Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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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.

Every fully faithful functor reflects isomorphisms

Statement

If F:CDF:\mathcal C\to\mathcal D is fully faithful and F(f)F(f) is an isomorphism, then ff is an isomorphism.

Facts & Assumptions

Given: A fully faithful functor FF and a morphism f:ABf:A\to B such that F(f)F(f) is invertible.

[L1]

Fullness lifts every morphism between FAFA and FBFB, while faithfulness reflects equality between parallel morphisms (Faithful, full, fully faithful, essentially surjective, and split essentially surjective functors).

[L2]

An isomorphism has a two-sided inverse (Isomorphism, groupoid, and connected category).

Proof

technique · direct
1.1

Let h:FBFAh:FB\to FA be the inverse of F(f)F(f); fullness gives g:BAg:B\to A with F(g)=hF(g)=h.

givenL1L2
2.1

Then F(gf)=hF(f)=1FA=F(1A)F(g\circ f)=h\circ F(f)=1_{FA}=F(1_A) and F(fg)=F(f)h=1FB=F(1B)F(f\circ g)=F(f)\circ h=1_{FB}=F(1_B).

step 1.1L1L2
3.1

Faithfulness gives gf=1Ag\circ f=1_A and fg=1Bf\circ g=1_B, so ff is an isomorphism.

step 2.1L1L2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 9 results over 5 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources