UNIVERSAL COEFFICIENT THEOREM IN TRIANGULATED CATEGORIES

TEIMURAZ PIRASHVILI AND MARÍA JULIA REDONDO

We assume that the reader is familiar with triangulated categories (see [2], [1]). If 𝒯 is a triangulated category, the shifting of an object X ∈ 𝒯 is denoted by X[1]. Assume an abelian category 𝒜 is given, which is equipped with an auto-equivalence x ↦ x[1]. Objects of 𝒜 are denoted by the small letters x, y, z, etc, while objects of 𝒯 are denoted by the capital letters X, Y, Z, etc. A homology theory on 𝒯 with values in 𝒜 is a functor h : 𝒯 → 𝒜 such that h commutes with shifting (up to an equivalence) and for any distinguished triangle X → Y → Z → X[1] in 𝒯 the induced sequence h(X) → h(Y) → h(Z) is exact. It follows that then one has the following long exact sequence

⋯ → h(Z)[−1] → h(X) → h(Y) → h(Z) → h(X)[1] → ⋯

In what follows Ext_{𝒜}^{1} (x, y) denotes the equivalence classes of extensions of x by y in the category 𝒜 and we assume that these classes form a set.

In this paper we prove the following result:

Theorem 1. Let h : 𝒯 → 𝒜 be a homology theory. Assume the following conditions hold

i) h reflects isomorphisms, ii) h is full, iii) for any object x ∈ 𝒜 there is an object X ∈ 𝒯 and an isomorphism

h(X) ≅ x in 𝒜. Then 𝒜 is a hereditary abelian category, the ideal

Ker(h) = {f ∈ Hom_{𝒯} (X, Y) | h(f) = 0}

is a square zero ideal, which as a bifunctor on 𝒯 is isomorphic to

(X, Y) ↦ Ext_{𝒜}^{1} (h(X)[1], h(Y)).

In particular for any X, Y ∈ 𝒯 one has the following short exact sequence

0 → Ext_{𝒜}^{1} (h(X)[1], h(Y)) → 𝒯 (X, Y) → Hom_{𝒜} (h(X), h(Y)) → 0.

Thus this is a sort of ”universal coefficient theorem” in triangulated categories.

Our result is a one step generalization of a well-known result which claims that if h is an equivalence of categories then 𝒜 is semi-simple meaning that Ext_{𝒜}^{1} = 0 (see for example [1, p. 250]).

Before we give a proof of the Theorem, let us explain notations involved on it. The functor h reflects isomorphisms, this means that f ∈ Hom_{𝒯} (X, Y) is an isomorphism provided h(f) is an isomorphism in 𝒜. Moreover h is full, this means that the homomorphism 𝒯 (X, Y) → Hom_{𝒜} (h(X), h(Y)) given by f ↦ h(f) is surjective for all X, Y ∈ 𝒯 . Furthermore an abelian category 𝒜 is hereditary provided for any two-fold extension

(1)  \xymatrix{0 \ar[r] & u \ar[r]^{α̂} & v \ar[r]^{β̂} & w \ar[r]^{γ̂} & x \ar[r] & 0}

there exists a commutative diagram with exact rows

\xymatrix{ & 0 \ar[r] & v \ar[r] \ar[d]_{Id} & z \ar[r] \ar[d] & x \ar[r] \ar[d]_{Id} & 0 \\ 0 \ar[r] & u \ar[r] & v \ar[r] & w \ar[r] & x \ar[r] & 0}

This exactly means that Ext_{𝒜}^{2} = 0, where Ext is understood a lá Yoneda. Let us also recall that an ideal 𝕀 in an additive category 𝔸 is a sub-bifunctor of the bifunctor Hom_{𝔸}(−, −) : 𝔸^{op} × 𝔸 → 𝔸𝕓. In this case one can form the quotient category 𝔸/𝕀 in an obvious way. One says that I^{2} = 0 provided gf = 0 as soon as f ∈ 𝕀(A, B) and g ∈ 𝕀(B, C). In this case the bifunctor 𝕀 : 𝔸^{op} × 𝔸 → 𝔸𝕓 factors through the quotient category 𝔸/𝕀 in a unique way.

Proof. It is done in several steps.

