LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-06 (claude-opus-5)

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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 $A$

Statement

Let $f : A \to B$ be a function. Then

$\ker f \;:=\; \{\,(a,a') \in A \times A : f(a) = f(a')\,\}$

is a set, is a relation on $A$ , and is an equivalence relation on $A$ .

Facts & Assumptions

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

[L1]

We write $f : A \to B$ , and say $f$ is a function from $A$ to $B$ , when $f$ is a function with $\operatorname{dom} f = A$ and $\operatorname{ran} f \subseteq B$ (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).

[L2]

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

[L3]

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

[L4]

$R$ is a relation on $A$ when $R \subseteq A \times A$ (Relation, $\operatorname{dom} R$ , $\operatorname{ran} R$ , $\operatorname{fld} R$ , and the specialisations "relation from $A$ to $B$ " and "relation on $A$ ").

[L5]

reflexive: $a \sim a$ for every $a \in A$ (Equivalence relation, equivalence class, and the quotient set $A/{\sim}$ ).

[L6]

symmetric: $a \sim b$ implies $b \sim a$ , for all $a, b \in A$ (Equivalence relation, equivalence class, and the quotient set $A/{\sim}$ ).

[L7]

transitive: $a \sim b$ and $b \sim c$ imply $a \sim c$ , for all $a, b, c \in A$ (Equivalence relation, equivalence class, and the quotient set $A/{\sim}$ ).

[L8]

$R$ is transitive when $(a,b) \in R$ and $(b,c) \in R$ imply $(a,c) \in R$ , for all $a, b, c \in A$ (Reflexive, irreflexive, symmetric, asymmetric, antisymmetric, transitive, and connex relations on a set).

Proof

technique · direct

1.1

$\ker f$ is a set and a relation on $A$ : it is obtained by separating inside $A \times A$ with the formula $\exists a\,\exists a'\,(z = (a,a') \wedge f(a) = f(a'))$ and the parameters $A$ and $f$ , so it is a subset of $A \times A$ and each of its elements is an ordered pair.

L1L2L3L4

2.1

It is reflexive on $A$ : for $a \in A$ the value $f(a)$ is a single set, so $f(a) = f(a)$ and $(a,a) \in \ker f$ .

L1L5step 1.1

2.2

It is symmetric: if $(a,a') \in \ker f$ then $f(a) = f(a')$ , hence $f(a') = f(a)$ and $(a',a) \in \ker f$ .

L6step 1.1

2.3

It is transitive: if $(a,a') \in \ker f$ and $(a',a'') \in \ker f$ then $f(a) = f(a')$ and $f(a') = f(a'')$ , hence $f(a) = f(a'')$ and $(a,a'') \in \ker f$ .

L7L8step 1.1

3.1

$\ker f$ is a relation on $A$ that is reflexive on $A$ , symmetric and transitive, which is what it means to be an equivalence relation on $A$ .

step 1.1step 2.1step 2.2step 2.3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 24 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)
Kernel (set theory) (Wikipedia) (standard reference, not scraped)

For a function f:A→Bf : A \to Bf:A→B the relation ker⁡f:={ (a,a′)∈A×A:f(a)=f(a′) }\ker f := \{\,(a,a') \in A \times A : f(a) = f(a')\,\}kerf:={(a,a′)∈A×A:f(a)=f(a′)} is an equivalence relation on AAA