elementary $g:C\cup acl(A\cup B)\rightarrow C\cup acl(A\cup B)$ with $g\vert C=id$.


Since $C\cup acl(A\cup B)\subset acl(CB)$ we have then that $C\cup acl(A\cup B)$ is independent of $D$ over $B$. So by stationarity again ($acl(B)=B$) we can extend $g$ to $h:D\cup C\cup acl(A\cup B)$ with $h\vert D=id$. 


