TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)

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.

$f : A \to B$ is a bijection if and only if there is a function $g : B \to A$ with $g \circ f = Δ_{A}$ and $f \circ g = Δ_{B}$ ; such a $g$ is unique, equals the inverse relation $f^{- 1}$ , and is itself a bijection

Statement

Let $f : A \to B$ . Then $f$ is a bijection if and only if there is a function $g : B \to A$ with $g \circ f = Δ_{A}$ and $f \circ g = Δ_{B}$ . When such a $g$ exists it is unique, it is the inverse relation $f^{- 1}$ , and it is itself a bijection $B \to A$ .

No choice principle is used: the value $g (b)$ is the unique $a$ with $f (a) = b$ , so it is determined rather than selected.

Facts & Assumptions

Given: a function $f : A \to B$ .

[L1]

$f$ is injective (one-to-one) if $f (x) = f (y)$ implies $x = y$ , for all $x, y \in A$ (Injection, surjection, bijection).

[L2]

$f$ is surjective (onto) if for every $b \in B$ there is some $x \in A$ with $f (x) = b$ (Injection, surjection, bijection).

[L3]

We write $f : A \to B$ , and say $f$ is a function from $A$ to $B$ , when $f$ is a function with $dom f = A$ and $ran f \subseteq B$ (A function is a relation $f$ with $(a, b) \in f$ and $(a, c) \in f$ implying $b = c$ ; $f : A \to B$ , the value $f (a)$ , domain and codomain).

[L4]

$(b, a) \in R^{- 1}$ holds if and only if $(a, b) \in R$ (The inverse relation $R^{- 1}$ , the composite $S \circ R$ , and the restriction $R ↾ A$ ).

[L5]

$dom (R^{- 1}) = ran R$ and $ran (R^{- 1}) = dom R$ ( $T \circ (S \circ R) = (T \circ S) \circ R$ , $(S \circ R)^{- 1} = R^{- 1} \circ S^{- 1}$ , $(R^{- 1})^{- 1} = R$ , $dom (R^{- 1}) = ran R$ , and $Δ_{B} \circ R = R = R \circ Δ_{A}$ for a relation $R$ from $A$ to $B$ ).

[L6]

$g \circ f$ is a function, $dom (g \circ f) = f^{- 1} [dom g]$ , and $(g \circ f) (x) = g (f (x))$ for every $x$ in that domain (If $f$ and $g$ are functions then $g \circ f$ is a function with domain $f^{- 1} [dom g]$ and $(g \circ f) (x) = g (f (x))$ there; $Δ_{A}$ is a function with $Δ_{A} (a) = a$ ; and $f \circ Δ_{A} = f = Δ_{B} \circ f$ for $f : A \to B$ ).

[L7]

$Δ_{A}$ is a function with $dom Δ_{A} = A$ and $Δ_{A} (a) = a$ for every $a \in A$ (If $f$ and $g$ are functions then $g \circ f$ is a function with domain $f^{- 1} [dom g]$ and $(g \circ f) (x) = g (f (x))$ there; $Δ_{A}$ is a function with $Δ_{A} (a) = a$ ; and $f \circ Δ_{A} = f = Δ_{B} \circ f$ for $f : A \to B$ ).

[L8]

$f = g$ if and only if $dom f = dom g$ and $f (x) = g (x)$ for every $x \in dom f$ (Functions $f$ and $g$ are equal if and only if $dom f = dom g$ and $f (x) = g (x)$ for every $x$ in that common domain).

[L9]

$ran R : = {b : \exists a (a, b) \in R}$ (Relation, $dom R$ , $ran R$ , $fld R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L10]

$(a, b) \in Δ_{A}$ holds if and only if $a = b$ and $a \in A$ (The identity relation $Δ_{A} = {(a, b) \in A \times A : a = b}$ and the membership relation $\in_{A} = {(a, b) \in A \times A : a \in b}$ ).

Proof

technique · direct

1.1

Suppose $f$ is a bijection. The inverse relation $f^{- 1}$ is a function: if $(b, a)$ and $(b, a^{'})$ lie in it then $f (a) = b = f (a^{'})$ , so $a = a^{'}$ by injectivity. Its domain is $ran f$ , which is $B$ by surjectivity, and its range is $dom f = A$ ; hence $f^{- 1} : B \to A$ .

L1L2L3L4L5L9

1.2

Conversely, suppose $g : B \to A$ satisfies $g \circ f = Δ_{A}$ and $f \circ g = Δ_{B}$ . If $f (a) = f (a^{'})$ then $a = g (f (a)) = g (f (a^{'})) = a^{'}$ , so $f$ is injective; and any $b \in B$ satisfies $b = Δ_{B} (b) = f (g (b))$ , so $b$ is a value of $f$ and $f$ is surjective. Hence $f$ is a bijection.

L1L2L6L7L10

1.3

Any two such functions agree: if $g$ and $g^{'}$ both satisfy the two identities then, for $b \in B$ , $f (g^{'} (b)) = b$ , so $g (b) = g (f (g^{'} (b))) = Δ_{A} (g^{'} (b)) = g^{'} (b)$ ; both have domain $B$ , so $g = g^{'}$ .

L6L7L8

2.1

For a bijection $f$ , the function $f^{- 1}$ of step 1.1 satisfies the two identities: $f^{- 1} \circ f$ and $Δ_{A}$ are functions with domain $A$ , and $f^{- 1} (f (a)) = a = Δ_{A} (a)$ for $a \in A$ ; likewise $f \circ f^{- 1}$ and $Δ_{B}$ are functions with domain $B$ agreeing at every point.

L4L6L7L8step 1.1

2.2

Such a $g$ is itself a bijection: $f$ is a function $A \to B$ with $f \circ g = Δ_{B}$ and $g \circ f = Δ_{A}$ , which is the hypothesis of step 1.2 applied to $g$ in place of $f$ .

step 1.2

3.1

Such a $g$ is the inverse relation: $f^{- 1}$ satisfies the two identities by step 2.1, and step 1.3 says there is only one function that does.

step 1.3step 2.1

4.1

The two directions, the uniqueness, the identification with $f^{- 1}$ and the bijectivity of $g$ are established, which is the statement.

step 1.1step 1.2step 1.3step 2.1step 2.2step 3.1∎

Depends on

Used by

Dependency tree · two levels

19 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

Bijection, injection and surjection (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §2.2 (standard reference, not scraped)
Inverse function (Wikipedia) (standard reference, not scraped)

f:A→Bf:A→B is a bijection if and only if there is a function g:B→Ag:B→A with g∘f=ΔAg∘f=ΔA​ and f∘g=ΔBf∘g=ΔB​; such a gg is unique, equals the inverse relation f−1f−1, and is itself a bijection