Free Objects in Functor Categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:31:42Z http://mathoverflow.net/feeds/question/114040 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114040/free-objects-in-functor-categories Free Objects in Functor Categories Brian Hepler 2012-11-21T08:55:57Z 2012-11-21T11:33:03Z <p>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. </p> <p>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 $[C,Ab]$. </p> <p>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 $[C,Ab]$. </p> <hr> <p>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"))</p> http://mathoverflow.net/questions/114040/free-objects-in-functor-categories/114042#114042 Answer by Qiaochu Yuan for Free Objects in Functor Categories Qiaochu Yuan 2012-11-21T09:43:43Z 2012-11-21T09:54:44Z <p>$[C, \text{Ab}]$ 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. </p> http://mathoverflow.net/questions/114040/free-objects-in-functor-categories/114043#114043 Answer by Eric Wofsey for Free Objects in Functor Categories Eric Wofsey 2012-11-21T09:56:14Z 2012-11-21T09:56:14Z <p>I think Qiaochu's answer is really the best way of thinking about it, but here's an alternative approach.</p> <p>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 $H$.</p> <p>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.</p> http://mathoverflow.net/questions/114040/free-objects-in-functor-categories/114049#114049 Answer by Zhen Lin for Free Objects in Functor Categories Zhen Lin 2012-11-21T11:33:03Z 2012-11-21T11:33:03Z <p>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:</p> <ul> <li><p>Each sort is an abelian group under the respective group operations.</p></li> <li><p>The functions between the sorts are group homomorphisms.</p></li> <li><p>The functions between the sorts compose the way we expect them to, so that we get a functor.</p></li> <li><p>The functions between the sorts can be added pointwise the way we expect them to, so that we get an <em>additive</em> functor.</p></li> </ul> <p>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 $\operatorname{ob} \mathcal{C} \hookrightarrow \mathcal{C}$.)</p>