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Triangulated Categories. (AM-148) available in Paperback

- ISBN-10:
- 0691086869
- ISBN-13:
- 9780691086866
- Pub. Date:
- 01/03/2001
- Publisher:
- Princeton University Press

## Paperback

## Overview

The first two chapters of this book offer a modern, self-contained exposition of the elementary theory of triangulated categories and their quotients. The simple, elegant presentation of these known results makes these chapters eminently suitable as a text for graduate students. The remainder of the book is devoted to new research, providing, among other material, some remarkable improvements on Brown's classical representability theorem. In addition, the author introduces a class of triangulated categories"--the "well generated triangulated categories"--and studies their properties. This exercise is particularly worthwhile in that many examples of triangulated categories are well generated, and the book proves several powerful theorems for this broad class. These chapters will interest researchers in the fields of algebra, algebraic geometry, homotopy theory, and mathematical physics.

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## Product Details

ISBN-13: | 9780691086866 |
---|---|

Publisher: | Princeton University Press |

Publication date: | 01/03/2001 |

Series: | Annals of Mathematics Studies Series |

Pages: | 449 |

Product dimensions: | 6.00(w) x 9.25(h) x 1.10(d) |

## About the Author

**Amnon Neeman** holds a Ph.D. in algebraic geometry from Harvard University. He has taught at Princeton University and the University of Virginia and is currently Senior Visiting Fellow at the Australian National University in Canberra. He has published widely on derived and triangulated categories.

## Read an Excerpt

#### Triangulated Categories

**By Amnon Neeman**

**PRINCETON UNIVERSITY PRESS**

**Copyright © 2001 Princeton University Press**

All rights reserved.

ISBN: 978-0-691-08686-6

All rights reserved.

ISBN: 978-0-691-08686-6

CHAPTER 1

**Definition and elementary properties of triangulated categories**

**1.1. Pre–triangulated categories**

Definition 1.1.1. *Let C be an additive category and [summation] : C -> C be an additive endofunctor of C. Assume throughout that the endofunctor [summation] is invertible. A* candidate triangle *in C (with respect to [summation]) is a diagram of the form:*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*such that the composites v o u, w o v and [summation]u o w are the zero morphisms.*

A morphism of candidate triangles is a commutative diagram

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where each row is a candidate triangle.

Definition 1.1.2. A pre–triangulated category *T* is an additive category, together with an additive automorphism [summation], and a class of candidate triangles (with respect to [summation]) called *distinguished* triangles. The following conditions must hold:

**TR0:** Any candidate triangle which is isomorphic to a distinguished triangle is a distinguished triangle. The candidate triangle

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

is distinguished.

**TR1:** For any morphism *f: X -> Y* in *T* there exists a distinguished triangle of the form

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

**TR2:** Consider the two candidate triangles

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

If one is a distinguished triangle, then so is the other.

**TR3:** For any commutative diagram of the form

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where the rows are distinguished triangles, there is a morphism *h : Z -> Z',* not necessarily unique, which makes the diagram

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

commutative.

Remark 1.1.3. Parts of Definition 1.1.2 are known to be redundant. For instance, it is not necessary to assume that distinguished triangles are candidate triangles. In other words, we can assume that the distinguished triangles are sequences

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

without necessarily postulating that the composites *v o u, w o v* and [summation]*u o w* vanish. It follows from the other axioms that the composites *v o u, w o v* and [summation]u o w must be zero. Just consider the diagram

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

The bottom row is a distinguished triangle by hypothesis, the top by [TR0]. But by [TR3] the diagram may be completed to a commutative

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and we deduce that *v o u* = 0. The vanishing of *w o v* and [summation]*u o w* follows from the above and axiom [TR2].

Similarly, it is not necessary to assume that the category *T* is additive; something slightly less suffices. It suffices to assume that the category *T* is pointed (there is a zero object), and that the *Hom* sets are abelian groups. The fact that finite coproducts and products exist and agree follows from the other axioms.

Notation 1.1.4. Let *T* be a pre–triangulated category. If we speak of "triangles" in *T,* we mean distinguished triangles. When we mean candidate triangles, the adjective will always be explicitly used.

Remark 1.1.5. If *T* is a pre–triangulated category, then clearly so is its dual *Top.* For *Top*, the functor [summation] gets replaced by [summation]-1.

