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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.

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  • $\begingroup$ Sorry for the confusion over terminology. $\endgroup$ – Monroe Eskew Nov 4 '14 at 12:38
  • $\begingroup$ I'm sorry, it was my mistake. $\endgroup$ – Mohammad Golshani Nov 4 '14 at 13:17
  • $\begingroup$ This comment chain looks very strange now. $\endgroup$ – Paul McKenney Nov 4 '14 at 16:17
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    $\begingroup$ @PaulMcKenney I agree. I thought Monroe was talking to himself. $\endgroup$ – Sean Cox Nov 4 '14 at 21:01
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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.

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  • $\begingroup$ For readers- We use the fact that the finite support product of at most continuum $\sigma$-centered posets is $\sigma$-centered. $\endgroup$ – Monroe Eskew Nov 7 '14 at 2:53
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It's consistent that the answer is no. Bartoszynski and Judah prove the following on page 26 in their book, Set Theory: On the Structure of the Real Line:

Assume $MA_\kappa$. Then a partial order of size $\leq \kappa$ is ccc iff it is $\sigma$-centered.

Since a Cohen real adds a Suslin tree, there is a ccc iteration $Add(\omega) * \dot{T}$ of size $\aleph_1$ where $\dot{T}$ is forced to be a Suslin tree. Under $MA_{\aleph_1}$, this is $\sigma$-centered ($\omega$-centered), but the quotient forcing is not because Aronszajn trees are not $\sigma$-centered.

Mohammad was on the right track in the (deleted) comments. Maybe someone can construct a ZFC counterexample.

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