# Kernel (set theory)

Short description: Equivalence relation expressing that two elements have the same image under a function

In set theory, the kernel of a function $\displaystyle{ f }$ (or equivalence kernel[1]) may be taken to be either

• the equivalence relation on the function's domain that roughly expresses the idea of "equivalent as far as the function $\displaystyle{ f }$ can tell",[2] or
• the corresponding partition of the domain.

An unrelated notion is that of the kernel of a non-empty family of sets $\displaystyle{ \mathcal{B}, }$ which by definition is the intersection of all its elements: $\displaystyle{ \ker \mathcal{B} ~=~ \bigcap_{B \in \mathcal{B}} \, B. }$ This definition is used in the theory of filters to classify them as being free or principal.

## Definition

Kernel of a function

For the formal definition, let $\displaystyle{ f : X \to Y }$ be a function between two sets. Elements $\displaystyle{ x_1, x_2 \in X }$ are equivalent if $\displaystyle{ f\left(x_1\right) }$ and $\displaystyle{ f\left(x_2\right) }$ are equal, that is, are the same element of $\displaystyle{ Y. }$ The kernel of $\displaystyle{ f }$ is the equivalence relation thus defined.[2]

Kernel of a family of sets

The kernel of a family $\displaystyle{ \mathcal{B} \neq \varnothing }$ of sets is[3] $\displaystyle{ \ker \mathcal{B} ~:=~ \bigcap_{B \in \mathcal{B}} B. }$ The kernel of $\displaystyle{ \mathcal{B} }$ is also sometimes denoted by $\displaystyle{ \cap \mathcal{B}. }$ The kernel of the empty set, $\displaystyle{ \ker \varnothing, }$ is typically left undefined. A family is called fixed and is said to have non-empty intersection if its kernel is not empty.[3] A family is said to be free if it is not fixed; that is, if its kernel is the empty set.[3]

## Quotients

Like any equivalence relation, the kernel can be modded out to form a quotient set, and the quotient set is the partition: $\displaystyle{ \left\{\, \{w \in X : f(x) = f(w)\} ~:~ x \in X \,\right\} ~=~ \left\{f^{-1}(y) ~:~ y \in f(X)\right\}. }$

This quotient set $\displaystyle{ X /=_f }$ is called the coimage of the function $\displaystyle{ f, }$ and denoted $\displaystyle{ \operatorname{coim} f }$ (or a variation). The coimage is naturally isomorphic (in the set-theoretic sense of a bijection) to the image, $\displaystyle{ \operatorname{im} f; }$ specifically, the equivalence class of $\displaystyle{ x }$ in $\displaystyle{ X }$ (which is an element of $\displaystyle{ \operatorname{coim} f }$) corresponds to $\displaystyle{ f(x) }$ in $\displaystyle{ Y }$ (which is an element of $\displaystyle{ \operatorname{im} f }$).

## As a subset of the square

Like any binary relation, the kernel of a function may be thought of as a subset of the Cartesian product $\displaystyle{ X \times X. }$ In this guise, the kernel may be denoted $\displaystyle{ \ker f }$ (or a variation) and may be defined symbolically as[2] $\displaystyle{ \ker f := \{(x,x') : f(x) = f(x')\}. }$

The study of the properties of this subset can shed light on $\displaystyle{ f. }$

## Algebraic structures

If $\displaystyle{ X }$ and $\displaystyle{ Y }$ are algebraic structures of some fixed type (such as groups, rings, or vector spaces), and if the function $\displaystyle{ f : X \to Y }$ is a homomorphism, then $\displaystyle{ \ker f }$ is a congruence relation (that is an equivalence relation that is compatible with the algebraic structure), and the coimage of $\displaystyle{ f }$ is a quotient of $\displaystyle{ X. }$[2] The bijection between the coimage and the image of $\displaystyle{ f }$ is an isomorphism in the algebraic sense; this is the most general form of the first isomorphism theorem.

## In topology

If $\displaystyle{ f : X \to Y }$ is a continuous function between two topological spaces then the topological properties of $\displaystyle{ \ker f }$ can shed light on the spaces $\displaystyle{ X }$ and $\displaystyle{ Y. }$ For example, if $\displaystyle{ Y }$ is a Hausdorff space then $\displaystyle{ \ker f }$ must be a closed set. Conversely, if $\displaystyle{ X }$ is a Hausdorff space and $\displaystyle{ \ker f }$ is a closed set, then the coimage of $\displaystyle{ f, }$ if given the quotient space topology, must also be a Hausdorff space.

A space is compact if and only if the kernel of every family of closed subsets having the finite intersection property (FIP) is non-empty;[4][5] said differently, a space is compact if and only if every family of closed subsets with F.I.P. is fixed.