triangulated category in nLab

Context

Homological algebra

Stable homotopy theory

• Idea
• History
• Definition
• Properties
• Long fiber-cofiber sequences
• From stable model categories and stable ∞\infty-categories
• Examples
• Related concepts
• References

Definition

A triangulated category is

• an additive category;

• with (additive) translation;

• equipped with a collection of triangles called distinguished triangles (dts)

• such that the following axioms hold

TR0: every triangle isomorphic to a distinguished triangle is itself a distinguished triangle;

TR1: the triangle

X→Id XX→0→TX X \stackrel{Id_X}{\to} X \to 0 \to T X

is a distinguished triangle;

TR2: for all f:X→Yf : X \to Y, there exists a distinguished triangle

X→fY→Z→TX; X \stackrel{f}{\to} Y \to Z \to T X \,;

TR3: a triangle

X→fY→gZ→hTX X \stackrel{f}{\to} Y \stackrel{g}{\to} Z \stackrel{h}{\to} T X

is a distinguished triangle precisely if

Y→gZ→hTX→−T(f)TY Y \stackrel{g}{\to} Z \stackrel{h}{\to} T X \stackrel{-T(f)}{\to} T Y

is a distinguished triangle;

TR4: given two distinguished triangles

X→fY→gZ→hTX X \stackrel{f}{\to} Y \stackrel{g}{\to} Z \stackrel{h}{\to} T X

and

X′→f′Y′→g′Z′→h′TX′X' \stackrel{f'}{\to} Y' \stackrel{g'}{\to} Z' \stackrel{h'}{\to} T X'

and given morphisms α\alpha and β\beta in

X →f Y ↓ α ↓ β X′ →f′ Y′\array{ X &\stackrel{f}{\to}& Y \\ \downarrow^\alpha && \downarrow^\beta \\ X' &\stackrel{f'}{\to}& Y' }

there exists a morphism γ:Z→Z′\gamma : Z \to Z' extending this to a morphism of distinguished triangles in that the diagram

X →f Y →g Z →h TX ↓ α ↓ β ↓ ∃γ ↓ T(α) X′ →f′ Y′ →g′ Z′ →h′ TX′ \array{ X &\stackrel{f}{\to}& Y &\stackrel{g}{\to}& Z &\stackrel{h}{\to}& T X \\ \downarrow^\alpha && \downarrow^{\mathrlap{\beta}} && \downarrow^{\mathrlap{\exists \gamma}} && \downarrow^{\mathrlap{T(\alpha)}} \\ X' &\stackrel{f'}{\to}& Y' &\stackrel{g'}{\to}& Z' &\stackrel{h'}{\to}& T X' }

commutes;

TR5 (octahedral axiom): given three distinguished triangles of the form

X→fY→hY/X→TX Y→gZ→kZ/Y→TY X→g∘fZ→lZ/X→TX \begin{aligned} & X \stackrel{f}{\to} Y \stackrel{h}{\to} Y/X \stackrel{}{\to} T X \\ & Y \stackrel{g}{\to} Z \stackrel{k}{\to} Z/Y \stackrel{}{\to} T Y \\ & X \stackrel{g \circ f}{\to} Z \stackrel{l}{\to} Z/X \stackrel{}{\to} T X \end{aligned}

there exists a distinguished triangle

Y/X→uZ/X→vZ/Y→wT(Y/X) Y/X \stackrel{u}{\to} Z/X \stackrel{v}{\to} Z/Y \stackrel{w}{\to} T (Y/X)

such that the following big diagram commutes

X →g∘f Z →k Z/Y →w T(Y/X) f↘ ↗ g ↘ l ↗ v ↘ ↗ T(h) Y Z/X TY ↘ h ↗ u ↘ ↗ T(f) Y/X → TX \array{ X &&\stackrel{g \circ f}{\to}&& Z &&\stackrel{k}{\to}&& Z/Y &&\stackrel{w}{\to}&& T (Y/X) \\ & {}_{f}\searrow && \nearrow_{g} && \searrow^{l} && \nearrow_{v} && \searrow^{} && \nearrow_{T(h)} \\ && Y &&&& Z/X &&&& T Y \\ &&& \searrow^{h} && \nearrow_{u} && \searrow^{} && \nearrow_{T(f)} \\ &&&& Y/X &&\stackrel{}{\to}&& T X }

If TR5 is not required, one speaks of a pretriangulated category.

