This is taken from Pillay's highly minimalistic book on stability theory.
Let $T$ be stable, $\mathcal{M} \prec \mathcal{N}$ models, and $a$ a tuple in the big model such that its type over $N$ inherits its type over $M$. The situation is as follows:
- $p' \in S(M\cup a)$ is the heir of $p \in S(M)$
- $p \subset q \subset q'$ where $q,q'$ are in $S(N)$ and $S(N\cup a)$, respectively. The type $q'$ inherits $q$, but $q$ need not necessarily inherit $p$.
The problem is to show that every formula in $p'$ is also in $q'$.
The assumption of stability is necessary. To see this, give $\mathcal{M} = \mathbb{Q}$ and $\mathcal{N} = \mathbb{R}$ the usual ordering. Let $a = \sqrt{2} - \varepsilon^{\ast}$ where $\varepsilon^{\ast}$ is positive but smaller than any positive real number. Let $p$ be the cut corresponding to $\sqrt{2}$ and $p'$ the below $a$. Finally, let $q'$ be the cut above $\sqrt{2}$ and $q$ the restriction of $q'$ to $\mathcal{N}$.
Now, $p'$ is a heir of $p$, as an unrealized son of an undefinable type. Also, $q'$ is a heir of $q$. But $q'$ does not contain $p'$ since they relate to $a$ differently: the formula $x > a$ is in $q'$ but not in $p'$.
I've tried to assume that some $\phi(x,a)$ is in $p'$ but not in $q'$. If $d$ is a defining schema for $p$, then the fact that $\models \neg{d\phi (a)}$ gives a contradiction, but I could not make much further progress. I also think one could translate the problem to the settings of algebraic geometry and equate the types with irreducible varieties to gain some intuition as for why it should hold, but also could not succeed.
Suppose $\phi(x,a)\in p'$, but $\phi(x,a)\notin q'$. Since $p'$ is the heir of $p$, this means $\models d_p\phi(a)$. Similarly, $\models \lnot d_q\phi(a)$.
So $d_p\phi(x)\land \lnot d_q\phi(x) \in \text{tp}(a/N)$. But since $\text{tp}(a/N)$ is the heir (hence also the coheir) of $\text{tp}(a/M)$, there is some $b\in M$ such that $\models d_p\phi(b)\land \lnot d_p\phi(b)$, i.e. $\phi(x,b)\in p$ but $\phi(x,b)\notin q$.
This contradicts the assumption that $q$ extends $p$.