Suppose $A$ is a complete subalgebra of a complete boolean algebra $B$, and $B$ is $\kappa$-centered. Let $G \subset A$ be a generic ultrafilter. Is $B/G$ $\kappa$-centered in $V[G]$?


Naively, we might attempt to prove it as follows. Let $\{ F\_\alpha : \alpha < \kappa \}$ be a collection of filters and in $V[G]$ let $F^\*\_\alpha = \{ [b]\_G : b \in F\_\alpha, b \not=\_G 0 \}$. But for $b\_1, b\_2 \in F\_\alpha$, maybe $b\_1 \wedge b\_2 =\_G 0$, so that $F^\*\_\alpha$ is not a filter.


An example where the centeredness cardinal goes up when modding out by a filter is given by comparing $\scr P(\omega)$ and $\scr P(\omega)/ \mathrm{fin}$. But maybe the generic filter case is different.



centeredness in forcing iterations
Let $P$ be the forcing for adding a Suslin tree $\mathring{T}$ by finite conditions. Then both $P$ and $P \star \mathring{T}$ are forcing isomorphic to adding $\omega\_1$ Cohen reals (so $P \star \mathring{T}$ is sigma-centered). See lemma 5.6 [here](http://arxiv.org/pdf/math/9506208v1.pdf).


