Consider the category of abstract $\sigma$-algebras ${\mathcal B} = (0, 1, \vee, \wedge, \bigvee_{n=1}^\infty, \bigwedge_{n=1}^\infty, \overline{\cdot})$ (Boolean algebras in which all countable joins and meets exist), with the morphisms being the $\sigma$-complete Boolean homomorphisms (homomorphisms of Boolean algebras which preserve countable joins and meets). If a morphism $\phi: {\mathcal A} \to {\mathcal B}$ between two $\sigma$-algebras is surjective, then it is certainly an epimorphism: if $\psi_1, \psi_2: {\mathcal B} \to {\mathcal C}$ are such that $\psi_1 \circ \phi = \psi_2 \circ \phi$, then $\psi_1 = \psi_2$. But is the converse true: is every epimorphism $\phi: {\mathcal A} \to {\mathcal B}$ surjective?

Setting ${\mathcal B}_0 := \phi({\mathcal A})$, the question can be phrased as following non-unique extension problem. If ${\mathcal B}_0$ is a proper sub-$\sigma$-algebra of ${\mathcal B}$, do there exist two $\sigma$-algebra homomorphisms $\psi_1, \psi_2: {\mathcal B} \to {\mathcal C}$ into another $\sigma$-algebra ${\mathcal C}$ that agree on ${\mathcal B}_0$ but are not identically equal on ${\mathcal B}$?

In the case that ${\mathcal B}$ is generated from ${\mathcal B}_0$ and one additional element $E \in {\mathcal B} \backslash {\mathcal B}_0$, then all elements of ${\mathcal B}$ are of the form $(A \wedge E) \vee (B \wedge \overline{E})$ for $A, B \in {\mathcal B}_0$, and I can construct such homomorphisms by hand, by setting ${\mathcal C} := {\mathcal B}_0/{\mathcal I}$ where ${\mathcal I}$ is the proper ideal $$ {\mathcal I} := \{ A \in {\mathcal B}_0: A \wedge E, A \wedge\overline{E} \in {\mathcal B}_0 \}$$ and $\psi_1, \psi_2: {\mathcal B} \to {\mathcal C}$ are defined by setting $$ \psi_1( (A \wedge E) \vee (B \wedge \overline{E}) ) := [A]$$ and $$ \psi_2( (A \wedge E) \vee (B \wedge \overline{E}) ) := [B]$$ for $A,B \in {\mathcal B}_0$, where $[A]$ denotes the equivalence class of $A$ in ${\mathcal C}$, noting that $\psi_1(E) = 1 \neq 0 = \psi_2(E)$. However I was not able to then obtain the general case; the usual Zorn's lemma type arguments that one normally invokes to give Hahn-Banach type extension theorems don't seem to be available in the $\sigma$-algebra setting. I also played around with using the Loomis-Sikorski theorem but was not able to get enough control on the various null ideals to settle the question (some subtle issues reminiscent of "measurable selection theorems" seem to arise). However, Stone duality seems to settle the corresponding question for Boolean algebras.

Amalgamation and Epimorphisms in m-Complete Boolean Algebrasshows that, for any infinite cardinal $m$ or $m$=arbitrary, the category of $m$-complete Boolean algebras with $m$-complete morphisms has the strong amalgamation property, which always implies that epis are surjective. $\endgroup$ – Badam Baplan May 22 at 5:39