Inequality of Morley Ranks I'm reading theses notes on model theoretic stability and I got to an exercise on p. 43:

Show that if $X \subseteq \mathbb{M}_{x}, Y \subseteq \mathbb{M}_{y}$ and $f: X \rightarrow Y$ are definable, and $\operatorname{MR}\left(f^{-1}(y)\right) \leq 1$ for all $y \in Y$, then $\operatorname{MR}(X) \leq \operatorname{MR}(Y)+1$.

I think that to show it, I would need to adapt Lemma 10.1 above, but I do not know how to do so. I also do not know if I can assume saturation of $\mathbb{M}.$ I would appreciate any help!
 A: Let $A$ be a set of parameters such that $X$, $Y$, and $f$ are definable over $A$. Let $\mathrm{MR}(Y) = \alpha$ and $\mathrm{MD}(Y) = d$.
We proceed by induction on $\alpha$. For a fixed $\alpha$, we first handle the case $d = 1$. Let $(Z_i)_{i\in \omega}$ be an infinite family of pairwise disjoint definable subsets of $X$, defined over $A'\supseteq A$. To show $\mathrm{MR}(X)\leq \alpha+1$, it suffices to show that $\mathrm{MR}(Z_i)\leq \alpha$ for some $i$.
Let $y\in Y$ be generic over $A'$ in the sense that it is not contained in any $A'$-definable set of Morley rank $<\alpha$. Let $X_y = f^{-1}(y)$. By our hypothesis, $\mathrm{MR}(X_y)\leq 1$, so $X_y$ does not contain an infinite family of pairwise disjoint definable sets of Morley rank $\geq 1$. It follows that for some $i\in \omega$, $\mathrm{MR}(Z_i\cap X_y) = 0$, i.e., $Z_i\cap X_y$ is finite, say of size $n$.
Let $Y'$ be the set of all $y'\in Y$ such that $|Z_i\cap f^{-1}(y')|\leq n$, and note that $Y'$ is definable over $A'$. Let $Y'' = Y\setminus Y'$. Since $y\in Y'\subseteq Y$, $\mathrm{MR}(Y') = \alpha$, and since $\mathrm{MD}(Y) = 1$, $\mathrm{MR}(Y'')<\alpha$. Define $Z_i' = Z_i\cap f^{-1}(Y')$ and $Z_i'' = Z_i\cap f^{-1}(Y'')$, so $Z_i = Z_i'\cup Z_i''$. Then $f$ restricts to definable functions $f'\colon Z_i'\to Y'$ and $f''\colon Z_i''\to Y''$. All fibers of $f'$ are finite, so by Lemma 10.1 in the notes, $\mathrm{MR}(Z_i')\leq \mathrm{MR}(Y') = \alpha$. And by induction, $\mathrm{MR}(Z_i'') \leq \mathrm{MR}(Y'')+1 \leq \alpha$, since $\mathrm{MR}(Y'')<\alpha$. So $\mathrm{MR}(Z_i) = \max(\mathrm{MR}(Z_i'),\mathrm{MR}(Z_i'')) \leq \alpha$, as desired.
Now for the general case $\mathrm{MD}(Y) = d$, we can partition $Y$ into irreducible components $Y = Y_1\cup \dots \cup Y_d$, where $\mathrm{MR}(Y_i) = \alpha$ and $\mathrm{MD}(Y_i) = 1$ for all $i$. Let $X_i = f^{-1}(Y_i)$. Then $f$ restricts to definable functions $f_i\colon X_i\to Y_i$ for all $i$, and we have
\begin{align*}
\mathrm{MR}(X) &= \max_{1\leq i \leq d} \mathrm{MR}(X_i)\\
&\leq \max_{1\leq i \leq d} (\mathrm{MR}(Y_i)+1)\\
&\leq (\max_{1\leq i \leq d} \mathrm{MR}(Y_i))+1\\
&\leq \mathrm{MR}(Y)+1,
\end{align*}
where the first inequality uses the argument above, applied to each $f_i\colon X_i\to Y_i$ and $\mathrm{MD}(Y_i) = 1$.

You might hope that this argument generalizes to show that if $\mathrm{MR}(f^{-1}(y))\leq \beta$ for all $y\in Y$, then $\mathrm{MR}(X)\leq \mathrm{MR}(Y)+\beta$. But in fact, when $\beta>1$, all we get is the weaker result $$\mathrm{MR}(X)\leq \beta\cdot (\mathrm{MR}(Y)+1) = \beta\cdot \mathrm{MR}(Y)+\beta.$$
See Exercise 6.4.4 in A Course In Model Theory by Tent and Ziegler, as well as this related recent question. For an accessible proof that this bound is sharp, see the preprint A Remark on Morley Rank by Ziegler, currently available at this link (but warning: the proof of Theorem 1 written there is not correct). As the simplest surprising case, we can have a definable function $f\colon X\to Y$ with $\mathrm{MR}(Y) = 1$ and $\mathrm{MR}(f^{-1}(y)) = 2$ for all $y\in Y$, but $\mathrm{MR}(X) = 4$ (!).
Where does that extra factor of $\beta$ come from? Well, in generalizing the argument, it helps to argue by induction on $\alpha$ and $\beta_{\mathrm{gen}}$, where $\alpha = \mathrm{MR}(Y)$ and $\beta_{\mathrm{gen}}$ is an ordinal bounding the Morley rank of the generic fiber.
Here is a brief sketch: We proceed with the argument above, and we note that if the rank of the generic fiber is positive, then there is some $Z_i$ where the rank of the generic fiber intersected with $Z_i$ drops and we are done by induction. But if the rank of the generic fiber is already $\leq 0$, it cannot drop, and instead we partition $Y$ into definable subsets $Y'$ and $Y''$. Over $Y'$, all fibers are finite, and we can apply Lemma 10.1. And $Y''$ has Morley rank $<\alpha$, so we can apply the inductive hypothesis. Now over $Y''$, the ranks of the fibers are still bounded above by $\beta$, but we have no control on the rank of the generic fiber: it might jump all the way back up from $0$ to $\beta$. So in the worst case $\alpha$ has decreased by $1$ and $\beta_{\mathrm{gen}}$ has increased by $\beta$. To balance this out, we have to use the bound $\beta\cdot \alpha + \beta_{\mathrm{gen}}$.
