I am a little confused with the definition of an additive functor. In my notes, it says that if $R,S$ are two rings and $F: \text{Mod}_R \to \text{Mod}_S$ is a covariant functor, then we call $F$ additive if the induced map on Hom-sets $$\text{Hom}_R(M,N) \overset{F} \to \text{Hom}_S(F(M),F(N))$$ is a homomorphism of abelian groups.

First of all, under which operation is this? Is it under composition of function? If so, then isnt this precisly the definition of a covariant functor? Namely we have a map from the two Hom-sets that preserves composition?

Also, if $F$ is a covariant functor, how can I see that $F$ maps chain complexes to chain complexes? Do I have to require $F$ to be additive here?

Please help me to sort out my confusion.

  • 2
    $\begingroup$ Categories of modules are so-called "preadditive" categories: each homset $\text{Hom}(M, N)$ in fact has the structure of an abelian group (under pointwise addition) in such a way that composition is bilinear. $\endgroup$ – Qiaochu Yuan Oct 14 '15 at 20:24

A covariant functor $F\colon \mathrm{Mod}_R\to\mathrm{Mod}_S$ from the category of $R$-modules to the category of $S$-modules is just a map on objects and morphisms satisfying $F(f\circ g)=F(f)\circ F(g)$ for every pair of composable morphisms $f,g\in\mathrm{Mor}(\mathrm{Mod}_R)$ and $F(\operatorname{id}_X)=\operatorname{id}_{F(X)}$ for each object $X\in\mathrm{Mod_R}$.

This definition doesn't even care about $\operatorname{Hom}_R(M,N)$ being an abelian group. Covariant functors are defined between arbitrary categories and you often don't have a group structure on $\operatorname{Hom}_{\mathcal C}(X,Y)$ for objects $X$, $Y$ of a category $\mathcal C$.

For module homomorphisms however, you do have the structure of an abelian group on $\operatorname{Hom}_R(M,N)$. you add two such homomorphisms $f,g\colon M\to N$ pointwise (i.e. $(f+g)(x)= f(x)+g(x)$)) and the identity element is the zero map $z\colon M\to N$ defined by $z(x)=0$ for all $x\in M$.

Now since $F$ takes a homomorphism $M\to N$ to a homomorphism $F(M)\to F(N)$ and both $\operatorname{Hom}_R(M,N)$ and $\operatorname{Hom}_S(F(M),F(N))$ are abelian groups, you can ask wether all those maps on $\operatorname{Hom}$-sets are group homomorphisms. To be precise, you are asking if $$ F(f+g) = F(f) + F(g) $$ holds for all $M,N\in\operatorname{Mod}_R$ and $f,g\colon M\to N$. If this is the case, $F$ is an additive functor.

Here is an example of a functor fitting all your criteria except being additive: Let $R=S=\mathbb Z$, so we are looking at an endofunctor on the category $\mathrm{Ab}$ of abelian groups. Define $F\colon \mathrm{Ab}\to\mathrm{Ab}$ by letting $F(G) = \mathbb Z$ for every abelian group $G$ and $F(f)=\operatorname{id}_\mathbb Z$ for any homomorphism $f$ of abelian groups. Check that this is a covariant endofunctor on $\mathrm{Ab}$.

Can you see what $\operatorname{Hom}(G,H)\xrightarrow{F}\operatorname{Hom}(\mathbb Z,\mathbb Z)$ does as a map between abelian groups and why it isn't a homomorphism?

  • $\begingroup$ Perfect answer. Thanks. $\endgroup$ – Beru Oct 14 '15 at 21:03

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