Do coproducts in categories of algebras preserve monos? - MathOverflow

Do coproducts in categories of algebras preserve monos?

2011-04-02T18:59:22Z

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.

Answer by Todd Trimble for Do coproducts in categories of algebras preserve monos?

2011-04-02T23:01:06Z

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.

Answer by Rob Myers for Do coproducts in categories of algebras preserve monos?

2011-08-27T22:45:15Z

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.