TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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.
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The inclusion $Z ↪ Q$ is monic and epic but neither surjective nor an isomorphism in $R i n g$

Statement

In $R i n g$ , the canonical inclusion $i : Z ↪ Q$ is monic and epic, but its underlying function is not surjective and it is not an isomorphism.

Facts & Assumptions

Given: The canonical unital ring homomorphism $i : Z \to Q$ .

[L1]

The integers form a commutative ring (The integers form a commutative ring), the rationals form a field and hence a commutative ring (The rationals form a field, Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring), and The integers embed in the rationals identifies $i$ as an injective embedding.

[L2]

Morphisms of $R i n g$ are unit-preserving ring homomorphisms (Unital rings and unit-preserving ring homomorphisms form the large locally small category $R i n g$ ); monic, epic, and isomorphism mean cancellation and a two-sided inverse (Monomorphism and epimorphism by left and right cancellation, Isomorphism, groupoid, and connected category).

Proof

technique · direct

1.1

If $i \circ u = i \circ v$ , injectivity of $i$ gives $u = v$ pointwise, so $i$ is monic.

givenL1L2

2.1

If ring homomorphisms $f, g : Q \to R$ agree after $i$ , then for $q = a / b$ with $b \neq 0$ both send $q$ to the common image of $a$ times the inverse of the common image of $b$ ; hence $f (q) = g (q)$ for every $q$ , so $i$ is epic.

step 1.1L1L2

3.1

The rational $1 / 2$ is not an integer, so the underlying function of $i$ is not surjective; a categorical inverse would be an inverse function and would force surjectivity, so $i$ is not an isomorphism.

step 2.1L1L2∎

Depends on

Used by

Every morphism that is both monic and epic is an isomorphism False statement

Dependency tree · two levels

40 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, Chapter 1 (standard reference, not scraped)

The inclusion Z↪QZ↪Q is monic and epic but neither surjective nor an isomorphism in RingRing