Perhaps you should refine the question by asking what if $\sigma$ is defined using the $P$-name $\pi(\dot{G})$.

@DavidWhite I think my point is different. It’s not only cases when “there isn't any sensible location for a period” but also when maybe there is, but default TeX makes a poor choice.

@DmytroTaranovsky Why does it matter whether or not $A$ is definable in $V$? The model satisfies $\kappa$-covering for $\lambda$ since $A$ is cofinal in $P_\kappa(\lambda)^V$.

I still contend that your assertion, "$N \cap P_\kappa(\lambda) \in U$ requires so many sequences of ordinals to be present that there does not appear to be a canonical way to well-order them," is simply wrong. Despite the background structures in $V$ that enable this, we only add a certain $A \in U$ which is definably well-ordered. Any $B \subseteq A$ also in $U$ will suffice! And then when we add $U$ as a predicate, we get a supercompactness measure! So there appears to be a canonical way to well-order the sequences at a local level.

I don't know the answer to your question, but I think your intuition is misguided. Solovay proved that if $U$ is a supercompactness measure, then there is a set $A \in U$ such that the function $x \mapsto \sup(x)$ is injective on $A$. Therefore there is a canonical way to well-order enough sequences of ordinals to get a degree of supercompactness into a relatively small model. Just take $L(A)[U]$ for such an $A,U$. Now, is it "canonical"? The consensus is no.

Why not? If there is a set $X$ such that $\Phi(X)$, then take the one of least rank. Otherwise there are only classes $X$ such that $\Phi(X)$, so none of them can possess another.

@MihaHabič I’m not sure this works. Suppose we have a lottery sum of Cohen forcing, but in a weird way. For $n<2$ let $P_n$ be Cohen forcing but with the length of the conditions being $n$ mod 2. Take the lottery sum $P$ and let $\pi$ be the projection to Cohen via the identity map on each component. Now suppose $p \in P_0$ and $q\leq p$ is of odd length. Then there is no $p’\leq p$ such that $\pi(p’)=q$.

@MihaHabič Interesting. What is the dense subset? Is it just the image $\pi[Q]$?

Yes, there are provable counterexamples to preservation. See mathoverflow.net/questions/193522/preservation-of-properness

@MihaHabič Yes, here’s one kind of squirrelly example. Abraham projections are surjective. Let P be some poset whose Boolean completion is strictly larger (like adding one Cohen real). Then the standard embedding of P into its completion is a Cummings projection, but there’s no surjection.

I usually just write out the statement, $(\forall\alpha<\kappa)\alpha^\lambda<\kappa$.

There is no way to assign positive numbers to all stationary sets and get a measure, because of Solovay’s stationary partition theorem.

This comment will probably be removed as inappropriate, but I feel this question illustrates the problem with many of your posts. This is a legitimate question touching on the important Reflection Theorem of ZF, but it is an easy exercise for anyone who has gotten to that result in their first course on set theory. Could you please read a standard book on set theory like Jech, and start asking questions based on that? It seems like you have been trapped in a fascination with formal syntax and axioms for years, and you never get around to learning the actual content of set theory.

This seems impossible, since the forcing adds a cofinal $\omega$ sequence in $\omega_1$, so it adds reals. But also why would the Prikry property imply no new reals? The intersection of countably many stationary sets can be empty.

Do you mean $f:[\omega_1]^{<\omega} \to 2$?

Try Quora. It has some qualified people there answering lay math questions.