Question from Jech, Set Theory-
let $ M=Ult(V,U)$ be the Ultrapower of V by U (isomorphic to transitive collapse). And the embedding
$$
j:V→M ,j(x):=[c_x]
$$
s.t for all $a$ : $[c_x](a) = x$
Lemma 20.14 states that
the following defines an Ultrafilter on $P_\kappa(\lambda):$
$$
X\in U\iff j''\lambda\in j(X)
$$
Then it says: $U$ is normal: if $f(x)\in x$ for almost all $x$, then $(jf)(j''λ)\in j’’(\gamma)$. Hence $(jf)(j''\lambda)=j(\gamma)$ for some $\gamma<\lambda$,
And so $f(x)=\gamma$ for almost all $x$.
I don't understand the part that comes after the "bold". Maybe because I don't realize what function the expression $(jf)(j''\lambda)$ represents. If we denote $a:=j''\lambda$, $ $than it should be: $$(jf)(j''\lambda)=[c_f](a)$$
So, given $x\in P_\kappa(\lambda)$ ,Is this the correct operation? $$[c_f](a)(x) = f(a(x))$$
thanks
Best Answer
I think your confusion actually lies somewhere else since your set up defining the standard ultrapower embedding doesn't make much sense. The canonical embeding has nothing (directly) to do with the direction of the theorem you are asking about, which starts from the supposition of an elementary embedding satisfying certain properties (i.e. not necessarily an ultrapower embedding) and deduces the existence of a normal measure on $P_\kappa(\lambda).$ The other direction shows that the canonical embedding with respect to a normal measure has these properties.
As for how the part after the bold follows from the part before it, $f(x)=\gamma$ for almost all $x$ means that $$ \{x: f(x)=\gamma\}\in U,$$ which from the definition of $U$ means that $$ j''\lambda\in j\{x: f(x)=\gamma\}.$$ We have $$j\{x: f(x)=\gamma\}=\{x:jf(x)=j(\gamma)\},$$ so $f(x)=\gamma$ for almost all $x$ means exactly that $j''\lambda\in \{x:jf(x)=j(\gamma)\}$, i.e. $ jf(j''\lambda)=j(\gamma).$
Regarding your last part, yes, it is correct that for $[g]\in \operatorname{Ult}_U(V),$ that $[h]=(j_Uf)([g])$ iff $\{x:h(x)=f(g(x))\}\in U.$ As for which $g$ have $[g]=j_U''\lambda,$ if $U$ is a normal measure then $j_U''\lambda$ is represented by the diagonal function $d(x)=x.$ So we have $h(x)=f(d(x))=f(x)$ almost everywhere. So the fact that $[f]=[h]=(j_Uf)(j_U''\lambda)=j(\gamma)=[c_\gamma]$ means $f(x)=\gamma$ almost everywhere, which ties out.