This is an excellent and interesting question! You are asking whether the 2-step iteration of a normal measure μ on a measurable cardinal κ is uniquely factored by the steps of the iteration itself.

(I think I have just been exchanging email with you--or perhaps with one of your colleagues, with a different name?--about the one-dimensional version of this question.)

My answer is that No, you haven't quite got all the possibilities. The reason is that actually, there is a commuting square of elementary embeddings for the 2-step iteration, as follows. The ultrapower j₀₂ is just the ultrapower of V by the measure μ x μ, and it is a standard model-theoretic fact that the ultrapower of any structure by a product measure μ x ν is to first take the ultrapower by ν, and then take the ultrapower by μ. The lower half of the square is the iteration you have defined, with the diagonal being j₀₂. The upper half of the square is that j₀₂ is also equal to the composition j₀₁.j₀₁. This is not obvious, but it can be verified by proving the model-theoretic fact that I mentioned. Thus, we can take i = j₀₁:V-->M₁ and k = j₀₁ restricted to M₁, and this will map elementarily into M₂ with j₀₂ = k.i. But in this case, we have N = M₁ and k " N is just j₀₁ " M₁.

This is not equal to any of the sets on your list. It is not j₀₂ " V, since it includes j₀₁(κ). It is not j₁₂ " M₁ since it does not include κ. And it is not M₂ since it is not transitive.

A much more interesting follow-up question, is whether the factorings arising in the commuting square are exhaustive, and I will give this some thought. I think it may be the case that if you add my extra case, then the list is exhaustive...

This is an excellent and interesting question! You are asking whether the 2-step iteration of a normal measure μ on a measurable cardinal κ is uniquely factored by the steps of the iteration itself.

The answer is Yes.

Let me denote κ₀ just by κ and j₀₂ by j. Since μ₁ is a measure in M₁, it has the form j₀₁(m)(κ), where m = (ν_α | α < κ). Since you have said that μ₁ is not in ran(j₀₁), we may choose the ν_α to be all different, and different from μ₀. In this case, there is a partition of κ as the disjoint union of X_α, with X_α in ν_α and none in μ₀. Let x = (X_α | α < κ). Note that κ is not in j₀₁(X_α) for any α < κ, and similarly κ₁ is not in j(X_α). But κ is in j₀₁(x)(β) for some β < κ₁, since this is a partition of κ₁. Apply j₁₂ to conclude that κ₁ is in j(x)(β) for this β. Thus, there is some β in the interval [κ, κ₁) having the form β = j(f)(κ₁) for the function f that picks the index. From this, it follows from normality of μ₀ that we can write κ = j(g)(κ₁) for some function g, since any β < κ₁ generates κ via j₀₁. In my favored terminology, the seed κ₁ generates κ via j and in fact generates all β in [κ,κ₁) via j.

Similarly, suppose that δ is in the interval [κ₁,j(κ)). We know δ = j₁₂(f)(κ₁) for some function f on κ₁ in M₁. We also know f = j₀₁(F)(κ) for some F in V. Thus, δ = j(F)(κ, κ₁). In Y, let (α,β) be the smallest pair with δ = j(F)(α,β). It cannot be that both are below κ₁, since this would be inside ran(j₁₂) and so the least pair must have β = κ₁. Thus, δ generates κ₁, which we already observed generates κ.

To summarize, every ordinal in the interval [κ₁,j(κ)) generates κ₁, which generates all the ordinals β in [κ,κ₁), any of which generate κ and all the other such β.

This is enough to answer your question. The k " N in your question is just an arbitrary elementary substructure of M₂ containing ran(j), so suppose we have Y elementary in M₂ and ran(j) subset Y. The case Y = ran(j) is one of your cases. Otherwise, Y has something not in ran(j). Every object in M₂ has form j(h)(κ,κ₁) for some function h, so by looking at the smallest pair of ordinals to generate a given object with j(h), we see that there must be ordinals below j(κ) in Y. If Y contains any ordinal δ in the interval [κ₁,j(κ)), then it will contain both κ and κ₁, since we observed that any such δ generates these ordinals. In this case, Y = M₂, since those two ordinals generate everything. So we assume that Y contains no such δ. In this last case, Y must contain some ordinal β in the interval [κ,κ₁). Since any such β generates κ, Y contains all such ordinals. It follows that ran(j₁₂) subset Y and in fact = Y, since if Y contained anything more it would have to have an additional ordinal δ in [κ₁,j(κ)).

So we've seen that your three cases are the only possibilities. And like your previous question, there is no need to assume that Y or N is somehow internally definable.

By the way, this was a problem that I had solved many years ago for my dissertation, although perhaps other people had also thought about it. I was interested in understanding which pairs of ordinals (α,β) generate product measures via an embedding j, and this question is very much related to that.

(Click the edit history to see my previous answer, which was just about the case when μ₁ is in the range of j₀₁, a case for which the answer is no.)