Kernel (set theory)

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In set theory, the kernel of a function f may be taken to be either


For the formal definition, let X and Y be sets and let f be a function from X to Y. Elements x1 and x2 of X are equivalent if f(x1) and f(x2) are equal, i.e. are the same element of Y. The kernel of f is the equivalence relation thus defined.[1]


Like any equivalence relation, the kernel can be modded out to form a quotient set, and the quotient set is the partition:

This quotient set is called the coimage of the function , and denoted (or a variation). The coimage is naturally isomorphic (in the set-theoretic sense of a bijection) to the image, ; specifically, the equivalence class of in (which is an element of ) corresponds to in (which is an element of ).

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 X × X. In this guise, the kernel may be denoted "ker f" (or a variation) and may be defined symbolically as


The study of the properties of this subset can shed light on .

In algebraic structures

If X and Y are algebraic structures of some fixed type (such as groups, rings, or vector spaces), and if the function f from X to Y is a homomorphism, then ker f will be a subalgebra of the direct product X × X. Subalgebras of X × X that are also equivalence relations (called congruence relations) are important in abstract algebra, because they define the most general notion of quotient algebra.[1] Thus the coimage of f is a quotient algebra of X much as the image of f is a subalgebra of Y; and the bijection between them becomes an isomorphism in the algebraic sense as well (this is the most general form of the first isomorphism theorem in algebra). The use of kernels in this context is discussed further in the article Kernel (algebra).

In topological spaces

If X and Y are topological spaces and f is a continuous function between them, then the topological properties of ker f can shed light on the spaces X and Y. For example, if Y is a Hausdorff space, then ker f must be a closed set. Conversely, if X is a Hausdorff space and ker f is a closed set, then the coimage of f, if given the quotient space topology, must also be a Hausdorff space.


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