CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30‡ rests on unproved material (inherited)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The kernel-cokernel sequence of a composite is a snake

Statement

For composable morphisms $A \overset{f}{\to} B \overset{g}{\to} C$ in an abelian category, the exact sequence of The kernel-cokernel sequence of a composite is an instance of the snake sequence.

Facts & Assumptions

Given: Composable morphisms $A \overset{f}{\to} B \overset{g}{\to} C$ .

[L1]

The snake lemma gives an exact kernel-cokernel sequence for a morphism of short exact sequences (Snake lemma in an abelian category).

[L2]

The composite already has a kernel-cokernel exact sequence (The kernel-cokernel sequence of a composite).

[L3]

An abelian category is additive and has finite biproducts (Abelian category).

Proof

technique · direct

1.1

Consider the morphism between the canonical split short exact sequences tikzcd 0 \arrow[r] & A \arrow[r, "j_A"] \arrow[d, "f"'] & A\oplus B \arrow[r, "\pi_B"] \arrow[d, "m"'] & B \arrow[r] \arrow[d, "g"'] & 0 \\ 0 \arrow[r] & B \arrow[r, "j_B"'] & B\oplus C \arrow[r, "\pi_C"'] & C \arrow[r] & 0 where, in biproduct matrix notation, $m = (\begin{array}{cc} f & - 1_{B} \\ 0 & g \end{array}), m (a, b) = (f (a) - b, g (b)) .$ The two squares commute, so [L1] applies.

L1L3givenconstruct

2.1

The map $a \mapsto (a, f (a))$ identifies $\ker (g f)$ with $\ker (m)$ : the equations $m (a, b) = 0$ are exactly $b = f (a)$ and $g f (a) = 0$ . Dually, the map $B \oplus C ⟶ coker (g f), (b, c) ⟼ [c + g (b)]$ identifies $coker (m)$ with $coker (g f)$ . Under these identifications, the snake sequence of step 1.1 is exactly $0 \to \ker (f) \to \ker (g f) \to \ker (g) \to coker (f) \to coker (g f) \to coker (g) \to 0.$

L1L2L3step 1.1algebra

3.1

The maps in step 2.1 are the canonical comparison maps of [L2], so the kernel-cokernel sequence of a composite is a special case of the snake lemma.

L2step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

Saunders Mac Lane, Categories for the Working Mathematician, Exercise VIII.4.6 (standard reference, not scraped)