This is exercise 9 of Steel's notes on ultrapowers, found here. I have managed to progress through it, but I am stuck and I would appreciate any help. So this is the statement:
Let $M\models \mathsf{ZFC}^-+``U \mbox{ is a normal ultrafilter on } \kappa\mbox{''}$, with $M$ transitive. Also suppose that whenever $\pi:N \rightarrow M$ is elementary, and $N$ is countable with $\pi(W) = U$, there is a $\sigma$ such that
commutes. Then $U$ is $\omega_1$-complete.
What I have tried so far is fairly standard. First assume that $\langle A_n: n\in \omega \rangle$ is a sequence of elements of $U$, such that $\bigcap_nA_n=\emptyset$. Then choose some $\theta$ large enough and let $\pi:H \rightarrow H_\theta$ be elementary such that $H$ is countable and transitive, there is some $(N, W) \in H$, such that $\pi((N, W)) = (M, U)$ and also there is some sequence $\langle B_n: n\in \omega \rangle \in H$ such that $\pi(\langle B_n: n\in \omega \rangle) = \langle A_n: n\in \omega \rangle$.
Then $\pi|N:N \rightarrow M$ is elementary and $\pi|N(W) = U$ and so $\operatorname{Ult}(N, W)$ can be realized in $M$, via some $\sigma$. Now my strategy was to use $\bigcap_nB_n=\emptyset$(since $\bigcap_nA_n=\emptyset$), to say that $\operatorname{Ult}(N, W)$ is ill-founded and yet is injected into $M$ which is absurd. But the problem is that $\langle B_n: n\in \omega \rangle$ may not live in $N$ and so the regular argument for ill-foundedness won't go through. I would appreciate any fixes to this argument or hints for other approaches.
Remark. We are checking $\omega_1$-completeness in $V$, yet we take ultrapowers with respect to smaller models. And in this context $\omega_1$-completeness means that every countable sequence of members of $U$ has nonempty intersection.
EDIT I.
I am starting to doubt if this is even correct, because at least in my approach I think I am asking for too much information, which isn't there. This exercise is derived from a lemma which is claimed to be due to Jensen, which states the equivalence of the realizability and $\omega_1$-completeness of the ultrafilter. And in the above notes only the other direction is proven. Since this result is attributed to Jensen, I thought it would be best to ask for a reference if anybody knows of one.
After consulting with Dr.M. Golshani, we came to a proof for this exercise. Continuing from where I have left the proof above, since we have $B_n \in W$(because $A_n \in U$) for every $n \in \omega$, then $\xi \in i^N_W(B_n)$ for every $n$(Because $N \models ``W \mbox{ is normal''}$, as $M \models ``U \mbox{ is normal''}$.)(Also here $\xi$ is such that $\pi(\xi) = \kappa$.)
And so $$\sigma(\xi) \in \sigma(i^N_W(B_n)) = \pi(B_n) = A_n,$$ for every $n$, and so $$\sigma(\xi) \in \bigcap_n A_n,$$ which is a contradiction!
[The whole point here is that members of countable fragments of $U$ have these "typical" elements like $\sigma(\xi)$, just like the proof for the other direction.]