Lemma

Given a triangulated category, def. , and

A⟶fB⟶gB/A⟶hΣA A \overset{f}{\longrightarrow} B \overset{g}{\longrightarrow} B/A \overset{h}{\longrightarrow} \Sigma A

a distinguished triangle, then

g∘f=0 g\circ f = 0

is the zero morphism.

Proof

Consider the commuting diagram

A ⟶id A ⟶ 0 ⟶ ΣA ↓ id ↓ f A ⟶f B ⟶g B/A ⟶h ΣA. \array{ A &\overset{id}{\longrightarrow}& A &\overset{}{\longrightarrow}& 0 &\overset{}{\longrightarrow}& \Sigma A \\ \downarrow^{\mathrlap{id}} && \downarrow^{\mathrlap{f}} \\ A &\overset{f}{\longrightarrow}& B &\overset{g}{\longrightarrow}& B/A &\overset{h}{\longrightarrow}& \Sigma A } \,.

Observe that the top part is a distinguished triangle by axioms T1 and T2 in def. . Hence by T3 there is an extension to a commuting diagram of the form

A ⟶id A ⟶ 0 ⟶ ΣA ↓ id ↓ f ↓ ↓ Σid A ⟶f B ⟶g B/A ⟶h ΣA. \array{ A &\overset{id}{\longrightarrow}& A &\overset{}{\longrightarrow}& 0 &\overset{}{\longrightarrow}& \Sigma A \\ \downarrow^{\mathrlap{id}} && \downarrow^{\mathrlap{f}} && \downarrow && \downarrow^{\mathrlap{\Sigma \mathrlap{id}}} \\ A &\overset{f}{\longrightarrow}& B &\overset{g}{\longrightarrow}& B/A &\overset{h}{\longrightarrow}& \Sigma A } \,.

Now the commutativity of the middle square proves the claim.

Proposition

Consider a triangulated category, def. , with shift functor denoted Σ\Sigma and with hom-functor denoted [−,−] *:Ho op×Ho→Ab[-,-]_\ast \colon Ho^{op}\times Ho \to Ab. Then for XX any object, and for any distinguished triangle

A⟶fB⟶gB/A⟶hΣA A \overset{f}{\longrightarrow} B \overset{g}{\longrightarrow} B/A \overset{h}{\longrightarrow} \Sigma A

the sequences of abelian groups

1. (long cofiber sequence)

   [ΣA,X] *⟶[h,X] *[B/A,X] *⟶[g,X] *[B,X] *⟶[f,X] *[A,X] * [\Sigma A, X]_\ast \overset{[h,X]_\ast}{\longrightarrow} [B/A,X]_\ast \overset{[g,X]_\ast}{\longrightarrow} [B,X]_\ast \overset{[f,X]_\ast}{\longrightarrow} [A,X]_\ast

2. (long fiber sequence)

   [X,A] *⟶[X,f] *[X,B] *⟶[X,g] *[X,B/A] *⟶[X,h] *[X,ΣA] * [X,A]_\ast \overset{[X,f]_\ast}{\longrightarrow} [X,B]_\ast \overset{[X,g]_\ast}{\longrightarrow} [X,B/A]_\ast \overset{[X,h]_\ast}{\longrightarrow} [X,\Sigma A]_\ast

are long exact sequences.

Proof

Regarding the first case:

Since g∘f=0g \circ f = 0 by lemma , we have an inclusion im([g,X] *)⊂ker([f,X] *)im([g,X]_\ast) \subset ker([f,X]_\ast). Hence it is sufficient to show that if ψ:B→X\psi \colon B \to X is in the kernel of [f,X] *[f,X]_\ast in that ψ∘f=0\psi \circ f = 0, then there is ϕ:C→X\phi \colon C \to X with ϕ∘g=ψ\phi \circ g = \psi. To that end, consider the commuting diagram

A ⟶f B ⟶g B/A ⟶h ΣA ↓ ψ↓ 0 ⟶ X ⟶id X ⟶ 0, \array{ A &\overset{f}{\longrightarrow}& B &\overset{g}{\longrightarrow}& B/A &\overset{h}{\longrightarrow}& \Sigma A \\ \downarrow && {}^{\mathllap{\psi}}\downarrow \\ 0 &\overset{}{\longrightarrow}& X &\overset{id}{\longrightarrow}& X &\overset{}{\longrightarrow}& 0 } \,,

where the commutativity of the left square exhibits our assumption.

