# Do coproducts in categories of algebras preserve monos?

Even if the answer is no, I am interested in a more specific question.

Let $\Sigma$ be a set of operations of finite arity, $E$ be a set of equations over $\Sigma$ and $\mathcal{A}(\Sigma,E)$ be the respective category of algebras and algebra morphisms. Also denote the free algebra functor by $F: \mathsf{Set} \to \mathcal{A}(\Sigma,E)$.

If $f : A \to B$ is a monomorphism in such a category i.e. an injective algebra morphism, and also $X$ is set, does it follow that $FX + f : FX + A \to FX + B$ is also injective?

Any help much appreciated.

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The answer to the title question is 'no'; am I right that your question in the penultimate sentence is what you meant by the more specific one?

A counterexample to the title question might be: take the category of commutative rings where the coproduct is tensor product, let $C = \mathbb{Z}/2$, and let $i: A \to B$ be the inclusion of $\mathbb{Z}$ in $\mathbb{Q}$. Then $C + B$ is the terminal ring and $C + i: C + A \to C + B$ is not injective.

My guess is that the answer to the more specific question is 'yes', but I don't have a proof. Hopefully this can be settled soon.

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Thanks a lot, that answers the title question. In fact it is the more specific question that I have been trying to prove, but have only succeeded in checking some specific cases e.g. distributive lattices, boolean algebras, modules over rings. –  Rob Myers Apr 2 '11 at 23:51
I was trying to subject my somewhat offhand guess to some more acidic tests today. Still no conclusion, but I was led to the category of relation algebras where a pushout of a pair of monos need not be a pair of monos (that is, the coprojections to the pushout need not be monic). Could you describe in your question what your motivations might be? Are they connected with mathematical logic? –  Todd Trimble Apr 4 '11 at 1:53
The motivation comes from coalgebras for a finitary endofunctor T on a variety. T-coalgebras are algebra morphisms A --> TA, where elements of A have a 'behaviour' via a unique map to the final coalgebra. Coalgebras dualise algebras for a functor. It is somehow natural to consider FX + T-coalgebras for a fixed set X, since A --> FX + TA may be viewed as a T-coalgebra plus labels'. Such labels can be used to define operations over coalgebras, like a label one jumps to via a goto'. Coalgebra often assumes the functor preserves monos, so if T preserves monos, FX + T should also preserve them. –  Rob Myers Apr 4 '11 at 12:05

Well, I have a counterexample.

Let $\Sigma$ contain two unary operations $a$ and $b$.

Further suppose $E$ contains the equations $a(p) = a(q)$ and $b(p) = b(q)$.

Then $F0 = \emptyset$ because $\Sigma$ contains no nullary operations and we also have the terminal algebra $\mathsf{1}$ with singleton carrier $\{*\}$. Furthermore the free algebra on omega generators $F\omega$ has carrier $\omega + 1 + 1$, the omega corresponding to the generators, the remaining two elements being the equivalence class containing $a$-prefixed terms and the equivalence class containing $b$-prefixed terms.

Now we certainly have the injective algebra morphism $\iota : \emptyset \hookrightarrow \mathsf{1}$.

However $F\omega + \emptyset \cong F\omega$ so it has carrier $\omega + 1 + 1$, whereas $F\omega + \mathsf{1}$ has carrier $\omega + 1$. This follows because in $F\omega + \mathsf{1}$ we can deduce: $a(x) = a(*) = * = b(*) = b(x)$ for any $x \in U(F\omega + \mathsf{1})$.

Since $F\omega + \iota$ is surjective but not injective, the respective functor $F\omega + Id$ does not preserves monos.

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That's a very interesting example. The algebras are sets that are either empty or have two named points $a$ and $b$, which may be equal; and (provided $A\neq\emptyset$) $A+1$ is the quotient of $A$ in which we force $a=b$, whereas $A+\emptyset$ is just $A$. That's bizarrely different from any variety I have seen before! –  Neil Strickland Aug 28 '11 at 9:59