Suppose $\kappa$ is an inaccessible cardinal. Consider the termspace forcing for adding a Cohen subset of $\kappa$ after $\mathbb P= Col(\omega,<\kappa)$. Members are Levy names for countable partial functions from $\kappa$ to 2, ordered by $\tau \leq \sigma$ iff $1 \Vdash \tau \leq \sigma$. Call this $T(\mathbb P, Add(\kappa))$.


It is well known that the identity map is a projection from $\mathbb P \times T(\mathbb P, Add(\kappa))$ to the iteration $\mathbb P \* \dot Add(\kappa)$. Furthermore in this situation it is not hard to show that $T(\mathbb P, Add(\kappa)) \cong Add(\kappa)$. So if $G \times H$ is $\mathbb P \times Add(\kappa)$-generic, then in the extension there is $I$ such that $G \* \dot I$ is $\mathbb P \* \dot Add(\kappa)$-generic.


My question is what about the opposite. In $V[G\* \dot I]$ is there an $Add(\kappa)^V$-generic?



Reverse of a termspace forcing fact
The answer is no. Forcing with
$\mathbb{P}\*\dot{\text{Add}}(\kappa,1)$ adds no fresh subsets to
$\kappa$, that is, a new subset of $\kappa$ all of whose initial
segments are in $V$. Thus, it cannot add a $V$-generic filter for
$\text{Add}(\kappa,1)^V$.


To see this, note that $\mathbb{P}$ is productively $\kappa$-c.c.,
meaning that $\mathbb{P}\times\mathbb{P}$ is $\kappa$-c.c., since
$\mathbb{P}$ remains $\kappa$-c.c. after forcing with
$\mathbb{P}$, since in the extension it amounts to
$\text{Add}(\omega,\kappa)$, which is c.c.c. there.


It now follows from Spencer Unger's improved version of my lemma
on [the approximation and cover properties](http://jdh.hamkins.org/approximation-and-cover-properties/), which go back to results implicit in 
Mitchell's dissertation. Specifically, it follows from lemma 1.3
in his paper [Spencer Unger, FRAGILITY
AND INDESTRUCTIBILITY II](http://www.math.ucla.edu/~sunger/Indestructible2.pdf) that $\mathbb{P}\*\dot{\text{Add}}(\kappa,1)$ has
the $\kappa$-approximation property, meaning this forcing adds no
new sets of ordinals all of whose small approximations are in the
ground model. Thus, it can add no $V$-generic subsets of $\kappa$
using $\text{Add}(\kappa,1)^V$.


