CorollaryStatement: AI-adaptedProof: 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.

For $f : A \to B$ with $A \neq \emptyset$ : $f$ is injective if and only if there is $g : B \to A$ with $g \circ f = Δ_{A}$ ; for $A = \emptyset$ the empty function is injective and has a left inverse if and only if $B = \emptyset$

Statement

Let $f : A \to B$ .

(i) If $A \neq \emptyset$ , then $f$ is injective if and only if there is a function $g : B \to A$ with $g \circ f = Δ_{A}$ .
(ii) If $A = \emptyset$ , then $f = \emptyset$ and $f$ is injective; and there is a function $g : B \to A$ with $g \circ f = Δ_{A}$ if and only if $B = \emptyset$ .

The hypothesis $A \neq \emptyset$ in (i) is not removable: for $A = \emptyset$ and $B \neq \emptyset$ the empty function is injective and has no left inverse at all.

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]

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).

[L3]

$g \circ f$ 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 $f : A \to B$ ).

[L4]

$Δ_{A}$ 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 $f : A \to B$ ).

[L5]

$(a, b) \in Δ_{A}$ 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 $\in_{A} = {(a, b) \in A \times A : a \in b}$ ).

[L6]

For any parameters $\overset{ˉ}{p}$ and any set $x$ , there is a set $y$ whose elements are exactly the elements $z$ of $x$ for which $φ (z, \overset{ˉ}{p})$ holds (The Axiom Schema of Separation: for each formula $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ ).

[L7]

$b \in R [A]$ holds if and only if $(a, b) \in R$ for some $a \in A$ (The image $R [A]$ and the preimage $R^{- 1} [B]$ of a set under a relation).

[L8]

There is exactly one set with no elements, written $\emptyset$ (There is exactly one set with no elements, written $\emptyset$ ).

[L9]

$z \in A \times B$ holds if and only if $z = (a, b)$ for some $a \in A$ and some $b \in B$ (The Cartesian product $A \times B : = {z \in P (P (A \cup B)) : \exists a \in A \exists b \in B z = (a, b)}$ ).

[L10]

$A \times B = \emptyset$ if and only if $A = \emptyset$ or $B = \emptyset$ ( $A \times (B \cup C) = (A \times B) \cup (A \times C)$ , $A \times (B \cap C) = (A \times B) \cap (A \times C)$ , $A \times (B ∖ C) = (A \times B) ∖ (A \times C)$ , $(A \cap B) \times (C \cap D) = (A \times C) \cap (B \times D)$ ; $A \times B = \emptyset$ if and only if $A = \emptyset$ or $B = \emptyset$ ; and for nonempty $A$ and $B$ , $A \times B \subseteq C \times D$ if and only if $A \subseteq C$ and $B \subseteq D$ ).

[L11]

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

Proof

technique · direct

1.1

Claim (i), from right to left: if $g \circ f = Δ_{A}$ and $f (a) = f (a^{'})$ for $a, a^{'} \in A$ , then $a = Δ_{A} (a) = g (f (a)) = g (f (a^{'})) = Δ_{A} (a^{'}) = a^{'}$ .

L1L3L4

1.2

Claim (i), from left to right: assume $f$ injective and $A \neq \emptyset$ , and fix $a_{0} \in A$ . Separating inside $B \times A$ gives the set $g : = {(b, a) \in B \times A : (a, b) \in f or (b \notin f [A] and a = a_{0})}$ . For $b \in f [A]$ the first alternative supplies exactly one $a$ , by injectivity, and the second supplies none; for $b \in B$ with $b \notin f [A]$ the first supplies none, since $ran f = f [A]$ , and the second supplies $a_{0}$ alone. Hence $g$ is a function with domain $B$ and range inside $A$ , so $g : B \to A$ .

L1L2L6L7L8L9L11

1.3

Claim (ii): if $A = \emptyset$ then $dom f = \emptyset$ , so $f$ has no element and $f = \emptyset$ ; the injectivity condition quantifies over elements of $A$ and holds vacuously.

L1L2L8L11

2.1

Claim (i) concluded: with $g$ as in step 1.2, $g \circ f$ and $Δ_{A}$ are functions with domain $A$ , and $g (f (a)) = a$ for every $a \in A$ , since $f (a) \in f [A]$ selects the first alternative; so the two functions are equal.

L3L4L5L7step 1.2

2.2

Claim (ii) concluded: a function $g : B \to \emptyset$ has $ran g \subseteq \emptyset$ , so $g \subseteq B \times \emptyset = \emptyset$ , giving $g = \emptyset$ and $B = dom g = \emptyset$ . Conversely, if $B = \emptyset$ then $g : = \emptyset$ is a function $\emptyset \to \emptyset$ , and $g \circ f$ and $Δ_{\emptyset}$ are both $\emptyset$ , so the identity holds.

L2L3L5L8L9L10L11step 1.3

3.1

Claims (i) and (ii) are established, which is the statement.

step 1.1step 1.2step 1.3step 2.1step 2.2∎

Remarks

Why the surjective analogue is missing. The corresponding statement, that every surjection has a right inverse, is not proved on this page: it is equivalent to the Axiom of Choice, which this page states at The Axiom of Choice and assumes in no proof, and choosing one preimage for each point of the codomain is exactly the simultaneous selection no proof here makes. The Axiom of Choice is stated on this page and assumed by no proof on it; the two statements that would need it are identified and left unsettled records where that boundary falls.

Depends on

Used by

The Axiom of Choice is stated on this page and assumed by no proof on it; the two statements that would need it are identified and left unsettled Remark

Dependency tree · two levels

24 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)
Inverse function (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §2.2 (standard reference, not scraped)

For f:A→Bf:A→B with A≠∅A=∅: ff is injective if and only if there is g:B→Ag:B→A with g∘f=ΔAg∘f=ΔA​; for A=∅A=∅ the empty function is injective and has a left inverse if and only if B=∅B=∅