# Free Objects in Functor Categories

As a bit of background, consider the category of all covariant, additive functors from a small Abelian category $C$ to Abelian groups, which I'll denote $[C,Ab]$. First, $[C,Ab]$ is an Abelian category, and an easy argument shows that representable functors are projective objects in this category. Indeed, if $T = \text{Hom}_C(X,-)$ is a representable functor and \begin{equation*} 0 \to F_1 \to F_2 \to F_3 \to 0 \end{equation*} is a short exact sequence in $[C,Ab]$ (where exactness is checked on objects of $C$), then Yoneda's lemma gives $\text{Nat}(T,F_i) \cong F_i(X)$, and the result is immediate. I think that the reverse implication also holds (projective implies representable), but I don't remember the proof being as apparent. We even have the Eilenberg-Watts theorems that give criteria for when additive functors (from $R$-Mod to $Ab$) are representable.

Anyway, this is nice, but I find it a bit lacking compared to the tools we have in, say $R$-Mod for some ring $R$. In that setting, we have such results as "A module $P$ is projective iff it is a direct summand of a free module, etc." Or in $Grp$, we have $G$ is projective iff it is free. My point is that we have a notion of "free object" since these are all nice concrete categories, and such a notion seems to have no nice analogue in $[C,Ab]$.

Of course, we have results like $[C,Ab]$ being concrete over $[C,Set]$ which gives a "locally free Abelian group" type example (where $C^{op}$ here would be the category of open sets of a topological space, and $[C,Ab]$ would be presheaves of Abelian groups). However, these are not free objects in $[C,Ab]$.

My question is then "Can $[C,Ab]$ be reasonably thought of as concrete over another category so that we can construct free object?" As an example, is $[R-Mod,Ab]$ concrete over the functor category $[Set,Set]$? (of course, we'd have to juggle Grothendieck universes for this to make any sense; The standard way being to fix some universe $\mathfrak{U}$, and say $Set$ is the category of $\mathfrak{U}$-sets, and let $\mathfrak{U}'$ be the smallest Grothendieck universe containing $\mathfrak{U}$ as an element, so that $Set$ is now $\mathfrak{U}'$-small (see Schuberts "Categories"))

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$[C, \text{Ab}]$ admits a forgetful functor to $[\text{Ob}(C), \text{Set}]$ (where $\text{Ob}(C)$ denotes the category with the same objects as $C$ but no non-identity morphisms). This is a direct generalization of the module case, which corresponds to taking $C$ to have one object. The corresponding free objects are coproducts of representables.

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This observation is relevant in the context of, for example, the acyclic models theorem. – Qiaochu Yuan Nov 21 '12 at 9:54
This relates to a great way of thinking of objects of $[C,Ab]$, which is as "modules" over $C$. That is, if you think of a ring as an additive category with one object, a more general additive category is like a "ring with multiple objects" in which multiplications are only defined when domains and codomains coincide, and then a functor from $C$ to $Ab$ is like a "module" over $C$. The natural notion of "generators" for a module over such a generalized ring should then specify which object each one of the generators lives it. That is, it should exactly be a functor from $Ob(C)$ to Set. – Eric Wofsey Nov 21 '12 at 9:55
Is this forgetful functor monadic? – S. Carnahan Nov 21 '12 at 9:58
@Scott: yes it is. – Todd Trimble Nov 21 '12 at 12:27

I think Qiaochu's answer is really the best way of thinking about it, but here's an alternative approach.

Let $H$ be the direct sum of all representable functors in $[C,Ab]$. Then $H$ is a projective generator for the abelian category $[C,Ab]$, and so $[C,Ab]$ is equivalent to a full exact subcategory of the category of modules over the ring $R=End(H)$ (and if $C$ has only finitely many objects, $[C,Ab]$ will actually be equivalent to $R$-modules). You can think of $R$ as the ring of endomorphisms of the direct sum of all objects of $C$, if such a direct sum were to actually exist. This makes $[C,Ab]$ concrete over $Set$ itself. You can thus say that an element of $[C,Ab]$ is free if it's isomorphic to a direct sum of copies of $H$. As we would like, every representable functor is a direct summand of the free module $H$.

This might be a bit unsatisfying since this $H$ is rather noncanonical; if we replace $C$ by an equivalent category that has a different number of isomorphic copies of each object, then $H$ will get bigger. However, we shouldn't really expect to be able to do better than this because the notion of "free module" is not Morita-invariant--even when we restrict to modules over a ring, you can have equivalent abelian categories with different collections of free objects.

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To amplify Qiaochu's answer (and answer S. Carnahan's question), I'd like to add that the forgetful functor $[\mathcal{C}, \textbf{Ab}]_{\textbf{Ab}} \to [\operatorname{ob} \mathcal{C}, \textbf{Set}]$ is (finitary) monadic. The easiest way to see this is to treat functors $\mathcal{C} \to \textbf{Ab}$ as a many-sorted finitary algebraic theory, with one copy of the group operation symbols for each object of $\mathcal{C}$ and one function symbol for each morphism of $\mathcal{C}$. Of course, the axioms of this algebraic theory simply express the following:

• Each sort is an abelian group under the respective group operations.

• The functions between the sorts are group homomorphisms.

• The functions between the sorts compose the way we expect them to, so that we get a functor.

• The functions between the sorts can be added pointwise the way we expect them to, so that we get an additive functor.

This presentation essentially corresponds to factorising $[\mathcal{C}, \textbf{Ab}] \to [\operatorname{ob} \mathcal{C}, \textbf{Set}]$ as the composite $[\mathcal{C}, \textbf{Ab}]_{\textbf{Ab}} \to [\mathcal{C}, \textbf{Ab}] \to [\operatorname{ob} \mathcal{C}, \textbf{Ab}] \to [\operatorname{ob} \mathcal{C}, \textbf{Set}]$. The free models of this theory can then be constructed by applying the left adjoints of each of these functors in turn, because left adjoints are unique up to unique isomorphism. (Note: the left adjoint of $[\mathcal{C}, \textbf{Ab}] \to [\operatorname{ob} \mathcal{C}, \textbf{Ab}]$ is left Kan extension along the inclusion $\operatorname{ob} \mathcal{C} \hookrightarrow \mathcal{C}$.)

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