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A Boolean algebra is a ring with unity all of whose elements are idempotent. We regard a zero ring $0$ as a Boolean algebra. Let $\mathcal{B}$ be the category of Boolean algebras. A morphism in $\mathcal{B}$ is a homomorphism of rings preserving unity. Is a monomorphism in $\mathcal{B}$ always injective? Is an epimorphism in $\mathcal{B}$ always surjective?

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Monomorphisms are always injective, because the forgetful functor $\mathcal{B} \to \mathbf{Set}$ has a left adjoint. – Zhen Lin Aug 11 '13 at 23:33

In the category of Boolean algebras, epimorphisms are surjective. See B. Banaschewski, "On the strong amalgamation of Boolean algebras," Algebra universalis 63, pp. 235-238 (2010)

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In any algebraic category monomorphisms are injective, because the forgetful functor to the category of sets has a left adjoint (the free algebra on a given set) and therefore preserves monomorphisms. In the case of boolean rings, the free boolean ring on one generator is $\mathbb{F}_2[x]/(x^2-x) \cong \mathbb{F}_2 \times \mathbb{F}_2$, and hence the free boolean ring on a set $B$ is $\bigotimes_{b \in B} \mathbb{F}_2[x]/(x^2-x)$, where this is the (possibly infinite) tensor product as $\mathbb{F}_2$-algebras, which is here also the coproduct in the category of boolean rings.

I don't know right now if epimorphisms are surjective, but using Stone duality this is equivalent to a pure topological question: If $A$ is a closed subspace of a totally disconnected compact Hausdorff space $X$, does every open-closed subset of $A$ lift to a open-closed subset of $X$? If I remember correctly, there are (exotic) counterexamples.

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