PlanetMath: functor

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (213)
Orphanage
Unclass'd (1)
Unproven (473)
Corrections (39)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

functor

(Definition)

Given two categories $\mathcal{C}$ and $\mathcal{D}$ , a covariant functor $T:\mathcal{C}\to\mathcal{D}$ consists of an assignment for each object of $\mathcal{C}$ an object of $\mathcal{D}$ (i.e. a “function” $T:{\rm Ob}(\mathcal{C})\to{\rm Ob}(\mathcal{D})$ ) together with an assignment for every morphism $f\in{\rm Hom}_{\mathcal{C}}(A,B)$ , to a morphism $T(f)\in{\rm Hom}_{\mathcal{D}}(T(A),T(B))$ , such that:

$T(1_A) = 1_{T(A)}$ where denotes the identity morphism on the object (in the respective category).
$T(g \circ f) = T(g)\circ T(f)$ , whenever the composition $g\circ f$ is defined.

A contravariant functor $T :\mathcal{C}\to\mathcal{D}$ is just a covariant functor $T:\mathcal{C}^{\rm op}\to\mathcal{D}$ from the opposite category. In other words, the assignment reverses the direction of maps. If $f\in{\rm Hom}_{\mathcal{C}}(A,B)$ , then $T(f)\in{\rm Hom}_{\mathcal{D}}(T(B),T(A))$ and $T(g\circ f) = T(f)\circ T(g)$ whenever the composition is defined (the domain of is the same as the codomain of ).

Given a category $\mathcal{C}$ and an object we always have the functor $T : \mathcal{C}\to{\bf Sets}$ to the category of sets defined on objects by $T(A) = {\rm Hom}(X, A)$ . If $f : A \to B$ is a morphism of $\mathcal{C}$ , then we define $T(f) : {\rm Hom}(X,A)\to {\rm Hom}(X,B)$ by $g\mapsto f\circ g$ . This is a covariant functor, denoted by ${\rm Hom}(X,-)$ .

Similarly, one can define a contravariant functor ${\rm Hom}(-,X) :\mathcal{C}\to{\bf Sets}$ .

"functor" is owned by nerdy2.

(view preamble)