Elementary embeddings and measurable cardinals Given a measurable cardinal $\kappa$ we can consider its associated embedding $j:V\longrightarrow M\cong Ult_U(V)$ where $U$ is a $\kappa-$complete non principal normal ultrafilter on $\kappa$. In Lemma 17.9 Jech proves that the ultrafilter $U$ does not lie in the transitive collapse $M$. The strategy of the proof for me it is clear but I would like to ask regarding to some of its technicall details. The following are my explanation to Jech's assertions:


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*$\kappa^\kappa\in M$: This is due to $\kappa\in M$ (because $\kappa<j(\kappa)\in M$) and by absoluteness of function.

*$F\in M$: $f\mapsto \pi[f]$ is a function from $\kappa^\kappa\in M$ in $M$, then by replacement $b=\{\pi[f]:f\in \kappa^\kappa\}\in M$. Now, since $M$ is a model in particular $F=\kappa^\kappa\times b\in M$.


Are that details well proved?
Thank you in advance.
 A: Your reasoning for $\kappa^\kappa \in M$ is flawed: You are right, that any function $f \in M$ such that $M \models f \colon \kappa \to \kappa$ is really (i.e. from $V$'s point of view) a function $f \colon \kappa \to \kappa$, but that doesn't tell you that every function $g \colon \kappa \to \kappa$ (in $V$) is an element of $M$. 
To see that this holds, let $g \colon \kappa \to \kappa$ be a function in $V$. Then $j(g) \colon j(\kappa) \to j(\kappa) \in M$. $j(g)$ is a subset of $\pi(\kappa) \times \pi(\kappa)$ and I claim that $j(g) \cap \kappa \times \kappa = g$. (If we know that, then obviously $g \in M$)
So let $\xi, \eta < \kappa$. Then $(\xi, \eta) \in g) \leftrightarrow j((\xi, \eta))= (j(\xi), j(\eta)) \in j(g)$. Since $\xi, \eta < \kappa = \operatorname{crit}(j)$ we have $j(\xi)= \xi$ and $j(\eta)=\eta$ and thus $(\xi, \eta) \in g \leftrightarrow (\xi, \eta) \in j(g)$, i.e. $j(g) \cap \kappa \times \kappa = g$. (Note however that it isn't true that $g = j(g)$)
Your second statement is a bit fuzzy, but it seems to me that you want to show that $b = \{ [f]_U \mid f \in ^\kappa \kappa \} \in M$, where $[f]_U = \{ g \in ^\kappa \kappa \mid \{\xi < \kappa \mid f(\xi)=g(\xi)\} \in U \}$, provided that $U \in M$. (Leading to the desired contradiction)
If I'm right about what you're asking, then this is essentially trivial. We already know that $^\kappa \kappa \in M$ and we assume that $U \in M$. Then, for every $f \in ^\kappa \kappa$, $[f]_U$ is a subset of $^\kappa \kappa$ that can be formed by applying the comprehension axiom to $^\kappa \kappa$ using the $\Sigma_0$-formula 
$$g \in ^\kappa \kappa \wedge \{ \xi < \kappa \mid f(\xi) = g(\xi) \} \in U$$ 
(we use $^\kappa \kappa$ and $U$ as parameters). So, there is a single $\Sigma_0$-formula that yields (inside $M$!) for any $f \in ^\kappa \kappa$ the correct $[f]_U$ and we may now apply replacement to this formula in order to obtain that $b = \{ [f]_U \mid f \in ^\kappa \kappa \} \in M$.
