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A chain map is a quasi-isomorphism exactly when its cone is acyclic

Statement

Let $f : C_{∙} \to D_{∙}$ be a chain map in an abelian category. Then $f$ is a quasi-isomorphism if and only if $Cone (f)$ is acyclic.

Facts & Assumptions

Given: A chain map $f : C_{∙} \to D_{∙}$ in an abelian category and an integer $n$ .

[L1]

A quasi-isomorphism induces isomorphisms on all homology objects (Quasi-isomorphism, A chain map induces a well-defined map on homology).

[L2]

The cone has degree- $n$ object $D_{n} \oplus C_{n - 1}$ and differential $(y, x) \mapsto (d_{D} y + f_{n - 1} x, - d_{C} x)$ ; this is biproduct matrix notation for morphisms (The mapping cone of a chain map).

[L3]

The homology map $q_{E, n} : Z_{n} (E) \to H_{n} (E)$ is the cokernel of the boundary inclusion. Acyclicity means $H_{n} (E) = 0$ for every $n$ (Homology object of a chain complex, Exactness of a complex at a degree and acyclic complexes).

[L4]

Every boundary inclusion is the kernel of its homology cokernel, and epimorphisms remain epic under pullback. Hence if $q_{E, n} t = 0$ for a morphism $t : T \to Z_{n} (E)$ , an epic pullback cover $e : T^{'} \to T$ admits $v : T^{'} \to E_{n + 1}$ with $d_{n + 1} v = z_{E, n} t e$ (Every monomorphism is the kernel of its cokernel, and dually every epimorphism is the cokernel of its kernel, The pullback of an epimorphism is an epimorphism).

[L5]

In an abelian category, zero kernel detects monicity, zero cokernel detects epicity, and a monic epic is an isomorphism (In an abelian category, monic means zero kernel and epic means zero cokernel, An abelian category is balanced).

Proof

technique · direct

1.1L3L4

By [L3], $H_{n} (E) = 0$ exactly when the boundary map $E_{n + 1} \to Z_{n} (E)$ is epic. By [L4], this is equivalent to the following arrow condition: every $t : T \to Z_{n} (E)$ becomes a boundary after precomposition with some epimorphism $e : T^{'} \to T$ . We use this condition for $E = Cone (f)$ , and use the analogous condition $q_{E, n} t = 0$ for boundaries in $C$ and $D$ .

1.2L1L2L4assume-hypconstructalgebra

Conversely, assume $f$ is a quasi-isomorphism. Let $t : T \to Z_{n} (Cone (f))$ be any morphism, and write the components of its composite with the cycle inclusion as $y : T \to D_{n}$ and $x : T \to C_{n - 1}$ . The cone-cycle equation gives $d_{C} x = 0$ and $d_{D} y + f_{n - 1} x = 0$ . Thus $x = z_{C, n - 1} x_{0}$ for a unique $x_{0} : T \to Z_{n - 1} (C)$ , and $H_{n - 1} (f) q_{C, n - 1} x_{0} = q_{D, n - 1} f_{n - 1} x = 0$ . Since $H_{n - 1} (f)$ is monic, $q_{C, n - 1} x_{0} = 0$ . By [L4], after an epic cover $e_{0} : T_{0} \to T$ there is $w : T_{0} \to C_{n}$ with $d_{C} w = x e_{0}$ .