Proposition 1.1.6. *Let T be a pre–triangulated category. Then the functor [summation] preserves products and coproducts. Let us state this precisely. Suppose {Xλ, λ [member of] Λ} is a set of objects of T, and suppose the categorical coproduct Πλ[member of]Λ Xλ exists in T. Then the natural map*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*is an isomorphism. In other words, the natural maps*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*give [summation] {Πλ[member of]Λ Xλ} the structure of a coproduct in the category T. Similarly, if Πλ[member of]Λ Xλ exists in 'J, then the natural maps*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*give [summation]{Πλ[member of]Λ Xλ} the structure of a product in the category T.*

**Proof:** The point is that [summation], being invertible, has both a right and a left adjoint, namely [summation]-1. There are natural isomorphisms

Hom ([summation]X, Y) [equivalent] Hom (X, [summation]-1Y)

and

Hom (X, [summation]Y) [equivalent] Hom ([su,,ation]-1 X, Y),

induced by [summation]-1. A functor possessing a left adjoint respects products, a functor possessing a right adjoint respects coproducts. Thus [summation] respects both.

Because *Top* is a pre–triangulated category with [summation]-1 playing the role of [summation], it follows that [summation]-1 also respects products and coproducts.

Definition 1.1.7. *Let T be a pre–triangulated category. Let H be a functor from T to some abelian category A. The functor H is called* homological *if, for every (distinguished} triangle*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*the sequence*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*is exact in the abelian category A.*

Remark 1.1.8. Because of axiom [TR2], it follows that the sequence above can be continued indefinitely in both directions. In other words, the infinite sequence

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

is exact everywhere.

Remark 1.1.9. A homological functor on the pre–triangulated category *Top* is called a *cohomological* functor on *T*. Thus, a cohomological functor is a contravariant functor *H : T -> A* such that, for any triangle

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

the sequence

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

is exact in the abelian category *A.*

Lemma 1.1.10. *Let T be a pre–triangulated category, U be an object of T. Then the representable functor Hom(U, –) is homological.*

**Proof:** Suppose we are given a triangle

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

We need to show the exactness of the sequence

*Hom(U, X) -> Hom(U, Y) -> Hom(U, Z)*

We know in any case that the composite is zero. Let *f [member of] Hom(U, Y)* map to zero in *Hom(U, Z).* That is, let *f: U -> Y* be such that the composite

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

is zero. Then we have a commutative diagram

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

The bottom row is a triangle by [TR2], the top row by [TRO] and [TR2]. There is therefore, by [TR3], a map *h : U -> X* such that [summation]*h* : [summation]*U* -> [summation]*X* makes the diagram above commute. But this means in particular that the square

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

commutes, and hence *f = uoh.* Thus, we have produced an *h* [member of] *Hom(U, X)* mapping to *f* [member of] *Hom(U, Y).*

Remark 1.1.11. Recall that the dual of a pre–triangulated category is pre–triangulated. It follows from Lemma 1.1.10, applied to the dual of *T*, that the functor *Hom(-, U)* is cohomological.

Definition 1.1.12. *Let H : T -> A be a homological functor. The functor H is called decent if*

1.1.12.1. *The abelian category A satisfies AB*4*; *that is products exist, and the product of exact sequences is exact.*

1.1.12.2. *The functor H respects products. For any collection {Xλ, λ [member of] Λ} of objects Xλ [member of] T whose product exists in T, the natural map*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*is an isomorphism.*

Example 1.1.13. The functor *Hom(U, -) : T -> Ab* is a decent homological functor. It is homological by Lemma 1.1.10, the abelian category *Ab* of all abelian groups satisfies *AB*4*, and *Hom(U, -)* preserves products.

Definition 1.1.14. *Let T be a pre–triangulated category. A candidate triangle*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*is called a pre–triangle if, for every decent homological functor H : T -> A, the long sequence*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*is exact*.

Example 1.1.15. Every (distinguished) triangle is a pre–triangle. A direct summand of a pre–triangle is a pre–triangle. The next little lemma will show that an arbitrary product of pre–triangles is a pre–triangle.

Caution 1.1.16. There are pre–triangles which are not distinguished. See for example the discussion of Case 2, pages 232-234 of [**22**]. An example of a pre–triangle which is not a triangle is the mapping cone on the map of triangles in the middle of page 234, loc. cit.

