What does it mean to compose two Functors?

Question

Exercise 5 of the Haskell Typeclassopedia Section 3.2 asks for a proof or counterexample on the statement: *The composition of two Functors is also a Functor.* I thought at first that this was talking about composing the `fmap` methods defined by two separate instances of a `Functor`, but that doesn't really make sense, since the types wouldn't match up as far as I can tell. For two types `f` and `f'`, the types of `fmap` would be `fmap :: (a -> b) -> f a -> f b` and `fmap :: (a -> b) -> f' a -> f' b`, and that doesn't really seem composable. So what does it mean to compose two `Functors`?

Answer 1

Please read the wikipedia article.

It says that the composition of two functors is also a functor.

functor F: C-->D and functor G:- D--->E, where C,D,E are categories.

H = G comp F

let x be an object in C.

Then y=F(x) is an object in D.

So then z=G(y) is an object in E.

so H(x) = G(F(x)) = G(y) = z which proves closure.

Since F and G are functors, they preserve the identity and composition of morphisms...

That is F(id(x)) = F(x) = y = id(y) = id(F(x)) for all x in C and

G(id(y)) = G(y) = z = id(z) = if(G(y)) for all y in D

and

for morphisms T1 and T2 of F and U1 and U2 of G, on C and D , respectively

F(T1 comp T2) = F(T1) comp F(T2) and G(U1 comp U2) = G(U1) comp G(u2)

H(T1 comp T2)=

G(F( T1 comp T2) ) =

G( F(T1) comp F(T2)) = <--- F preserves the composition of morphisms

G(F(T1)) comp G(F(T2)) = <--- G preserves the composition of morphisms

H(T1) comp H(T2) <--- the entire composition preserves the composition of

morphisms

Also,

H(id(x)) = G(F(id(x)) = G(F(x)) = H(x) = id(H(x) , so the identity morphism is preserved

by the entire composition

[end of proof]