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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.

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.

I would be deeply grateful for any associations here!

My examples are:

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).

In particular, S is always closed with respect to compositions and direct sums of morphisms.

In fact, I am interested in all aspects of this setup!

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Presumably Ore conditions are not satisfied in your example? It would help (me, at least) if this is made explicit. –  Sheikraisinrollbank Oct 29 '10 at 11:45
    
I have updated the question. –  Mikhail Bondarko Oct 29 '10 at 14:35
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4 Answers

up vote 16 down vote accepted

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 $n\geq 0$.

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

$$(S\times S')^{-1}(C\times C')\to S^{-1}C\times {S'}^{-1}C'$$

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.

Another elementary exercise is that, given any adjunction

$$L:C\rightleftarrows D:R$$

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

$$L:S^{-1}C\rightleftarrows T^{-1}D:R$$

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.

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...

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Thank you! My only question is: if this is so easy, why nobody had formulated this before?:) –  Mikhail Bondarko Oct 29 '10 at 17:53
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Well, even if I don't know any precise reference, I cannot believe that no one has formulated this before! Did you try the books/papers of Peter Gabriel for instance? –  Denis-Charles Cisinski Oct 29 '10 at 20:37
    
Not yet; I will try, thank you! –  Mikhail Bondarko Oct 29 '10 at 23:10
    
At least, in "Gabriel, M. Zisman, Calculus of Fractions and Homotopy Theory" the additivity of the localization is proved only when $S$ satisfies the Ore condition. –  Mikhail Bondarko Oct 29 '10 at 23:58
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In a particular context, I have given a criterion in section 6 of "On the 3-arrow calculus for homotopy categories" (available at http://www.math.rwth-aachen.de/~Sebastian.Thomas/publications/). 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.

Could you give more details of your example?

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Thank you; that's interesting! I am not sure that this paper will help, but this is quite possible. Actually, I am interested in a rather general family of examples. An important part of them can be constructed as follows: for an additive (pseudo-abelian) category B consider some full triangulated (thick) subcategory $D$ of $K(B)$; then my S for B is the set of morphisms in B that yield objects of D (if considered as complexes of length 1). In particular, S is always closed with respect to compositions and direct sums of morphisms (so there exists a certain addition of 'zig-zags'). –  Mikhail Bondarko Oct 29 '10 at 14:27
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Here is the statement of Exercise 8.4 p. 202 of

[KS] Categories and Sheaves by Kashiwara and Schapira:

(a) Let $\mathcal C$ be an additive category and $\mathcal S$ a right multiplicative system. Prove that the localization $\mathcal C_{\mathcal S}$ is an additive category and $Q:\mathcal C\to\mathcal C_{\mathcal S}$ is an additive functor.

I'll describe what I think is the argument Kashiwara and Schapira have in mind.

Statement (a) follows from

(b) there is a pre-additive category on $\mathcal C_{\mathcal S}$ making $Q$ additive,

(c) $Q$ is essentially surjective.

Statement (b) is easy, and is treated in a very detailed way at the beginning of the following text of Dragan Miličić:

http://www.math.utah.edu/~milicic/Eprints/dercat.pdf

As (b) is obvious, it only remains to explain why (b) and (c) imply (a). The key is Lemma 8.2.3 (ii) p. 169 of [KS], which says:

(d) Let $\mathcal C$ be a pre-additive category; let $X,X_1,$ and $X_2$ be objects of $\mathcal C$; and, for $a=1,2$, let $X_a\xrightarrow{i_a}X\xrightarrow{p_a}X_a$ be morphisms satisfying $$ p_a\circ i_b=\delta_{ab}\ \operatorname{id}_{X_a},\quad i_1\circ p_1+i_2\circ p_2=\operatorname{id}_X. $$ Then $X$ is a product of $X_1$ and $X_2$ by $p_1,p_2$ and a coproduct of $X_1$ and $X_2$ by $i_1,i_2$.

Proof. For any $Y$ in $\mathcal C$ we have $$ \operatorname{Hom}_\mathcal C(Y,p_a)\circ\operatorname{Hom}_\mathcal C(Y,i_b)=\delta_{ab}\ \operatorname{id}_{\operatorname{Hom}_\mathcal C(Y,X_a)}, $$ $$ \operatorname{Hom}_{\mathcal C}(Y,i_1)\circ\operatorname{Hom}_{\mathcal C}(Y,p_1)+\operatorname{Hom}_{\mathcal C}(Y,i_2)\circ\operatorname{Hom}_{\mathcal C}(Y,p_2)=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(Y,X)}. $$ This implies that $\operatorname{Hom}_{\mathcal C}(Y,X)$ is a product of $\operatorname{Hom}_{\mathcal C}(Y,X_1)$ and $\operatorname{Hom}_{\mathcal C}(Y,X_2)$ by $\operatorname{Hom}_{\mathcal C}(Y,p_1),\operatorname{Hom}_{\mathcal C}(Y,p_2)$, and thus, $Y$ being arbitrary, that $X$ is a product of $X_1$ and $X_2$ by $p_1,p_2$, and we conclude by applying this observation to the opposite category. q.e.d.

(d) implies:

(e) Let $F:\mathcal C\to\mathcal C'$ be an additive functor of pre-additive categories; let $X,X_1,$ and $X_2$ be objects of $\mathcal C$; and, for $a=1,2$, let $X_a\xrightarrow{i_a}X\xrightarrow{p_a}X_a$ be morphisms such that $X$ is a product of $X_1$ and $X_2$ by $p_1,p_2$ and a coproduct of $X_1$ and $X_2$ by $i_1,i_2$. Then $F(X)$ is a product of $F(X_1)$ and $F(X_2)$ by $F(p_1),F(p_2)$ and a coproduct of $F(X_1)$ and $F(X_2)$ by $F(i_1),F(i_2)$.

In view of (e), (b) and (c) imply (a).

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See the Book "Categories" by Horst Schubert (Springer 1972), Prop. 19.5, page 272.

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The Ore conditions is not fulfilled –  Mikhail Bondarko Nov 18 '10 at 18:25
    
You mean: Horst Schubert. –  Todd Trimble Jul 27 '13 at 22:03
    
Todd Trimble: sure, edited. –  Buschi Sergio Jul 28 '13 at 10:44
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