False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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.

FALSE: if coimage and image happen to be isomorphic as objects, then the canonical map is automatically an isomorphism

Statement

If the coimage and image of a morphism are isomorphic as objects, then the canonical map from the coimage to the image is automatically an isomorphism.

Facts & Assumptions

Given: The torsion-free abelian-group subcategory and the morphism $2 : Z \to Z$ .

[L1]

The torsion-free abelian-group subcategory is not abelian (Torsion-free abelian groups do not form an abelian category).

[L2]

Every morphism with kernels and cokernels has a canonical map from its coimage to its image (The canonical morphism from the coimage to the image exists and is unique).

[L3]

The cokernel of $0 \to A$ is $A$ , and dually the kernel of $A \to 0$ is $A$ (The cokernel of the zero map out of the zero object is the target, and dually for kernels).

Refutation

1.1

In the torsion-free abelian-group subcategory, the morphism $2 : Z \to Z$ has zero kernel. Its cokernel in that subcategory is also zero, because any homomorphism out of $Z$ that kills the even subgroup must send $1$ to torsion and hence to $0$ . Therefore [L3] identifies both $coim (2)$ and $im (2)$ with $Z$ .

L1L3

2.1

The canonical map of [L2] is still the morphism $2 : Z \to Z$ , which is not an isomorphism. So isomorphism of the endpoint objects does not force the canonical comparison map itself to be invertible.

L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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

Emily Riehl, Category Theory in Context, Example 4.5.13 (standard reference, not scraped)