First step. The equality (Ker h)^{2} = 0. Actually the proof of this fact does not use Assumption iii). To make notations easier we denote h(X), h(Y) simply by x, y, etc. Moreover, for a morphism α : X → Y, we let α̂ : x → y be the morphism h(α). Suppose α : X → Y and β : Y → Z are morphisms such that α̂ = 0 and β̂ = 0. We have to prove that γ := βα is the zero morphism. By the octahedron axiom there is a diagram of distinguished triangles

\xymatrix{X \ar[r]^{α} \ar[d]^{Id} & Y \ar[r] \ar[d]^{β} & U \ar[r] \ar[d] & X[1] \ar[d]^{Id} \\ X \ar[r]^{γ} & Z \ar[r]^{ω} \ar[d] & V \ar[r]^{ν} \ar[d] & X[1] \ar[d] \\  & W \ar[r]^{Id} \ar[d] & W \ar[r] \ar[d] & Y [1] \\  & Y [1] \ar[r] & U[1]}

Apply h to get

\xymatrix{0 \ar[r] & y \ar[r] \ar[d]^{0} & u \ar[r] \ar[d] & x[1] \ar[d]^{Id} \ar[r] & 0 \\ 0 \ar[r] & z \ar[r]^{ω̂} & v \ar[r]^{ν̂} & x[1] \ar[r] & 0}

It follows that there is a morphism μ̂ : x[1] → v in 𝒜 such that ν̂μ̂ = Id_{x[1]}. Thus (ω̂, μ̂) : z ⊕ x[1] → v is an isomorphism. Since h is full, we can find μ : X[1] → V which realizes μ̂, meaning that h(μ) = μ̂. The morphism (ω, μ) : Z⊕X[1] → V is an isomorphism, because h reflects isomorphisms. In particular ω is a monomorphism and therefore γ = 0 and first step is done.

For objects X, Y ∈ 𝒯 we put

𝕀(X, Y) := {α ∈ Hom_{𝒜} (X, Y) | h(α) = 0}.

We have just proved that 𝕀^{2} = 0. In particular 𝕀 as a bifunctor factors through the category 𝒯 /𝕀. The next step shows that it indeed factors through the category 𝒜 and a quite explicit description of this bifunctor is given.

Second step. Bifunctorial isomorphism 𝕀(X, Y) ≅ Ext_{𝒜}^{1} (h(X)[1], h(Y)). Also the proof of this fact does not use Assumption iii). We put as usual x = h(X), y = h(Y), etc. Let α : X → Y be an element of 𝕀(X, Y). Consider a distinguished triangle

(2)  X →^{α} Y →^{β} Z →^{γ} X[1].

By applying h one obtains the following short exact sequence

(3)  0 → y →^{β̂} z →^{γ̂} x[1] → 0

whose class in Ext_{𝒜}^{1} (x[1], y) is independent on the choice of the triangle in (2) and it is denoted by Ξ(α). In this way one obtains the binatural transformation Ξ : 𝕀 → Ext_{𝒜}^{1} ((−)[1], (−)). We claim that Ξ is an isomorphism. Indeed, if Ξ(α) = 0, then there exists a section μ̂ : x[1] → z of γ̂ in (3). Then (β̂, μ̂) : y ⊕ x[1] → z is an isomorphism. Since h is full, we can find μ : X[1] → Z which realizes μ̂. The morphism (β, μ) : Y ⊕ X[1] → Z is an isomorphism, because h reflects isomorphisms. In particular β is a monomorphism and therefore α = 0. Hence Ξ is a monomorphism. Let us take any element in Ext_{𝒜}^{1} (x[1], y), which is represented by a short exact sequence, say the sequence (3). Take any realization β : Y → Z of β̂ and consider a distinguished triangle

U →^{ν} Y →^{β} Z →^{η} U[1]

containing β. Apply h and use the fact that β̂ is a monomorphism to obtain an extension

0 → y →^{β̂} z →^{η̂} u[1] → 0.

Comparing this exact sequence with our original exact sequence (3) and using the fact that h is full, one concludes that there is a morphism ϵ : X → U such ϵ̂[1]∘γ̂ = η̂. Since ϵ̂ is an isomorphism we see that ϵ is an isomorphism as well. Thus one has a distinguished triangle (2) with α = ν ∘ ϵ and γ = ϵ^{−1}[1] ∘ η. It follows that Ξ(α) represents our original element in Ext_{𝒜}^{1} (x[1], y). Hence Ξ is an isomorphism.

Third step. 𝒜 is hereditary. Let (1) be a two-fold extension in 𝒜. We put y = Im(α̂). Thus the exact sequence (1) splits in the following two short exact sequences

0 → u ⟶^{α̂} v ⟶^{μ̂} y → 0

and

0 → y ⟶^{ν̂} w ⟶^{γ̂} x → 0

with β̂ = ν̂μ̂. Using Assumption iii) and without loss of generality we can assume that u, v, w, x as well as α̂ and γ̂ have realizations. By Lemma 2 below we obtain the following distinguished triangles

U ⟶^{α} V ⟶^{μ} Y ⟶^{ξ} U[1]

and

Y ⟶^{ν} W ⟶^{γ} X ⟶^{χ} Y [1].

