haskell - Understanding Identity functor -

I am working my way through this. As mentioned in the tutorial, I copy some of the code below to represent factor structure and identity factor

  (F (ga)) example (Show (f (ga)) = & gt; FComp fga = C {unC :: f (ga)} Example (Show (f (ga)) => Show (Fcomp FGA) where Show (C X) = "FComp" ++x Example (Factor F, Fencer G) = & gt; Functor (FComp fg) where fmap h (C x) = C (fmap (fmap h) x) newtype id a = identification {unId :: a} deriving For example show Functor Id where fmap fx = Identity (f (unId x ))   
 Now, this is what about the tutorial about the  identifier factor :  
  same Combined with the identity factor in the category was thought to be such. FQF = F IDAFF = F   
 What I'm stuck with is that it is trying to think about the factor formation as  FComp < / Code> Below is an example of the code given above:    $ a = C (identity (bus (5 :: int)) $: ta :: FComp ID may be $ int $ B = C (bus (identity (5 :: int)) $ $: tbb :: FComp probably ID int   
 I type that type of  a  and  B  as shown in the example above In the context of  fact factor      
 
 
 In the Haskell-applicable class theory, like many equations,  f  one ??? id  < Em> b   ≡ id   a   â  f  ≡  f  In fact, only  equivalences  should be read as  FCOMP ID may be IIT  not as much as  FComp probably ID difference  in type tester; Although you can easily write  
  idFunctorIso :: Functor f => FCMP F - ID -> F an idFunctorIso (CIDIDA) = FMAP UIID FIDCA IDFfactorISE :: Future F => F.A - & gt; The FComF id is an idFunctorIso 'fa = C $ fmap Identity fIdc   
 which means that both types of similar information are  1 . By saying this we mean that they are isomorphic.  
 
  1  No information is lost in any direction, because  idFunctorIso 'IdFunctor also ≡ id  (such as Factor Law  Fmap id ≡ id , with the fact that the inverse of new types of constructors is simpler than  unC  and  unId )