The top part of this diagram is a distinguished triangle by assumption, and the bottom part is by condition T1T1 in def. . Hence by condition T3 there exists ϕ\phi fitting into a commuting diagram of the form

A ⟶f B ⟶g B/A ⟶h ΣA ↓ ψ↓ ↓ ϕ ↓ 0 ⟶ X ⟶id X ⟶ 0. \array{ A &\overset{f}{\longrightarrow}& B &\overset{g}{\longrightarrow}& B/A &\overset{h}{\longrightarrow}& \Sigma A \\ \downarrow && {}^{\mathllap{\psi}}\downarrow && \downarrow^{\mathrlap{\phi}} && \downarrow \\ 0 &\overset{}{\longrightarrow}& X &\overset{id}{\longrightarrow}& X &\overset{}{\longrightarrow}& 0 } \,.

Here the commutativity of the middle square exhibits the desired conclusion.

This shows that the first sequence in question is exact at [B,X] *[B,X]_\ast. Applying the same reasoning to the distinguished traingle (g,h,−Σf)(g,h,-\Sigma f) provided by T2 yields exactness at [C,X] *[C,X]_\ast.

Regarding the second case:

Again, from lemma it is immediate that

im([X,f] *)⊂ker([X,g] *) im([X,f]_\ast) \subset ker([X,g]_\ast)

so that we need to show that for ψ:X→B\psi \colon X \to B in the kernel of [X,g] *[X,g]_\ast, hence such that g∘ψ=0g\circ \psi = 0, then there exists ϕ:X→A\phi \colon X \to A with f∘ϕ=ψf \circ \phi = \psi.

To that end, consider the commuting diagram

X ⟶ 0 ⟶ ΣX ⟶−id ΣX ↓ ψ ↓ B ⟶g B/A ⟶h ΣA ⟶−Σf ΣB, \array{ X &\longrightarrow& 0 &\longrightarrow& \Sigma X &\overset{- id}{\longrightarrow}& \Sigma X \\ \downarrow^{\mathrlap{\psi}} && \downarrow \\ B &\overset{g}{\longrightarrow}& B/A &\overset{h}{\longrightarrow}& \Sigma A &\overset{-\Sigma f}{\longrightarrow}& \Sigma B } \,,

where the commutativity of the left square exhibits our assumption.

Now the top part of this diagram is a distinguished triangle by conditions T1 and T2 in def. , while the bottom part is a distinguished triangle by applying T2 to the given distinguished triangle. Hence by T3 there exists ϕ˜:ΣX→ΣA\tilde \phi \colon \Sigma X \to \Sigma A such as to extend to a commuting diagram of the form

X ⟶ 0 ⟶ ΣX ⟶−id ΣX ↓ ψ ↓ ↓ ϕ˜ ↓ Σψ B ⟶g B/A ⟶h ΣA ⟶−Σf ΣB, \array{ X &\longrightarrow& 0 &\longrightarrow& \Sigma X &\overset{- id}{\longrightarrow}& \Sigma X \\ \downarrow^{\mathrlap{\psi}} && \downarrow && \downarrow^{\mathrlap{\tilde \phi}} && \downarrow^{\mathrlap{\Sigma \psi}} \\ B &\overset{g}{\longrightarrow}& B/A &\overset{h}{\longrightarrow}& \Sigma A &\overset{-\Sigma f}{\longrightarrow}& \Sigma B } \,,

At this point we appeal to the condition in def. that Σ:Ho→Ho\Sigma \colon Ho \to Ho is an equivalence of categories, so that in particular it is a fully faithful functor. It being a full functor implies that there exists ϕ:X→A\phi \colon X \to A with ϕ˜=Σϕ\tilde \phi = \Sigma \phi. It being faithful then implies that the whole commuting square on the right is the image under Σ\Sigma of a commuting square

X ⟶−id X ϕ↓ ↓ ψ A ⟶−f B. \array{ X &\overset{-id}{\longrightarrow}& X \\ {}^{\mathllap{\phi}}\downarrow && \downarrow^{\mathrlap{\psi}} \\ A &\underset{-f}{\longrightarrow}& B } \,.

This exhibits the claim to be shown.