1
$\begingroup$

I am working with modules, but I guess this question is valid with any abelian category.

Let $R$ be a ring, and $F$ a covariant functor. Let $M$ be an $R$-module and $N$ a submodule of $M$. Looking at certain examples (like the $\cdot \oplus A$ functor, or the $\operatorname{Hom}_R(A, \cdot)$ functor, for a fixed $R$-module $A$), I am tempted to say that $FN$ is a submodule of $FM$, or at least that $FN$ embeds into $FM$ in a very natural way: with the homomorphism $F \iota$, where $\iota : N \longrightarrow M$ is the inclusion. This is true if $F$ is left exact, or more generally if it preserves monomorphisms, but it is always true?

If $F$ is instead contravariant, there seems to be no sensible notion of how $FN$ and $FM$ relate to each other in terms of inclusion.

I believe there is no reason for my first question to be true, but I cannot think of a counterexample. Also, more informally, I was wondering why it doesn't work. What I mean is that we define a functor with some "structure-preserving" properties, so that we can use it to study our objects. Then why isn't such a natural relation as inclusion be preserved by a functor?

$\endgroup$
3
  • 3
    $\begingroup$ Tensoring with a fixed module is a functor (from the category of $R$-modules to itself) which doesn't necessarily respect injectivity. In fact, a module for which tensoring with that module respects injectivity is called "flat", and it is a rather important concept. $\endgroup$
    – Arthur
    Jan 5, 2018 at 14:33
  • 1
    $\begingroup$ I know of the counterexample with $M = \mathbb{Z}$, $N = 2 \mathbb{Z}$ and $F$ that is tensoring with $\mathbb{Z} / 2 \mathbb{Z}$. Then the inclusion becomes the zero map $FN \longrightarrow FM$. However I feel like this is a sort of half-counterexample. The inclusion doesn't work, but clearly $FN$ embeds into $FM$. Do you know of any counterexample where there is no room for doubt? Sorry if I'm picky... $\endgroup$
    – frafour
    Jan 5, 2018 at 14:58
  • 3
    $\begingroup$ @user404944 Take the inclusion $\mathbb{Z}\to \mathbb{Q}$ with the same $F$. $\endgroup$
    – Arnaud D.
    Jan 5, 2018 at 15:28

2 Answers 2

2
$\begingroup$

I assume $F$ is a functor from modules to modules. A typical example is $M \otimes_R (-)$ where $M$ is a right $R$-module: this is a functor from left $R$-modules to abelian groups. It preserves monomorphisms / injectivity iff it is exact (for this we need to know that the functor is already additive and right exact) iff $M$ is flat, and there are many examples of modules which are not flat.

Perhaps the simplest one is $\mathbb{Z}/2\mathbb{Z}$, or more generally any abelian group with nontrivial torsion, as a module over $\mathbb{Z}$. The corresponding functor sends an abelian group $A$ to the quotient $A/2A$; the quotient map $A \to A/2A$ is the universal map from $A$ to a $\mathbb{Z}/2\mathbb{Z}$-module, or equivalently an $\mathbb{F}_2$-vector space. And this functor does not preserve the inclusion $\mathbb{Z} \subsetneq \mathbb{Z}[1/2]$, because the latter is sent to zero.

Then why isn't such a natural relation as inclusion be preserved by a functor?

The practical case is simply that there are many useful and interesting functors arising in practice, such as the one described above, which don' t preserve monomorphisms, so we don't want preserving monomorphisms to be part of the definition of a functor. There's more to say here but mostly it's a matter of getting more experience with how category theory works.

$\endgroup$
2
$\begingroup$

Let me first address your last question. It's not correct to think of a functor as an operation that preserves all the structure of modules. Rather, a functor is an operation that preserves certain very specific structure: namely, homomorphisms and their composition. There is no reason to expect a general functor to preserve anything more than that, because that's not what functors are designed to do. You may find it enlightening to consider group homomorphisms as an analogy. The property of being a normal subgroup is extremely important in group theory, but if $f:G\to H$ is a group homomorphism and $K\subseteq G$ is a normal subgroup, then $f(K)$ is not necessarily a normal subgroup of $H$. This isn't a flaw in the definition of group homomorphisms; it's just not what they're designed to do.

So, in particular, if you want a functor to preserve submodules, you first need to relate submodules to the structure that functors do preserve, namely homomorphisms. The natural way to express submodules in terms of homomorphisms is via the inclusion homomorphism, and up to isomorphism, inclusion homomorphisms are the same thing as monomorphisms. So it's not likely to be fruitful to talk about functors which preserve the relation of being a submodule literally; instead, the natural thing to hope for is for functors to preserve monomorphisms. Of course, this is not true of all functors, but it is true of many important functors, and this is the "right" way to talk about functors which "respect inclusion". This is even the right way to talk about "respecting inclusion" for contravariant functors (which you should think of as covariant functors to the opposite category), or more generally functors to more exotic categories where "monomorphisms" are not the same thing as inclusions in any direct sense.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .