Localizing an arbitrary additive category - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:48:58Z http://mathoverflow.net/feeds/question/44047 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44047/localizing-an-arbitrary-additive-category Localizing an arbitrary additive category Mikhail Bondarko 2010-10-29T02:30:57Z 2010-11-18T18:25:40Z <p>Under which conditions localizing an additive category by some class S of morphisms yields and additive category? It seems easy to define certain addition on morphisms if we fix their representatives as zig-zags (i.e. compositions of 'old' morphisms with inverses of morphisms in S; here I use the fact that 'my' S is closed with respect to direct sums of morphisms), but I am not sure at all that this addition will not depend on the choice on representatives. Is there any reasonable condition that will ensure this? I definitely do not want to restrict myself to abelian or triangulated categories.</p> <p>It seems that in the situations I am interested in, any morphism is a composition of the embedding of a direct summand, an inverse of a morphism from S, and an 'old' morphism (i.e. it is 'almost a fraction'). The Ore conditions are not fulfilled (in general, probably); yet some weakening of them could hold.</p> <p>I would be deeply grateful for any associations here!</p> <p>My examples are:</p> <p>For an additive (pseudo-abelian) category B consider some full triangulated (thick) subcategory D of $K^b(B)$; then my S for B is the set of morphisms in B that yield objects of D (if considered as complexes of length 1). </p> <p>In particular, S is always closed with respect to compositions and direct sums of morphisms. </p> <p>In fact, I am interested in all aspects of this setup!</p> http://mathoverflow.net/questions/44047/localizing-an-arbitrary-additive-category/44104#44104 Answer by Sebastian Thomas for Localizing an arbitrary additive category Sebastian Thomas 2010-10-29T10:58:12Z 2010-10-29T10:58:12Z <p>In a particular context, I have given a criterion in section 6 of "On the 3-arrow calculus for homotopy categories" (available at <a href="http://www.math.rwth-aachen.de/~Sebastian.Thomas/publications/" rel="nofollow">http://www.math.rwth-aachen.de/~Sebastian.Thomas/publications/</a>). In this context, every morphism is represented by a 3-arrow, that is, a formal inverse followed by an morphism in the original category followed by a formal inverse.</p> <p>Could you give more details of your example?</p> http://mathoverflow.net/questions/44047/localizing-an-arbitrary-additive-category/44155#44155 Answer by Denis-Charles Cisinski for Localizing an arbitrary additive category Denis-Charles Cisinski 2010-10-29T16:52:17Z 2010-10-29T16:52:17Z <p>This is a very elementary problem. To solve it, it is better not to try to understand the localization explicitely, but to work only with the universal property of the localization (there is no need for any calculus of zig-zags of any kind). You should also think of finite sums not as something defined on each family of objects, but as a left adjoint to the diagonal functor $C\to C^n$ for each $n\geq 0$.</p> <p>Let us look at a more general situation first. Let $C$ and $C'$ be categories, and $S$ and $S'$ be a class of maps in $C$ and $C'$ respectively, which contains all the identities. Then, it is an easy exercise to check that the canonical functor</p> <p>$$(S\times S')^{-1}(C\times C')\to S^{-1}C\times {S'}^{-1}C'$$</p> <p>is an equivalence of categories (or, if you prefer, an isomorphism, depending on whether you prefer to consider the localized category $S^{-1}C$ as the solution of a universal problem in the $2$-category of categories, or in the $1$-category of categories, respectively). Hint: just check that the two categories have the same universal property.</p> <p>Another elementary exercise is that, given any adjunction</p> <p>$$L:C\rightleftarrows D:R$$</p> <p>if $S$ (resp. $T$) is a class of maps in $C$ (resp. in $D$), such that $L(S)\subset T$ and $R(T)\subset S$, then we get a canonical adjunction</p> <p>$$L:S^{-1}C\rightleftarrows T^{-1}D:R$$</p> <p>It follows rather immediately from this that, if $C$ admits finite sums (resp. finite products), and if the class $S$ is closed under finite sums (resp. finite products), then the localized category $S^{-1}C$ admits finite sums (resp. finite products), and the canonical functor $\gamma: C\to S^{-1}C$ commutes with them.</p> <p>It is now obvious that, if $C$ is an additive category, and if $S$ is a class of maps which contains the identities and which is closed under finite sums (hence also under finite products), then the category $S^{-1}C$ is additive, and the canonical functor $\gamma$ is additive. Indeed, an additive category is nothing but a category with finite sums as well as finite products, such that, the initial and terminal object coincide, such that $X\amalg Y\simeq X\times Y$ for any objects $X$ and $Y$, and such that any object has the structure of an internal group object (which is necessarily unique). As the functor $\gamma$ preserves finite products, it preserves group objects, and, as $\gamma$ is essentially surjective, any object of $S^{-1}C$ has a canonical structure of group object...</p> http://mathoverflow.net/questions/44047/localizing-an-arbitrary-additive-category/44311#44311 Answer by Buschi Sergio for Localizing an arbitrary additive category Buschi Sergio 2010-10-31T08:15:56Z 2010-10-31T08:15:56Z <p>See the Book "CAtegories" by Hors Shubert (Springer 1972) Prop.19.5 pag. 272.</p>