About subposet of Levy collapse

Question

Let $\lambda>\kappa$ and $\operatorname{Coll}(\kappa, \lambda)$ be the poset collapsing $\lambda$ to $\kappa$. Pick a subposet $P$ which is $\lt\kappa$-closed and of size $\lambda$. Can we say that $P$ is forcing equivalent to $\operatorname{Coll}(\kappa,\lambda)$? Or, more generally, which are the minimal conditions to make $P$ equivalent to $\operatorname{Coll}(\kappa,\lambda)$?

When $\lambda=\kappa$ (i.e., $\kappa$-Cohen forcing) I am pretty sure that this be true, so I was interested in understanding whether it generalizes.

Answer 1

The result is not true in the generality you mentioned. For a counterexample, consider the case $\kappa=\omega$: the forcing to collapse $\lambda$ to $\omega$ is isomorphic to the forcing $\text{Add}(\omega,\lambda)*\text{Coll}(\omega,\lambda)$ to add $\lambda$ many Cohen reals and then collapse $\lambda$ to $\omega$. The first factor $\text{Add}(\omega,\lambda)$ is $\lt\omega$-closed (since any poset is) and has size $\lambda$, but it is not forcing equivalent to $\text{Coll}(\omega,\lambda)$.

So at the very least, you will want to insist that $\mathbb{P}$ also collapses all the cardinals up to $\lambda$.