1.3. Functors 9 Let’s use these ideas to prove something topological.4 Problem 1.13. Let i : S1 D2 be the inclusion of the circle into the disk. An early success of algebraic topology concerned the existence of a continuous function r : D2 S1 such that the composite r i : S1 S1 is the identity (in other words: is S1 a retract of D2?). The fundamental group is a covariant functor π1 : Top −→ G from the category Top of topological spaces and continuous functions to the category G of groups and homomorphisms. You may have seen, and we will show later, that π1(S1) = Z and π1(D2) = 0. Using this functor, show that there can be no such function r. The second kind of functor is just the same, except that it reverses the direction of arrows. A contravariant functor F : C D consists of a function F : ob(C) −→ ob(D) and, for each X, Y ob(C), a function F : morC(X, Y ) −→ morD(F (Y ),F (X)). These must satisfy the following conditions: (1) F (g f) = F (f) F (g). (2) F (idX) = idF (X) for any X ob(C). In contrast to the covariant functors, if F : C D is a contravariant functor and f : X Y in C, then F (f) : F (Y ) F (X). Thus a contravari- ant functor carries the domain of f to the target of F (f) and carries the target of f to the domain of F (f)—it ‘reverses the direction’ of arrows. Exercise 1.14. Let F : C D be a contravariant functor. (a) Show that if you apply F to a commutative diagram in C, the result is a commutative diagram in D. (b) Show that if f : X Y is an equivalence in C, then F (f) is an equiva- lence in D. Exercise 1.15. Show that the composite of two functors is a functor. What happens if one or both of the functors is contravariant? Exercise 1.16. Can there be a functor F : C D that is both covariant and contravariant? What special properties must such a functor have? 4 I.e., interesting!
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