Since μ̂ is an epimorphism and ν̂ is a monomorphism it follows that h(ξ) = 0 and h(χ) = 0. Thus ξ ∘ χ[−1] = 0 thanks to the fact that 𝕀^{2} = 0. Therefore there exists λ : X[−1] → V such that μ ∘ λ = χ[−1], in other words one has the following commutative diagram

\xymatrix{ &  & X[−1] \ar[dl]_{λ} \ar[d]^{χ[−1]} \\ U \ar[r]^{α} & V \ar[r]^{μ} & Y \ar[r]^{ξ} & U[1]}

We claim that one can always find λ with property h(λ) = 0. Indeed, for a given λ with μ ∘ λ = χ[−1] one obtains the following diagram after applying h:

\xymatrix{ &  &  & x[−1] \ar[dl]_{λ̂} \ar[d]^{0} \\ 0 \ar[r] & u \ar[r]^{α̂} & v \ar[r]^{μ̂} & y \ar[r] & 0}

Thus λ̂ = α̂ ∘ ϕ̂, for some ϕ : X[−1] → U. Now it is clear that λ′ = λ − α ∘ ϕ has the expected properties h(λ′) = 0 and μ ∘ λ′ = χ[−1], and the claim is proved.

One can use the octahedral axiom to conclude that there exists a commutative diagram

\xymatrix{X[−1] \ar[r]^{λ} \ar[d]^{Id} & V \ar[r] \ar[d]^{μ} & Z \ar[r] \ar[d] & X \ar[d]^{Id} \\ X[−1] \ar[r]^{χ[−1]} & Y \ar[r]^{ν} \ar[d] & W \ar[r]^{γ} \ar[d] & X \\  & U[1] \ar[r]^{Id} \ar[d] & U[1] \ar[d] \\  & V [1] \ar[r] & Z[1]}

Since h(λ) = 0, by applying h one obtains the following commutative diagram

\xymatrix{0 \ar[r] & v \ar[r] \ar[d]_{μ̂} & z \ar[r] \ar[d] & x \ar[r] \ar[d]_{Id} & 0 \\ 0 \ar[r] & y \ar[r]^{ν̂} & w \ar[r]^{γ̂} & x \ar[r] & 0}

which shows that one has a commutative diagram with exact rows

\xymatrix{ & 0 \ar[r] & v \ar[r] \ar[d]_{Id} & z \ar[r] \ar[d] & x \ar[r] \ar[d]_{Id} & 0 \\ 0 \ar[r] & u \ar[r]^{α̂} & v \ar[r]^{β̂} & w \ar[r]^{γ̂} & x \ar[r] & 0}

Thus 𝒜 is hereditary.

□

Lemma 2. Let h : 𝒯 → 𝒜 be a homology theory. Assume h reflects isomorphisms and is full. Suppose there is given a morphism α : U → V, an object W in 𝒯 and a short exact sequence

0 → u ⟶^{α̂} v ⟶^{β̂} w → 0

in 𝒜, where as usual u = h(U), v = h(V), w = h(W) and α̂ = h(α). Then there exists a distinguished triangle

U ⟶^{α} V ⟶^{β} W ⟶^{γ} U[1]

such that h(β) = β̂. The dual statement is also true: Suppose there is given a morphism β : V → W, an object U in 𝒯 and a short exact sequence

0 → u ⟶^{α̂} v ⟶^{β̂} w → 0

in 𝒜, where β̂ = h(β). Then there exists a distinguished triangle

U ⟶^{α} V ⟶^{β} W ⟶^{γ} U[1]

such that h(α) = α̂.

Proof. Take any distinguished triangle containing α,

U ⟶^{α} V ⟶^{η} Z ⟶^{ϵ} U[1].

Apply h to get a short exact sequence

0 → u ⟶^{α̂} v ⟶^{η̂} z → 0.

Then we get the following commutative diagram

\xymatrix{0 \ar[r] & u \ar[r]^{α̂} \ar[d]^{Id} & v \ar[r]^{η̂} \ar[d]^{Id} & z \ar[d]^{δ̂} \ar[r] & 0 \\ 0 \ar[r] & u \ar[r]^{α̂} & v \ar[r]^{β̂} & w \ar[r] & 0}

with δ̂ an isomorphism. By assumption one can realize δ̂ to obtain an isomorphism δ : Z → W, h(δ) = δ̂. Then we have an isomorphism of triangles

\xymatrix{U \ar[r]^{α} \ar[d]^{Id} & V \ar[r]^{η} \ar[d]^{Id} & Z \ar[r]^{ϵ} \ar[d]^{δ} & U[1] \ar[d]^{Id} \\ U \ar[r]^{α} & V \ar[r]^{β} & W \ar[r]^{γ} & U[1]}

where β = δη and γ = ϵ ∘ δ^{−1}. It follows that the triangle

U ⟶^{α} V ⟶^{β} W ⟶^{γ} U[1]

is also a distinguished triangle. Thus the first statement is proved. The dual argument gives the second result.

□

References

[1] S. I. Gelfand and Y. I. Manin. Methods of homological algebra. Second edition. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2003. xx+372 pp.

[2] J. L. Verdier Des Catégories derivées. des catégories abéliennes. Astérisque v. 239 (1996). 253 pp.