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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Every function $f : A \to B$ factors as $f = ι \circ \overset{ˉ}{f} \circ π$ with $π : A \to A / \ker f$ the surjective quotient map, $\overset{ˉ}{f} : A / \ker f \to f [A]$ the bijection $[a] \mapsto f (a)$ , and $ι : f [A] \to B$ the inclusion injection; $\overset{ˉ}{f}$ is the only function making the factorisation hold

Statement

Let $f : A \to B$ , write $\sim$ for $\ker f$ and let $π : A \to A / \sim$ be the quotient map. Put $ι : = Δ_{f [A]}$ , the identity relation on $f [A]$ . Then

(i) $π$ is surjective;
(ii) there is a function $\overset{ˉ}{f} : A / \sim \to f [A]$ with $\overset{ˉ}{f} ([a]) = f (a)$ for every $a \in A$ , and it is a bijection;
(iii) $ι$ is an injective function $f [A] \to B$ with $ι (y) = y$ ;
(iv) $ι \circ \overset{ˉ}{f} \circ π = f$ ;
(v) $\overset{ˉ}{f}$ is the only function $A / \sim \to f [A]$ satisfying (iv).

Facts & Assumptions

Given: a function $f : A \to B$ , with $\sim$ denoting $\ker f$ .

[L1]

$\ker f$ is a set, is a relation on $A$ , and is an equivalence relation on $A$ (For a function $f : A \to B$ the relation $\ker f : = {(a, a^{'}) \in A \times A : f (a) = f (a^{'})}$ is an equivalence relation on $A$ ).

[L2]

$π$ is a surjective function $A \to A / \sim$ with $π (a) = [a]$ (Let $\sim$ be an equivalence relation on $A$ with quotient map $π : A \to A / \sim$ , and let $f : A \to B$ . There is a function $g : A / \sim \to B$ with $g \circ π = f$ if and only if $a \sim a^{'}$ implies $f (a) = f (a^{'})$ ; and such a $g$ is then unique).

[L3]

there is a function $g : A / \sim \to B$ with $g \circ π = f$ if and only if $f (a) = f (a^{'})$ whenever $a \sim a^{'}$ (Let $\sim$ be an equivalence relation on $A$ with quotient map $π : A \to A / \sim$ , and let $f : A \to B$ . There is a function $g : A / \sim \to B$ with $g \circ π = f$ if and only if $a \sim a^{'}$ implies $f (a) = f (a^{'})$ ; and such a $g$ is then unique).

[L4]

when such a $g$ exists it is unique, and it satisfies $g ([a]) = f (a)$ for every $a \in A$ (Let $\sim$ be an equivalence relation on $A$ with quotient map $π : A \to A / \sim$ , and let $f : A \to B$ . There is a function $g : A / \sim \to B$ with $g \circ π = f$ if and only if $a \sim a^{'}$ implies $f (a) = f (a^{'})$ ; and such a $g$ is then unique).

[L5]

$A / \sim : = {[a] : a \in A}$ (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[L6]

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

[L7]

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

[L8]

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).

[L9]

$b \in R [A]$ 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).

[L10]

$Δ_{A}$ 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 $f : A \to B$ ).

[L11]

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

[L12]

$f = g$ 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).

[L13]

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

[L14]

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

[L15]

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

[L16]

$x \subseteq y$ means that every element of $x$ is an element of $y$ (Subset $x \subseteq y$ , proper subset $x ⊊ y$ , and the separation notation ${z \in x : φ (z)}$ ).

[L17]

$a \sim b$ if and only if $[a] = [b]$ (The equivalence classes of an equivalence relation are nonempty, cover $A$ , and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation).

Proof

technique · direct

1.1

$\sim$ is an equivalence relation on $A$ and $a \sim a^{'}$ says exactly $f (a) = f (a^{'})$ ; so the hypothesis of the universal property is satisfied, and clause (i) is the surjectivity already recorded for the quotient map.

L1L2L3

1.2

Claim (iii): $ι = Δ_{f [A]}$ is a function with domain $f [A]$ and $ι (y) = y$ ; its range is $f [A]$ , which is contained in $B$ because every element of $f [A]$ is a value of $f$ and $ran f \subseteq B$ . It is injective, since $ι (y) = ι (y^{'})$ reads $y = y^{'}$ .

L6L8L9L10L13L14L16

2.1

Claim (ii): the universal property gives a function $g : A / \sim \to B$ with $g \circ π = f$ and $g ([a]) = f (a)$ for every $a \in A$ . Every value of $g$ is therefore a value of $f$ , so $ran g \subseteq f [A]$ , and the same set $g$ is a function $A / \sim \to f [A]$ ; call it $\overset{ˉ}{f}$ . It is onto $f [A]$ , because each element of $f [A]$ is $f (a) = \overset{ˉ}{f} ([a])$ for some $a \in A$ ; and it is injective, because $\overset{ˉ}{f} ([a]) = \overset{ˉ}{f} ([a^{'}])$ gives $f (a) = f (a^{'})$ , that is $a \sim a^{'}$ , and therefore $[a] = [a^{'}]$ . Hence $\overset{ˉ}{f}$ is a bijection.

L3L4L5L6L7L8L9L13L17step 1.1

3.1

Claim (iv): $ι \circ \overset{ˉ}{f} \circ π$ is unambiguous by associativity, it is a function with domain $A$ , and for $a \in A$ its value is $ι (\overset{ˉ}{f} (π (a))) = ι (f (a)) = f (a)$ ; so it equals $f$ .

L11L12L15step 1.2step 2.1

3.2

Claim (v): if $h : A / \sim \to f [A]$ also satisfies $ι \circ h \circ π = f$ , then for $C \in A / \sim$ choose $a$ with $C = [a]$ and compute $h (C) = ι (h (π (a))) = f (a) = \overset{ˉ}{f} (C)$ ; both have domain $A / \sim$ , so $h = \overset{ˉ}{f}$ .

L5L10L11L12step 1.2step 2.1

4.1

Claims (i) to (v) are established, which is the statement.

step 1.1step 1.2step 2.1step 3.1step 3.2∎

Depends on

Used by

A two-to-one function on a four-element set, its kernel, the quotient, and the canonical factorisation computed explicitly Example

Dependency tree · two levels

27 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

Equivalence relation (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §3.2 (standard reference, not scraped)
Image (mathematics) (Wikipedia) (standard reference, not scraped)