Galois types, factorization 2 I have asked a related question elsewhere but I still have a problem with this proof of the Proposition $6.2$ on page $13$: in the $-3$rd paragraph in the first snippet why
exactly do we enumerate $U(N)\setminus U(h)(U(M))=\langle a_i\ |\ i<\alpha\rangle$ and
what do we intuitively construct by induction at places $1.$ and $2.$ from the last paragraphs of the first
snippet? Also, what shows on page $14$ that $$f_\alpha\bar{t}_\alpha h=f_\alpha m_{0\alpha}=f$$ ?


 A: The intuition is as follows. The set up is that we have $M \preceq N$ with $|N| < \lambda$ and $M \preceq K$ with $K$ $\lambda$-saturated. As we often do in model theory, we may view $M$ as a subset of $N$ and $K$. The goal is to construct an embedding $N \to K$ that keeps $M$ fixed. The general proof strategy is as follows.


*

*Enumerate the remainder of $N$, that is $N - M$ as $(a_i)_{i < \alpha}$. Note that this means that $\alpha < \lambda$.

*Inductively we will construct partial embeddings $f_i: M \cup \{a_k : k < i\} \to K$, such that $f_j$ extends $f_i$ for $i < j$.

*We set $f_0 = f$.

*Having constructed $f_i$, we let $b$ realise $\operatorname{tp}(a_i / M f_i(\{a_k : k < i\}))$ in $K$. This we can do by $\lambda$-saturation. Extend $f_i$ to $f_{i+1}$ by setting $f_{i+1}(a_i) = b$.

*At limit stages we set $f_\ell = \bigcup_{i < \ell} f_i$, where $\ell$ is the limit.

*Having constructed $f_i$ for $i < \alpha$, we can once more take the union to obtain $f_\alpha: N \to K$, which is then the required embedding.


I have stayed as close as possible to the original notation from the proof so you can see what everything matches to.
This was the general proof strategy, but obviously some things become a bit more complicated in the paper. One problem is that $\lambda$-Galois type saturation only makes sense if the parameter set (so "$M f_i(\{a_k : k < i\})$" in step 4 above) is actually a model. This is essentially solved by the $t_i$ that are constructed as well. In the above steps, my $f_i$ actually play to role of the $t_i$ and $f_i$ from the proof at the same time. The idea here is that $M_i$ will contain what was $M \cup f_i(\{a_k : k < i\})$ above. This embedding of a concrete set into $M_i$ (technically into $U(M_i)$) is done by $t_i$. Then we send $M_i$ into $K$ via $f_i$. So $U(f_i) t_i$ is really what my $f_i$ is above.
The reason we can find such $M_i$ is essentially Löwenheim-Skolem. We just find a small enough $M_{i+1}$ containing both $M_i$ and the realisation which I called $b$. Of course, this $M_{i+1}$ will contain some other noise, but that does not really matter. In particular, in the end $M_\alpha$ will contain more than just a copy of $N$. However, that is no problem, because we are only interested in this copy of $N$ landing in the right spot. That last fact is ensured by $t_\alpha$.
That brings us to your final question, why do we have
$$
f_\alpha \bar{t}_\alpha h = f?
$$
This is important, because it is witnessing that indeed $f_\alpha \bar{t}_\alpha$ is the required embedding $N \to K$ (the above equation shows that $M$ remains fixed). Well, by definition of $\bar{t}_\alpha$ we have $U(\bar{t}_\alpha) = t_\alpha$. So
$$
U(f_\alpha \bar{t}_\alpha h) = U(f_\alpha) t_\alpha U(h) = U(f_\alpha m_{0\alpha}).
$$
Then faithfulness of $U$ gives us that $f_\alpha \bar{t}_\alpha h = f_\alpha m_{0\alpha}$. The right hand side of that equation is by construction equal to $f$ (point 1 of the induction hypothesis).

I have not said anything about $L$ or $g_1$ and $g_2$ and the $u_i$. The role of $L$ is that of a monster model, as is also discussed around 2.3 and 2.4 in the paper. They also mention (page 5) that these can be "can typically be written out of proofs, albeit at some cost in length and comprehensibility". I agree, there is no essential reliance on the monster here (this should never really happen). If you would want to get rid of it, and thus of the additional assumption about the existence of certain cardinals, then you would end up constructing a chain $(L_i)_{i < \alpha}$ during your induction as well. Because every time we get an isomorphism $s: L \to L$ now that witnesses equality of Galois types, we would only get some extension $L_i \to L_{i+1}$. It is not hard to make this work, just messy.
