Equivalence relation expressing that two elements have the same image under a function
In set theory, the kernel of a function f {\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 f {\displaystyle f} can tell",[2] or
An unrelated notion is that of the kernel of a non-empty family of sets B , {\displaystyle {\mathcal {B}},} which by definition is the intersection of all its elements: ker B = ⋂ B ∈ B B . {\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.
Kernel of a function
For the formal definition, let f : X → Y {\displaystyle f:X\to Y} be a function between two sets. Elements x 1 , x 2 ∈ X {\displaystyle x_{1},x_{2}\in X} are equivalent if f ( x 1 ) {\displaystyle f\left(x_{1}\right)} and f ( x 2 ) {\displaystyle f\left(x_{2}\right)} are equal, that is, are the same element of Y . {\displaystyle Y.} The kernel of f {\displaystyle f} is the equivalence relation thus defined.[2]
Kernel of a family of sets
The kernel of a family B ≠ ∅ {\displaystyle {\mathcal {B}}\neq \varnothing } of sets is[3] ker B := ⋂ B ∈ B B . {\displaystyle \ker {\mathcal {B}}~:=~\bigcap _{B\in {\mathcal {B}}}B.} The kernel of B {\displaystyle {\mathcal {B}}} is also sometimes denoted by ∩ B . {\displaystyle \cap {\mathcal {B}}.} The kernel of the empty set, ker ∅ , {\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]
Like any equivalence relation, the kernel can be modded out to form a quotient set, and the quotient set is the partition: { { w ∈ X : f ( x ) = f ( w ) } : x ∈ X } = { f − 1 ( y ) : y ∈ f ( X ) } . {\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 X / = f {\displaystyle X/=_{f}} is called the coimage of the function f , {\displaystyle f,} and denoted coim f {\displaystyle \operatorname {coim} f} (or a variation). The coimage is naturally isomorphic (in the set-theoretic sense of a bijection) to the image, im f ; {\displaystyle \operatorname {im} f;} specifically, the equivalence class of x {\displaystyle x} in X {\displaystyle X} (which is an element of coim f {\displaystyle \operatorname {coim} f} ) corresponds to f ( x ) {\displaystyle f(x)} in Y {\displaystyle Y} (which is an element of im f {\displaystyle \operatorname {im} f} ).
Like any binary relation, the kernel of a function may be thought of as a subset of the Cartesian product X × X . {\displaystyle X\times X.} In this guise, the kernel may be denoted ker f {\displaystyle \ker f} (or a variation) and may be defined symbolically as[2] ker f := { ( x , x ′ ) : f ( x ) = f ( x ′ ) } . {\displaystyle \ker f:=\{(x,x'):f(x)=f(x')\}.}
The study of the properties of this subset can shed light on f . {\displaystyle f.}
If X {\displaystyle X} and Y {\displaystyle Y} are algebraic structures of some fixed type (such as groups, rings, or vector spaces), and if the function f : X → Y {\displaystyle f:X\to Y} is a homomorphism, then ker f {\displaystyle \ker f} is a congruence relation (that is an equivalence relation that is compatible with the algebraic structure), and the coimage of f {\displaystyle f} is a quotient of X . {\displaystyle X.} [2]The bijection between the coimage and the image of f {\displaystyle f} is an isomorphism in the algebraic sense; this is the most general form of the first isomorphism theorem.
If f : X → Y {\displaystyle f:X\to Y} is a continuous function between two topological spaces then the topological properties of ker f {\displaystyle \ker f} can shed light on the spaces X {\displaystyle X} and Y . {\displaystyle Y.} For example, if Y {\displaystyle Y} is a Hausdorff space then ker f {\displaystyle \ker f} must be a closed set. Conversely, if X {\displaystyle X} is a Hausdorff space and ker f {\displaystyle \ker f} is a closed set, then the coimage of f , {\displaystyle f,} if given the quotient space topology, must also be a Hausdorff space.
