Given fmap id = id, fmap (f . g) = fmap f . fmap g follows from the free
theorem for fmap.
This was published as an aside in a paper a long time back, but I forget
where.
-Edward Kmett
On Mon, Jan 4, 2010 at 5:14 PM, Paul Brauner paul.brau...@loria.fr wrote:
Hi,
I'm trying to get a deep
On Mon, Jan 4, 2010 at 3:01 PM, Derek Elkins derek.a.elk...@gmail.com wrote:
So without doing funky stuff involving bottoms and/or seq, I believe
that fmap id = id implies the other functor law (in this case, not in
the case of the general categorical notion of functor.)
So lets play with
Brent Yorgey wrote:
On Mon, Jan 04, 2010 at 11:49:33PM +0100, Steffen Schuldenzucker wrote:
[...]
As others have pointed out, this doesn't typecheck; but what it DOES
show is that if we had a type class
class Endofunctor a where
efmap :: (a - a) - f a - f a
then it would be
Hi,
I'm trying to get a deep feeling of Functors (and then pointed Functors,
Applicative Functors, etc.). To this end, I try to find lawless
instances of Functor that satisfy one law but not the other.
I've found one instance that satisfies fmap (f.g) = fmap f . fmap g
but not fmap id = id:
Hi Paul,
Paul Brauner wrote:
Hi,
I'm trying to get a deep feeling of Functors (and then pointed Functors,
Applicative Functors, etc.). To this end, I try to find lawless
instances of Functor that satisfy one law but not the other.
I've found one instance that satisfies fmap (f.g) = fmap f
On Tue, Jan 5, 2010 at 7:49 AM, Steffen Schuldenzucker
sschuldenzuc...@uni-bonn.de wrote:
Hi Paul,
Paul Brauner wrote:
Hi,
I'm trying to get a deep feeling of Functors (and then pointed Functors,
Applicative Functors, etc.). To this end, I try to find lawless
instances of Functor that
Steffen Schuldenzucker wrote:
data Foo a = Foo a
instance Functor Foo where
fmap f (Foo x) = Foo . f . f $ x
I think this doesn't typecheck.
Cheers, Jochem
--
Jochem Berndsen | joc...@functor.nl | joc...@牛在田里.com
___
Haskell-Cafe mailing
On Tue, Jan 5, 2010 at 7:14 AM, Paul Brauner paul.brau...@loria.fr wrote:
Hi,
I'm trying to get a deep feeling of Functors (and then pointed Functors,
Applicative Functors, etc.). To this end, I try to find lawless
instances of Functor that satisfy one law but not the other.
I've found one
On Mon, Jan 4, 2010 at 3:01 PM, Derek Elkins derek.a.elk...@gmail.com wrote:
Ignoring bottoms the free theorem for fmap can be written:
If h . p = q . g then fmap h . fmap p = fmap q . fmap g
When I play with http://haskell.as9x.info/ft.html I get examples that
look more like:
If fmap' has
On Mon, Jan 04, 2010 at 11:49:33PM +0100, Steffen Schuldenzucker wrote:
data Foo a = Foo a
instance Functor Foo where
fmap f (Foo x) = Foo . f . f $ x
Then:
fmap id (Foo x) == Foo . id . id $ x == Foo x
fmap (f . g) (Foo x) == Foo . f . g . f . g $ x
fmap f . fmap g $
On Tue, Jan 5, 2010 at 8:15 AM, Dan Piponi dpip...@gmail.com wrote:
On Mon, Jan 4, 2010 at 3:01 PM, Derek Elkins derek.a.elk...@gmail.com wrote:
Ignoring bottoms the free theorem for fmap can be written:
If h . p = q . g then fmap h . fmap p = fmap q . fmap g
When I play with
On Tue, Jan 5, 2010 at 8:22 AM, Brent Yorgey byor...@seas.upenn.edu wrote:
On Mon, Jan 04, 2010 at 11:49:33PM +0100, Steffen Schuldenzucker wrote:
data Foo a = Foo a
instance Functor Foo where
fmap f (Foo x) = Foo . f . f $ x
Then:
fmap id (Foo x) == Foo . id . id $ x == Foo x
fmap
On Mon, Jan 4, 2010 at 3:26 PM, Derek Elkins derek.a.elk...@gmail.com wrote:
Yes, I have the same problem...Basically, I'm
pretty sure the construction of that free theorem doesn't rely on any
of the actual details...
For a long time I've thought such a higher order free theorem must
exist,
Thanks. I was wondering if the free theorem of fmap entailed that
implication, but I'm not used enough to decipher the output of the tool,
neither am I able to generate it by hand.
However, if this holds (law1 = law2), I wonder why this isn't some
classic category theory result (maybe it is ?). I
On Mon, Jan 4, 2010 at 4:15 PM, Paul Brauner paul.brau...@loria.fr wrote:
I wonder why this isn't some classic category theory result (maybe it is ?)
It doesn't hold in category theory in general.
Haskell (or at least a certain subset) is special - many things that
just *look* category
G'day all.
Quoting Derek Elkins derek.a.elk...@gmail.com:
Ignoring bottoms the free theorem for fmap can be written:
If h . p = q . g then fmap h . fmap p = fmap q . fmap g
Setting p = id gives
h . id = h = q . g fmap h . fmap id = fmap q . fmap g
Using fmap id = id and h = q . g we get,
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