ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Euler class of a two-term mapping cone

Example

Let $f : P \to Q$ be a homomorphism of finitely generated projective left $A$ -modules over a unital associative ring $A$ , regarded as degree-zero cochain complexes. Then $Cone (f)$ has $P$ in degree $- 1$ and $Q$ in degree $0$ . Hence $[Cone (f)] = [Q [0]] - [P [0]]$ in $K_{0}^{t r i} (D_{p e r f} (A))$ and $χ (Cone (f)) = [Q] - [P]$ in $K_{0}^{s p l i t} ({Proj}_{f g} (A))$ . In particular the cone of right multiplication $r_{a} : A \to A$ , $x \mapsto x a$ , on the left regular module has Euler class zero for every $a \in A$ , regardless of its kernel or cokernel.

Facts & Assumptions

Given: A unital associative ring $A$ ; a homomorphism $f : P \to Q$ of finitely generated projective left $A$ -modules, viewed as cochain complexes concentrated in degree $0$ ; and an element $a \in A$ .

[F1]

The mapping cone of a chain map $f : C_{∙} \to D_{∙}$ has $Cone (f)_{n} = D_{n} \oplus C_{n - 1}$ with differential $d_{n} (y, x) = (d_{n}^{D} y + f_{n - 1} x, - d_{n - 1}^{C} x)$ (The mapping cone of a chain map).

[F2]

In the cochain convention of the derived category, $Cone (f)^{n} = Y^{n} \oplus X^{n + 1}$ with $d (y, x) = (d_{Y} y + f x, - d_{X} x)$ for a chain map $f : X \to Y$ of cochain complexes, the cone triangle ends in $X [1]$ , and the degree-zero stalk complex $S^{0} (M)$ has $M$ in degree $0$ and zero elsewhere (Derived category of an abelian category, Zero complex and stalk complex).

[F3]

In $K_{0}^{t r i}$ , $[X [n]] = (- 1)^{n} [X]$ (Shift signs and exact-functor maps on triangulated K0).

[F4]

For a bounded complex $P$ of finitely generated projective left modules, $χ (P) = \sum_{n} (- 1)^{n} [P^{n}]$ defines a class depending only on the represented perfect object (Euler class of a bounded projective complex is derived invariant and triangle additive).

[F5]

Degree-zero inclusion gives the isomorphism $K_{0}^{s p l i t} ({Proj}_{f g} (A)) \to K_{0}^{t r i} (D_{p e r f} (A))$ with $[P] \mapsto [P [0]]$ (Triangle K0 of perfect complexes equals split K0 of finite projectives).

Verification

technique · direct

1.1F1F2algebra

Under cochain reindexing $X^{i} = X_{- i}$ , the chain-cone terms $D_{n} \oplus C_{n - 1}$ of [F1] become $Q^{i} \oplus P^{i + 1}$ , agreeing with the cochain formula of [F2]. For degree-zero stalk complexes, the only nonzero terms are $Q$ in degree $0$ and $P$ in degree $- 1$ , with differential $f : P \to Q$ . Thus $Cone (f)$ is bounded with finitely generated projective terms.

2.1F2F3F4F5step 1.1algebra

The cone triangle $P [0] \to Q [0] \to Cone (f) \to P [1]$ of [F2] is a distinguished triangle of $D_{p e r f} (A)$ , since all three terms are bounded complexes of finitely generated projectives; its relation and the shift sign [F3] give $[Cone (f)] = [Q [0]] + [P [1]] = [Q [0]] - [P [0]]$ in $K_{0}^{t r i} (D_{p e r f} (A))$ . Independently, the Euler class formula of [F4] on the two-term complex of step 1.1 gives $χ (Cone (f)) = (- 1)^{- 1} [P] + (- 1)^{0} [Q] = [Q] - [P]$ in $K_{0}^{s p l i t} ({Proj}_{f g} (A))$ , and the comparison isomorphism of [F5] carries this class to $[Q [0]] - [P [0]]$ , so the two computations agree as promised.

3.1F4F5step 2.1algebra∎

Right multiplication $r_{a} (x) : = x a$ is left $A$ -linear: $r_{a} (b x) = (b x) a = b (x a) = b r_{a} (x)$ for all $b, x \in A$ . Taking $P = Q = A$ and $f = r_{a}$ in step 2.1 gives $[Cone (r_{a})] = [A [0]] - [A [0]] = 0$ and $χ (Cone (r_{a})) = [A] - [A] = 0$ for every $a \in A$ , whatever the kernel ${x : x a = 0}$ and cokernel $A / A a$ may be: the two projective terms cancel even when the cone is not acyclic, and no assertion that its cohomology modules are projective is used.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

43 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, More on Algebra, Lemma 15.121.1 (standard reference, not scraped)
Khovanov and Seidel, Quivers, Floer Cohomology, and Braid Group Actions, §2e.1 (standard reference, not scraped)