8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What is the free functor? I'm just curious. $\endgroup$ – Brian Rushton Jan 18 '14 at 1:12
  • 1
    $\begingroup$ @BrianRushton I think that he means a left adjoint to the forgetful functor $\mathcal{C} \longrightarrow \mathbf{Set}$. $\endgroup$ – user40276 Jan 18 '14 at 2:33
  • 1
    $\begingroup$ A free functor is a functor from $\mathsf{Set}$ to a concrete category $\mathsf{C}$ which assigns each set $S$ to an object in $\mathsf{C}$ 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 (en.wikipedia.org/wiki/Free_object#Free_functor). $\endgroup$ – ashpool Jan 18 '14 at 3:35
  • $\begingroup$ You should include this to the body of your question. $\endgroup$ – Martin Brandenburg Jan 18 '14 at 9:40
10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ When $U$ is conservative and $C$ has pullbacks, then $U$ reflects monomorphisms, right? $\endgroup$ – Martin Brandenburg Jan 18 '14 at 10:56
  • $\begingroup$ Yes. In fact, kernel pairs suffice. But there's an even easier criterion: $U$ reflects monomorphisms if it is faithful. $\endgroup$ – Zhen Lin Jan 18 '14 at 12:15
  • 1
    $\begingroup$ Ah, sure. So the claim is true for all concrete categories (because here one assumes that $U$ is faithful). $\endgroup$ – Martin Brandenburg Jan 18 '14 at 12:20
5
$\begingroup$

Yes, because monics in $\mathcal{Set}$ are split. Edit: This only holds with non-empty sets. That leaves this answer morally correct but not necessarily technically correct.

A monomorphism in set is an inclusion, and thus will have a left inverse (the inverse image of a point will either be a singleton or empty, and we can define the inverse arbitrarily when the inverse image is empty). Explicitly, let $m: X\to Y$ be monic. Then we can produce $r:Y\to X$ such that $r \circ m = \operatorname{id}_X$. Conversely, any map with a left inverse can be canceled on the left, so any such map must be monic. We have shown that in the category $\mathcal{Set}$, every monic is split monic, and that in an arbitrary category, a split monic is monic.

If $F$ is a functor $m$ is a split monic, then $F(r\circ m)=F(r)\circ F(m) = \operatorname{Id}_{F(X)}$, and so split monics are preserved by arbitrary functors (covariant) functors. In particular, free functors preserve monics.

$\endgroup$
  • 4
    $\begingroup$ This isn't quite right; if $Y$ is non-empty, then the unique function $m : \emptyset \rightarrow Y$ is monic, but fails to split. $\endgroup$ – goblin Jan 18 '14 at 6:02
  • $\begingroup$ Hmm...fair point. Although those maps are vacuously monic, as there are no maps $X\to \emptyset$ unless $X$ is empty (so you can't left cancel because you can never have it on the left). This answer at least reduces the problem to whether $F(\emptyset \to \{*\})$ is monic. Or, this works when we can restrict to the category of non-empty sets. $\endgroup$ – Aaron Jan 18 '14 at 6:25
5
$\begingroup$

Let $U : C \to \mathsf{Set}$ be a functor and $F : \mathsf{Set} \to C$ left adjoint to $U$. The question is if $F$ preserves monomorphisms.

If $f : X \to Y$ is a monomorphism in $\mathsf{Set}$ and $X \neq \emptyset$, then $f$ is a split monomorphism (just take preimages on $f(X)$, and on $Y \setminus f(X)$ map everything to a chosen element of $X$), hence also $F(f)$. Now let $X=\emptyset$. If $Y=\emptyset$, then $f$ is an isomorphism, and hence also $F(f)$. Otherwise $f$ factors over $\{\star\}$, and $\{\star\} \to Y$ is mapped to a monomorphism. Hence, the question reduces to the case $Y=\{\star\}$.

Is $F(\emptyset) \to F(\{\star\})$ a monomorphism? Note that $F(\emptyset)$ is the initial object of $C$ and $F(\{\star\})$ is the free $C$-object on one generator. If $U$ is monadic, then this turns out to be true, see the argument on p.89-90 in Linton, Coequalizers in categories of algebras. (Edit: As Zhen Lin points out, the same argument works when $U$ reflects monomorphisms). If $C$ is a category of algebraic structures in the sense of universal algebra, and $U$ is the forgetful functor, then $U$ is monadic. For this Beck's monadicity criterion is quite useful. This answers the question in the affirmative in a lot of cases.

It also holds in a lot of other cases, for example $\mathsf{Top}$ (here $F$ assigns to a set the corresponding discrete space).

I have tried to find counterexamples, but didn't succeed so far.

$\endgroup$
  • $\begingroup$ Unfortunately I don't have access to the Linton's book (or rather I can't afford to pay for it), and Wikipedia's definition ("an adjunction is monadic if it is equivalent to the Eilenberg-Moore category of its associated monad.") is far too involved for me (I'm just a beginner in category theory); can you give me a vague idea of what monadic means? $\endgroup$ – ashpool Jan 20 '14 at 2:05
  • $\begingroup$ Just read Zhen Lin's answer, one doesn't monadic functors here. If you want to learn this notion, see Mac Lane's book. $\endgroup$ – Martin Brandenburg Jan 20 '14 at 10:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.