Equivalence Relations · step 6 of 13

The kernel of a function as an equivalence

Every function f : A → B induces an equivalence relation on A by a ∼ b iff f(a) = f(b), called the kernel of f. Its classes are the fibres f⁻¹(y), and the quotient A/ker f is in bijection with the image of f. This is the first isomorphism theorem in its simplest form, and it means every equivalence relation arises as a kernel — take the projection map.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Define the relation from fa ∼ b iff f(a) = f(b). Reflexivity, symmetry and transitivity are inherited directly from equality.
Identify the classes as fibresThe class of a is f⁻¹(f(a)) — everything mapping to the same output.
Count classes via the imageThe number of classes equals |image of f|. Elements of B never hit contribute no class.
Read off injectivityf is injective exactly when every class is a singleton, i.e. when ker f is equality.

Worked example

For f : ℤ → ℤ given by f(n) = n², describe the classes of ker f.

n ∼ m iff n² = m², i.e. iff m = ±n.
So [n] = {n, −n} for every n.
For n = 0 this collapses: [0] = {0}, a singleton.
Every other class has exactly two elements.

Answer. The classes are {0} and the pairs {n, −n} for n ≥ 1 — so ℤ/ker f is in bijection with the perfect squares, the image of f.

Where marks get dropped

These are the specific errors that cost credit on the kernel of a function as an equivalence questions — QED's rubric penalises each of them separately.

Confusing this kernel with the group-theoretic kernel f⁻¹(0). For general functions there is no zero, and the equivalence-relation kernel is the right notion.
Counting classes as |B| rather than |image f|. Unhit elements of the codomain give empty fibres, which are not classes.
Forgetting the degenerate class. For f(n) = n² the class of 0 has one element, not two.

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The kernel of a function as an equivalence — frequently asked questions

Does every equivalence relation come from a function?

Yes — take the canonical projection π : A → A/∼. Its kernel is exactly ∼, so kernels and equivalence relations are the same thing viewed differently.

What is the link to the first isomorphism theorem?

The theorem says A/ker f is in bijection with im f, via [a] ↦ f(a). In algebra the same statement upgrades from sets to groups, rings and modules.

How does this help in practice?

It is the standard way to build a well-defined map out of a quotient: show your function is constant on classes, and it factors through the quotient automatically.

The rest of Equivalence Relations

Equivalence classes, partitions and quotient sets. Each subtopic below has its own method, worked example and mark-losing traps.

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