I'm not sure if this is research level, so feel free to vote to migrate.


Suppose we have a complete boolean algebra $A$, with a *dense*, $\sigma$-complete subalgebra $B$, and a $\sigma$-complete homomorphism $h : B \to C$, where $C$ is complete. Does $h$ have a $\sigma$-complete extension $h' : A \to C$?


EDIT: I forgot to say that I'm interested in the case that $A$ is atomless.



extending $\sigma$-complete boolean homomorphism
If the Boolean algebra $A$ is c.c.c., then of course every
$\sigma$-complete dense subalgebra $B$ is all of $A$, and so in this case the
answer is trivially affirmative.


If $A$ is not c.c.c., however, then fix an uncountable maximal antichain
$X\subset A$, which we may assume has size $\aleph\_1$, and proceed along Joseph's idea. Namely, let $B\subset A$ consist of the elements $a\in A$ that are below the
join of a countable subset of $X$, or above the complement of such
a join. This is a $\sigma$-complete subalgebra of $A$, which is dense because it has every element below any element of $X$. Let
$h:B\to C=\{0,1\}$ map $a$ to $0$ when it is below the join of a
countable subset of $X$, and otherwise $h(a)=1$. This is a
$\sigma$-complete homomorphism, which corresponds to the
co-countable filter on the subsets of $X$, as in Joseph's answer.


If we can extend this $h$ to a $\sigma$-complete homomorphism $h':A\to\{0,1\}$, however, then the
pre-image of $1$ will give a $\sigma$-complete nonprincipal
ultrafilter on $X$, which is impossible, since $\aleph\_1$ is not a
measurable cardinal.