2.1L1L2L3L4L5step 1.1assume-hypconstructalgebra

Assume the cone is acyclic. To test the kernel of $H_{n - 1} (f)$ , take a morphism $t : T \to Z_{n - 1} (C)$ with $H_{n - 1} (f) q_{C, n - 1} t = 0$ . By [L4], after an epic cover $e_{0} : T_{0} \to T$ there is $y : T_{0} \to D_{n}$ with $d_{D} y = f_{n - 1} z_{C, n - 1} t e_{0}$ . The column $(- y, z_{C, n - 1} t e_{0}) : T_{0} \to Cone (f)_{n}$ is a cycle by [L2]. By step 1.1, after another epic cover $e_{1} : T_{1} \to T_{0}$ it is the cone boundary of a column $(v, w) : T_{1} \to D_{n + 1} \oplus C_{n}$ . The second component of that boundary says $z_{C, n - 1} t e_{0} e_{1} = - d_{C} w$ , so $q_{C, n - 1} t e_{0} e_{1} = 0$ . Cancel the epimorphisms to obtain $q_{C, n - 1} t = 0$ . To apply this to the kernel $k : K \to H_{n - 1} (C)$ of $H_{n - 1} (f)$ , pull back the epic $q_{C, n - 1}$ along $k$ , obtaining an epic $e : T \to K$ and $t : T \to Z_{n - 1} (C)$ with $q_{C, n - 1} t = k e$ . The preceding argument gives $k e = 0$ , hence $k = 0$ . Thus $H_{n - 1} (f)$ is monic by [L5].

2.2L1L3L4step 1.2constructalgebra

The morphism $y e_{0} + f_{n} w : T_{0} \to D_{n}$ is a cycle, so it factors through $Z_{n} (D)$ as $r : T_{0} \to Z_{n} (D)$ . Since $H_{n} (f)$ is an isomorphism, there is $h : T_{0} \to H_{n} (C)$ with $H_{n} (f) h = q_{D, n} r$ . Pull back the epic $q_{C, n} : Z_{n} (C) \to H_{n} (C)$ along $h$ , obtaining an epic $e_{1} : T_{1} \to T_{0}$ and $s : T_{1} \to Z_{n} (C)$ with $q_{C, n} s = h e_{1}$ . Therefore $q_{D, n} (r e_{1} - Z_{n} (f) s) = 0$ . By [L4], after an epic cover $e_{2} : T_{2} \to T_{1}$ there is $v : T_{2} \to D_{n + 1}$ with $d_{D} v = (y e_{0} + f_{n} w) e_{1} e_{2} - f_{n} z_{C, n} s e_{2}$ .

3.1L1L2L3L5step 1.1step 2.1assume-hypconstructalgebra

To test epicity of $H_{n} (f)$ , take any $t : T \to Z_{n} (D)$ . The column $(z_{D, n} t, 0) : T \to Cone (f)_{n}$ is a cycle. By step 1.1, after an epic cover $e : T^{'} \to T$ it is the boundary of $(v, w) : T^{'} \to D_{n + 1} \oplus C_{n}$ . The second component says $d_{C} w = 0$ , so $w = z_{C, n} s$ for a unique $s : T^{'} \to Z_{n} (C)$ . The first component gives $q_{D, n} t e = H_{n} (f) q_{C, n} s$ . Hence $coker (H_{n} (f)) q_{D, n} t e = 0$ for every $t$ . Taking $t = 1_{Z_{n} (D)}$ and cancelling the epimorphisms $e$ and $q_{D, n}$ yields $coker (H_{n} (f)) = 0$ . Thus $H_{n} (f)$ is epic by [L5]. Since $n$ is arbitrary, this step and step 2.1 make every $H_{n} (f)$ an isomorphism; [L1] makes $f$ a quasi-isomorphism.

4.1L2L3step 1.1step 3.1step 1.2step 2.2constructalgebra∎

Put $u = - w e_{1} e_{2} + z_{C, n} s e_{2} : T_{2} \to C_{n}$ . The cone differential of $(v, u)$ is $(y e_{0} e_{1} e_{2}, x e_{0} e_{1} e_{2})$ by [L2] and steps 1.2 and 2.2. Thus $t e_{0} e_{1} e_{2}$ is a cone boundary. Since the composite cover is epic, step 1.1 gives $H_{n} (Cone (f)) = 0$ . This holds for every $n$ , so the cone is acyclic by [L3]. The two implications prove the equivalence.

A chain map is a quasi-isomorphism exactly when its cone is acyclic

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources