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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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Spectral sequence subquotient and local lifting calculus

Statement

In an abelian category, finite sums of subobjects are images of biproduct maps, intersections and inverse images are pullbacks. For subobjects U,V,W⊆X there are canonical isomorphisms (U+V)/V≅U/(U∩V),(X/V)/(U/V)≅X/U(V⊆U). If U⊆W, then W∩(U+V)=U+(W∩V). A map preserving a numerator and denominator induces a unique quotient map. A map into im⁡f lifts through f after an epimorphism onto its domain. Equality and factorization through a subobject can be tested after that epimorphism. These constructions are natural for maps preserving the displayed data.

Facts & Assumptions

Given: An abelian category, morphisms f,e and the subobjects in the statement.

[F1]

Finite biproducts, kernels, cokernels and the canonical isomorphism coim⁡f≅im⁡f exist (Abelian category).

[F2]

A monomorphism has zero kernel and an epimorphism has zero cokernel (The kernel of a monomorphism is zero and the cokernel of an epimorphism is zero).

[F3]

The cokernel of 0→A and the kernel of A→0 are represented by 1A (The cokernel of the zero map out of the zero object is the target, and dually for kernels).

Proof

technique · direct
1.1F1

A cokernel q is epic: if uq=vq, uniqueness in its universal property applied to that common composite gives u=v. A kernel is monic by the corresponding uniqueness. Thus [F1] factors every f:A→B as A→eim⁡f→mB with e epic, m monic and A/ker⁡f≅im⁡f.

1.2F1

For f:A→X and e:B→X, set D=fπA−eπB:A⊕B→X and P=ker⁡D. Its projections α:P→A, β:P→B satisfy fα=eβ. Any pair (a,b) with fa=eb factors uniquely through ker⁡D, proving the pullback property, also when one object is zero.

2.1F2F3step 1.1

If f is monic, [F2] and [F3] identify its coimage projection with an isomorphism; the factorization in step 1.1 therefore makes f a kernel of its cokernel. If f is epic, its image inclusion is an isomorphism by the kernel clause of [F3], so f is a cokernel of its kernel. In particular a map both monic and epic is an isomorphism.

3.1step 2.1step 1.2algebra

If e is epic, D is epic since DιB=−e. If q:A→Q satisfies qα=0, then qπA kills ker⁡D; step 2.1 gives v:X→Q with vD=qπA. Composing with ιB gives ve=0, hence v=0; composing with ιA gives q=0. Apply this to the difference of any two maps agreeing after α: α is epic.

3.2step 1.1step 2.1step 1.2

Define U+V=im⁡(U⊕V→X) and U∩V=U×XV. Both have their least-upper-bound and greatest-lower-bound properties: a common upper subobject contains the biproduct map and hence its image by normality, while a common lower subobject factors through the pullback. A map h:M→M′ with h(N)⊆N′ gives a unique hˉ:M/N→M′/N′ because the composite M→M′→M′/N′ kills N.

4.1step 1.1step 2.1step 3.1

For h:T→im⁡f, pull back the epic e of step 1.1 along h. Step 3.1 gives an epic t:T′→T and a lift a:T′→A with ea=ht. Equalities after t descend by epic cancellation. If ht factors through a subobject N↪Y, then (Y→Y/N)ht=0, hence (Y→Y/N)h=0, and normality in step 2.1 factors h through N. Conversely, a factorization of h gives one of ht by composition. Finite repetitions compose epimorphisms; no section or choice of global representatives is asserted.

5.1step 1.1step 2.1step 4.1step 3.2

The kernel of U→X/V is U∩V, since V=ker⁡(X→X/V) by step 2.1. Its image agrees with that of U+V→X/V: after the epic U⊕V→U+V the latter map is [uˉ,0], which has the same image as uˉ. The kernel of U+V→X/V is V, so step 1.1 identifies this image both with U/(U∩V) and with (U+V)/V, proving the first isomorphism.

6.1step 3.2step 5.1

For V⊆U, step 5.1 makes U/V→X/V monic. Maps X/V→T killing U/V correspond uniquely to maps X→T killing U: each direction follows by composing with the epic quotient from X and then applying cokernel descent. Thus X/U and (X/V)/(U/V) satisfy the same cokernel property, giving inverse compatible maps and the second isomorphism.

7.1step 4.1step 3.2step 5.1step 6.1∎

For U⊆W, let h:T→W∩(U+V). Step 4.1 locally writes its composite into X as u+v, with u through U and v through V. Since both h and u factor through W, their difference v factors through W∩V. Thus h locally, and by step 4.1 globally, factors through U+(W∩V). The reverse inclusion holds because both summands lie in W and in U+V. This proves modularity. All comparison maps above are characterized by the original inclusions and quotient projections; composing with a map preserving the subobjects yields the same characterizations. Uniqueness and epic cancellation prove naturality and identity/composition compatibility.

Source notes

Stacks, §12.5, Definition 12.5.1 and Lemmas 12.5.11–15. The finite categorical argument below is supplied locally.

Depends on

Dependency tree · 2 levels

Sources