Classifying functors of abelian categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:34:40Z http://mathoverflow.net/feeds/question/38793 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38793/classifying-functors-of-abelian-categories Classifying functors of abelian categories Martin Brandenburg 2010-09-15T08:21:45Z 2010-09-16T13:46:06Z <p>Let $AbCat$ denote the $2$-category of abelian categories with additive functors. Is the forgetful functor $AbCat \to Cat$ representable; i.e. is there an abelian category $T$ such that for every abelian category $A$, the category $Hom(T,A)$ is naturally isomorphic to (the category underlying) $A$?</p> <p>This would be nice, because then it is possible to reconstruct an abelian category which has only a universal property.</p> <p>I try to work out the structure of $T$: The isomorphism $Hom(T,T) \cong T$ maps $1_T$ to some object $x \in T$. If $A$ is arbitrary, then the isomorphism $Hom(T,A) \cong A$ is given by mapping a functor $F : T \to A$ to $F(x)$ and mapping a natural transformation $\eta : F \Rightarrow G$ to the morphism $\eta(x) : F(x) \to G(x)$.</p> <p>Let $T'$ be the smallest full abelian subcategory of $T$, which contains $x$ (Existence: Construct inductively subcategories $T_{n+3k}$, which contain direct sums $(n=0)$, kernels $(n=1)$ and cokernels $(n=2)$ from $T_{n+3k-1}$.). Then $T'$ has the same universal property as $T$. Thus $T'=T$, and we see that $T$ is generated by $x$. Now we have to find such a $T$, which has no additional relations. In particular, we have to ensure that a natural transformation between additive functors on $T$ is determined by the morphism at $x$, although the functors don't have to be exact.</p> http://mathoverflow.net/questions/38793/classifying-functors-of-abelian-categories/38963#38963 Answer by Laurent Moret-Bailly for Classifying functors of abelian categories Laurent Moret-Bailly 2010-09-16T12:17:25Z 2010-09-16T13:46:06Z <p>If we require an isomorphism (as opposed to an equivalence) the answer is no, for reasons having little to do with the (interesting aspect of) the question.</p> <p>Assume there exists an abelian category $T$ and an object $x$ of $T$ such that for every abelian category $A$ and object $a$ there is a unique functor $F_a:T\to A$ sending $x$ to $a$. Clearly, $x$ is nonzero, so the object $x^2$ has a nontrivial automorphism $\sigma$. Consider the functor $\Phi:T\to T$ defined as follows:</p> <p>$\Phi$=identity on objects,</p> <p>for a map $f:u\to v$, put:</p> <p>$\Phi(f)=f$ if $u\neq x^2\neq v$ or $u=x^2=v$,</p> <p>$\Phi(f)=\sigma f$ if $u\neq x^2$ and $v=x^2$,</p> <p>$\Phi(f)=f\sigma^{-1}$ if $u=x^2$ and $v\neq x^2$.</p> <p>Now $\Phi$ is indeed a functor (case-by-case inspection) sending $x$ to $x$. So it should be equal to the identity functor but isn't.</p>