Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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  • 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.
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Every morphism that is both monic and epic is an isomorphism

Statement

FALSE. Every morphism that is both monic and epic is an isomorphism.

Facts & Assumptions

Given: The unit-preserving ring inclusion j:Z↪Qj:\mathbb Z\hookrightarrow\mathbb Q.

[L1]

The map jj is monic and epic in Ring\mathbf{Ring} but is not an isomorphism (The inclusion Z↪Q\mathbb Z\hookrightarrow\mathbb Q is monic and epic but neither surjective nor an isomorphism in Ring\mathbf{Ring}).

Refutation

technique · direct
1.1

By [L1], jj satisfies both cancellation properties required of a monomorphism and an epimorphism.

L1
1.2

The same result proves that jj has no inverse ring homomorphism and hence is not an isomorphism.

L1
2.1

Thus jj is a morphism that is simultaneously monic and epic but not invertible, directly refuting the statement.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 41 results over 11 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