# Properness of quotient forcing

It is well known that if $P$ is a proper notion of forcing and $\Vdash_P \dot{Q} \text{ is proper}$ then the iteration $P \ast \dot{Q}$ is proper. Is the converse also true, i.e

Suppose that we have a two step iteration, such that $P$ and $P \ast \dot{Q}$ are proper forcings. Can we conclude that $\Vdash_P \dot{Q} \text{ is proper}$?

My intuition would tell me no, we could probably add in the first step a new stationary set, which gets killed after the second forcing. We have to ensure however that while doing this the iterated forcing remains proper, i.e. preserves stationary subsets of $[\lambda]^\omega$ of $V$ (for $\lambda$ an arbitrary uncountable cardinal).

If the answer to my question is negative, could at least the following be true: Suppose that $P$ and $P\times Q$ are proper. Does it follow that $\Vdash_P Q \text{ is proper}$?

Indeed, your remarks already very nearly provide a counterexample. Namely, consider the forcing $\mathbb{P}$ to add a Cohen subset of $\omega_1$ by initial segment. This forcing is countably closed and hence proper. Let $S$ be the generic set added by $\mathbb{P}$. It is easy to see that $S$ is both stationary and co-stationary. Let $\dot{\mathbb{Q}}$ be the forcing to shoot a club $C\subset S$. That is, conditions are closed bounded subsets of $S$, ordered by end-extension. This forcing is not proper in $V[S]$, because it destroys the stationarity of the complement of $S$.
Meanwhile, consider the combined forcing $\mathbb{P}\ast\dot{\mathbb{Q}}$. Every condition in this iteration has the form $(s,\tau)$, where $\tau$ is a $\mathbb{P}$-name for a closed bounded subset of $\dot S$. By extending this condition, we can decide a little more of $\tau$, and by iterating this $\omega$ many times, we find ourselves at a condition of the form $(s,c)$ where $s$ is a bounded subset of $\omega_1$, and $c$ is a closed subset of $s\cup\{\sup(s)\}$, containing this supremum. Thus, the collection of such conditions is dense in $\mathbb{P}\ast\dot{\mathbb{Q}}$. But that dense set of conditions is countably closed, and so $\mathbb{P}\ast\dot{\mathbb{Q}}$ is proper, as desired. In fact, it is forcing equivalent to $\text{Add}(\omega_1,1)$.