[Math] Does free functor preserve monomorphism

category-theorycommutative-algebra

The free functor is left adjoint to the forgetful functor so it preserves epimorphism. In the category of modules and algebras, it also preserves monomorphisms (the free functors being free modules and polynomial rings, respectively). Is it generally true that free functors preserve monomorphisms?

Edit: A free functor is a functor from Set to a concrete category C which assigns each set S to an object in C which is free over S. For example, for A-modules, it assigns a set S to a free A-module with basis S. More detailed definition can be found in Wikipedia.

Best Answer

Almost all monomorphisms in $\mathbf{Set}$ are split (hence are preserved by any functor whatsoever), the exceptions being maps with empty domain. So it's just a question of what happens with those maps.

We consider an adjunction $$F \dashv U : \mathcal{C} \to \mathbf{Set}$$ where the "forgetful" functor $U : \mathcal{C} \to \mathbf{Set}$ reflects monomorphisms. (If $U$ is faithful, then $U$ reflects monomorphisms. In particular this holds when $U$ is monadic, e.g. $U : R\mathbf{-Mod} \to \mathbf{Set}$, $U : \mathbf{CRing} \to \mathbf{Set}$ etc.) Now consider $U F(\emptyset \to X)$ for a non-empty set $X$. There are two cases:

  • If $U F \emptyset = \emptyset$, then $U F (\emptyset \to X)$ is an injective map (trivially), so $F (\emptyset \to X)$ is a monomorphism in $\mathcal{C}$.
  • If $U F \emptyset \ne \emptyset$, then there exists a map $X \to U F \emptyset$, hence a morphism $F X \to F \emptyset$ by adjoint transposition. But $F \emptyset$ is the initial object in $\mathcal{C}$, so there is a unique morphism $F \emptyset \to F X$; thus the composite $F \emptyset \to F X \to F \emptyset$ must be the identity, i.e. the morphism $F (\emptyset \to X)$ is split monic.

So $F : \mathbf{Set} \to \mathcal{C}$ indeed preserves all monomorphisms.

Related Question