# In the category of sigma algebras, are all epimorphisms surjective?

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.

• I don't think homomorphisms to $\mathbb{F}_2$ suffice. For instance if one takes the measure algebra ${\mathcal L}([0,1])/\sim$ (Lebesgue measurable subsets of $[0,1]$ modulo null sets) then there are no homomorphisms to $\mathbb{F}_2$ whatsoever. May 16 '20 at 23:59
• @Simon Henry Lagrange's paper Amalgamation and Epimorphisms in m-Complete Boolean Algebras shows 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. May 22 '20 at 5:39
• @BadamBaplan: you could answer the question properly (not in a comment) to collect the points and fame. May 22 '20 at 6:37
• @TerryTao is there a reason you're choosing the terminology "$\sigma$-algebra"? I'm not from this field, but I understand that this is widely referred as "$\sigma$-complete BAs", and hence could be called "category of $\sigma$-complete BAs". I think I learnt on this site that there are $\sigma$-complete BAs that cannot be "realized as $\sigma$-algebras", that is, $\sigma$-algebras usually refers to those $\sigma$-complete BAs consisting of subsets of a set, in a way compatible with countable supremums.
– YCor
May 22 '20 at 9:08
• I was not entirely serious about points and fame, but I was serious about answering the question because that would make the answer better documented and more visible. May 22 '20 at 12:00

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A 1974 paper of R. Lagrange, Amalgamation and epimorphisms in $$\mathfrak{m}$$-complete Boolean algebras (Algebra Universalis 4 (1974), 277–279, DOI link), settled this affirmatively. In the cited paper, Lagrange shows that for any infinite cardinal $$\mathfrak{m}$$, the category of $$\mathfrak{m}$$-complete Boolean algebras with $$\mathfrak{m}$$-complete morphisms has the strong amalgamation property, which implies that epimorphisms are surjective. He remarks that the proof works just as well for complete Boolean algebras, and I'd also add that it can be adapted for plain old Boolean algebras. If I understand your meaning of an abstract $$\sigma$$-algebra correctly, this is the result you are after.

Let $$\mathcal{C}$$ be a concrete category so that we can meaningfully talk about epimorphisms being surjective.

$$\mathcal{C}$$ is said to have the strong amalgamation property if for every span $$C \xleftarrow{f} A \xrightarrow{g} B$$ of monomorphisms (aka amalgam), there exists an object $$D \in \mathcal{C}$$, and a commutative diagram of monomorphisms $$\begin{CD} A @> g>> B\\ @VfVV @Vg'VV \\ C @>f'>> D, \end{CD}$$

such that $$g'(B) \cap f'(C) = g'g(A) = f'f(A)$$

Further restrict attention to a variety of algebraic structures, so that being a monomorphism is equivalent to mapping underlying sets injectively and every morphism canonically factors as a surjection followed by a monomorphism. Then the strong amalgamation property immediately implies that epis are surjective (see the corollary in Lagrange's paper).

I think this is a good — or at least thought-provoking — approach to addressing your question in light of your bounty comment that you are "looking for a canonical answer." For varieties, we see that the solution to the strong amalgamation problem always supplies a canonical non-unique extension: given a proper monomorphism $$A \rightarrow B$$, we get the strong amalgamation $$D$$ of $$B \leftarrow A \rightarrow B$$ together with distinct monomorphisms(!) $$B \rightarrow D$$ that agree on $$A$$.

Moreover, the solution to the strong amalgamation problem might be considered canonical in and of itself. Basically, Lagrange's method is a three-part construction: (1) Embed in the best coproduct available, which is in the ambient category of Boolean algebras (2) quotient that coproduct in order to force the desired intersection property of strong amalgamation (which has the awesome effect of restoring morphisms to our actual category) (3) Complete this quotiented coproduct, so that the whole embedding is now in-category. In other words, do the best you can with the coproduct you have... and then error-correct in the only sensible way. I guess that feels canonical.

On this last point, it might be interesting to compare this construction with solutions to the strong amalgamation problem in other varieties, in particular (finite) groups and Lie Algebras over fields.

• "A 1974 paper of Lagrange...": the start of this answer is already very exciting! :) May 22 '20 at 20:22
• @SamHopkins this is just a quite frequent surname, like Newton, Pascal, etc.
– YCor
May 22 '20 at 23:20
• trivial typo, I think: $f'(A)$ should be $f'(C)$, no? May 23 '20 at 7:32
• @YemonChoi Certainly yes, thanks to you for noticing and to YCor for editing. May 23 '20 at 14:36