Changing cofinalities above supercompact cardinals

Question

Question. Suppose $\kappa$ is a supercompact cardinal and $\lambda > \kappa$ is measurable (or even larger large cardinal if necessary). Is there a set generic extension of the universe in which $\kappa$ remains supercompact, $\lambda$ is preserved and $cf(\lambda)=\omega?$

Remark 1. By Gitik-Shelah indestructibility result, if supercompact cardinal is replaced with strong cardinal, then the answer is yes.

Remark 2. If we require that the forcing preserves $\lambda^+,$ then the answer is no, as it is shown by Yair.

Remark 3. In A note on sequences witnessing singularity - following Magidor-Sinapova, Gitik has conjectured the following:

Conjecture. Suppose that

1. $V ⊆ W$ models of ZFC with same ordinals,

2. $κ$ is a regular cardinal in $V$,

3. $cof(κ) = ω$ in $W$,

4. $\aleph_1^V=\aleph_1^W,$

5. $V, W$ agree about a final segment of cardinals.

Then there is a subclass $V′$ of $V$ which is a model of $ZFC$, agree with $V$ about a final segment of cardinals, and there is a sequence witnessing singularity of $κ$ (in $W$) which is generic over $V′$ for either Namba, Woodin tower or Prikry type forcing.

Assuming this conjecture, it seems quite plausible that the answer to the question might be no in general.

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Edition. I realized that the question has connection with recent work of Woodin:

Theorem. Assume $\kappa$ is an extendible cardinal. If Woodin's $HOD$-conjecture holds, then we can not change the cofinality of some large cardinal $\lambda > \kappa$, preserving the supercompactness of $\kappa,$ by set forcing without collapsing $\lambda.$

Answer 1

There is no such forcing that preserves $\lambda^+$. Since $\lambda$ is measurable, $2^{<\lambda} = \lambda$ and therefore $\square_{\lambda,\lambda}$ holds in $V$. Since $\lambda^{+}$ is preserved, the same sequence will witness that there is still a weak square at $\lambda$ in the generic extension. But weak square fails at singular cardinals of small cofinality above a supercompact cardinal, so $\kappa$ cannot be supercompact in the generic extension.

In fact, a theorem of Dzamonja and Shelah shows that if you change the cofinality of an inaccessible cardinal $\lambda$ while preserving $\lambda^+$, there is even a $\square_{\lambda,\omega}$ sequence in the generic extension.