Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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↪Q.

[L1]

The map j is monic and epic in Ring but is not an isomorphism (The inclusion Z↪Q is monic and epic but neither surjective nor an isomorphism in Ring).

Refutation

technique · direct
1.1

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

L1
1.2

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

L1
2.1

Thus j 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 · two levels

8 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