Lemma 1.1.17. *Let Λ be an index set, and suppose that for every λ [member of] Λ we are given a pre–triangle*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

*Suppose further that the three products*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

*exist in T. The sequence*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

*is identified, using Proposition 1.1. 6, with*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

*We assert that this candidate triangle is a pre–triangle. Thus, the product of pre–triangles is a pre–triangle.*

**Proof:** Let *H : T -> A* be a decent homological functor. Because for each λ [member of] Λ the sequence.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

is a pre–triangle, applying *H* we get a long exact sequence in *A*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and because *A* is assumed to satisfy *AB*4*, the product of these sequences is exact. But we are assuming *H* decent, and in particular by 1.1.12.2, the maps

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

are all isomorphisms. This means that the functor *H,* applied to the sequence

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

gives a long exact sequence. This being true for all decent *H,* we deduce that the sequence is a pre–triangle.

Lemma 1.1.18. *Let H be a decent homological functor T -> A. Let the diagram*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*be a morphism of pre–triangles. Suppose that for every n [member of] Z, H([summation]n f) and H([summation]ng) are isomorphisms. Then H([summation]nh) are all isomorphisms.*

**Proof:** Without loss, we are reduced to proving *H(h)* an isomorphism. But then the diagram

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

is a commutative diagram in the abelian category *A* with exact rows. By the 5-lemma, we deduce that *H(h)* is an isomorphism.

Lemma 1.1.19. *In the morphism of pre–triangles*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*if f and g are isomorphisms, then for any decent homological functor H, H(h) is an isomorphism.*

**Proof:** If *f* and *g* are isomorphisms, so are [summation]*n f* and [summation]*n g* for any *n.* Hence Lemma 1.1.18 allows us to deduce that *H(h)* is an isomorphism.

Proposition 1.1.20. *If in the morphism of pre–triangles*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*both f and g are isomorphisms, then so is h.*

**Proof:** By Lemma 1.1.19, we already know that for any decent homological functor *H : T -> A, H(h)* is an isomorphism. By Example 1.1.13, all representable functors *Hom(U, -)* are decent, for *U* [member of] *T*. We know therefore that the natural map

*Hom(U, h) : Hom(U, Z) -> Hom(U, Z')*

is an isomorphism for every *U*. But then the map

*Hom(-, h): Hom(-,Z) -> Hom(-, Z')*

is an isomorphism. It follows from Yoneda's Lemma that *h* is an isomorphism.

Remark 1.1.21. Let *u : X -> Y* be given. By [TR1] it may be completed to a triangle. Let

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

be two distinguished triangles "completing" *u.* We have a diagram

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

which by [TR3] may be completed to a morphism of triangles

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and since 1 : *X -> X* and 1 : *Y -> Y* are clearly isomorphisms, Proposition 1.1.20 says that his an isomorphism. It follows that *Z* is well defined up to isomorphism. In fact, the entire triangle is well defined up to isomorphism. But this isomorphism is not in general canonical.

**1.2. Corollaries of Proposition 1.1.20**

In this section, we will group together some corollaries of Proposition 1.1.20, which have in common that they concern products and coproducts.

Proposition 1.2.1. *Let T be a pre–triangulated category and Λ any index set. Suppose for every λ [member of] Λ we are given a (distinguished} triangle*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*in t. Suppose the three products*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*exist in T. We know by Lemma 1.1.17 that the product is a pre–triangle*

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

*We assert that it is a distinguished triangle.*

**Proof:** By [TR1], the map

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

can be completed to a triangle

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

For each λ [member of] Λ, we get a diagram where the rows are triangles

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

By [TR3] we may complete this to a morphism of triangles

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Taking the product of all these maps, we get a morphism

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Both rows are pre–triangles. The top row because it is a triangle, the bottom row by Lemma 1.1.17. It follows from Proposition 1.1.20 that this map is an isomorphism of the top row (a distinguished triangle) with the bottom, which is therefore a triangle.

Remark 1.2.2. Dually, the coproduct of distinguished triangles is distinguished.

*(Continues...)*

Excerpted fromTriangulated CategoriesbyAmnon Neeman. Copyright © 2001 Princeton University Press. Excerpted by permission of PRINCETON UNIVERSITY PRESS.

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## Table of Contents

- , pg. i
- Contents, pg. v
- 0. Acknowledgements, pg. 3
- 1. Introduction, pg. 3
- Chapter 1. Definition and elementary properties of triangulated categories, pg. 29
- APPENDIX E: Examples of non-perfectly-generated categories, pg. 427
- Bibliography, pg. 443
- Index, pg. 445