LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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.

A morphism is an isomorphism exactly when postcomposition, equivalently precomposition, induces bijections on every hom-collection

Statement

For a morphism $f : A \to B$ , the following are equivalent: $f$ is an isomorphism; for every object $X$ , postcomposition $f \circ - : C (X, A) \to C (X, B)$ is bijective; and for every $X$ , precomposition $- \circ f : C (B, X) \to C (A, X)$ is bijective.

Facts & Assumptions

Given: A morphism $f : A \to B$ in a category $C$ .

[L1]

Isomorphisms have two-sided inverses (Isomorphism, groupoid, and connected category), and a map is bijective exactly when it has a two-sided inverse (Injection, surjection, bijection).

[L2]

Reversing arrows exchanges postcomposition with precomposition (Every theorem about categories has a formal dual obtained by reversing morphisms and composition).

Proof

technique · direct

1.1

If $f$ has inverse $f^{- 1}$ , postcomposition by $f^{- 1}$ is a two-sided inverse to postcomposition by $f$ , and similarly precomposition by $f^{- 1}$ inverts precomposition by $f$ ; both maps are bijections.

givenL1

2.1

Conversely, suppose every postcomposition map is bijective. Surjectivity at $X = B$ gives $g : B \to A$ with $f \circ g = 1_{B}$ ; injectivity at $X = A$ applied to $f \circ (g \circ f) = f = f \circ 1_{A}$ gives $g \circ f = 1_{A}$ , so $f$ is an isomorphism.

step 1.1L1

3.1

The identical argument in $C^{o p}$ , using [L2], proves that bijectivity of every precomposition map also characterises isomorphisms.

step 2.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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)