[Haskell-beginners] Are functors in Haskell always injective?

[Brent Yorgey]

On Sat, Jul 05, 2014 at 11:51:10AM +0100, Julian Birch wrote:
> (Another beginner here)
> 
> That's my understanding too.  Functors in Haskell have to operate on the
> whole of Hask, which isn't true of functors in general.  

As stated, this is not true: a functor F : C -> D must be defined on
*all* the objects of the category C.  I.e., it is still true that
functors in general must operate on the whole of their domain
category.  I think what you are getting at is that there are (lowercase-f) functors
whose domain is a *sub*category of Hask (such as the category of all
types having an Ord instance, with order-preserving functions as
morphisms) which cannot be extended to functors on all of Hask, and so
they are not (capital-F) Functors.

-Brent

> Equally, they have
> to map to a subset of Haskell (although that mapping could be isomorphic to
> Hask e.g. Identity Functor).  It's the first restriction that's the most
> problematic, since it prevents set from being a Functor.  This is fixable (
> http://okmij.org/ftp/Haskell/RestrictedMonad.lhs) but not part of "standard
> Haskell".  The second restriction basically just means you've got to do
> some book-keeping to get back to the original type.
> 
> Julian
> 
> 
> On 5 July 2014 06:51, Dimitri DeFigueiredo <defigueiredo at ucdavis.edu> wrote:
> 
> >  Oops! Sorry for the typos! Fixing that below.
> >
> >
> > Hi All,
> >
> > My understanding is that a functor in Haskell is made of two "maps".
> > One that maps objects to objects that in Hask means types into types (i.e.
> > a type constructor)
> > And one that maps arrows into arrows, i.e. fmap.
> >
> > My understanding is that a functor F in category theory is required to
> > preserve the domain and codomain of arrows, but it doesn't have to be
> > injective. In other words, two objects X and Y of category C (i.e. two
> > types in Hask) can be mapped into the same object Z (same type) in category
> > D. As long as the "homomorphism" law holds:
> >
> > F(f:X->Y) = F(f):F(X)->F(Y)
> >
> > However, I don't think there is any way this mapping of types cannot be
> > injective in Haskell. It seems that a type constructor, when called with
> > two distinct type will always yield another two **distinct** types. (E.g.
> > Int and Double yield Maybe Int and Maybe Double) So, it seems that Functors
> > in Haskell are actually more restrictive than functors can be in general.
> > Is this observation correct or did I misunderstand something?
> >
> >
> > Thanks!
> >
> >
> > Dimitri
> >
> > Em 05/07/14 02:42, Dimitri DeFigueiredo escreveu:
> >
> > Hi All,
> >
> > My understanding is that a functor in Haskell is made of two "maps".
> > One that maps objects to objects that in Hask means types into types (i.e.
> > a type constructor)
> > And one that maps arrows into arrows, i.e. fmap.
> >
> > My understanding is that a functor F in category theory is required to
> > preserve the domain and codomain of arrows, but it doesn't have to be
> > injective. In other words, two objects X and Y of category C (i.e. two
> > types in Hask) can be mapped into the same object Z (same type) in category
> > Z. As long as the "homomorphism" law holds:
> >
> > F(f:X->Y) = F(f):F(X)->F(Y)
> >
> > However, I don't think there is any way this mapping of types cannot be
> > injective in Haskell. It seems that a type constructor, when called with
> > two distinct type will always yield another two *distinct* types. (E.g. Int
> > and Fload yield Maybe Int and Maybe) So, it seems that Functors in Haskell
> > are actually more restrictive than functors can be in general. Is this
> > observation correct or did I misunderstand something?
> >
> >
> > Thanks!
> >
> >
> > Dimitri
> >
> >
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> >
> >
> >
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> >

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