On Jun 9, 2004, at 9:39 AM, Jerzy Karczmarczuk wrote:
I have *nothing* to add, just a question.
Do you /anybody/ know of any edible work on ADJUNCTIONS in the context
of Haskell structures? Perhaps instead of searching for 'inverses' one
should think more about adjoints?...
Yes, I think this is t
On Wednesday 09 June 2004 17:20, Ron de Bruijn wrote:
> --- "Iavor S. Diatchki" <[EMAIL PROTECTED]> wrote:
> Only I still find it weird that join is called a
> multiplication, because according to the definition of
> multiplication, there should be an inverse. I think,
> thus that multiplication is
G'day all.
Quoting Ron de Bruijn <[EMAIL PROTECTED]>:
> I have thought a while about morphisms and although I
> had written down in my paper that a functor and also a
> natural transformation are also morphisms, but in a
> different category, I now am not sure anymore of this.
It's true. In par
G'day all.
Quoting Rik van Ginneken <[EMAIL PROTECTED]>:
> It is more even subtle if one considers the rotation group.
> The unit is keeping an object on its place.
> The multiplication is doing rotations sequently.
> Allright, one has an inverse here, but
> the rule (a*b)*c = a*(b*c) doesn't occ
Hello again,
I have thought a while about morphisms and although I
had written down in my paper that a functor and also a
natural transformation are also morphisms, but in a
different category, I now am not sure anymore of this.
If you see everything(objects and morphisms) as dots
and arrows, an
At 08:20 09/06/04 -0700, Ron de Bruijn wrote:
Only I still find it weird that join is called a
multiplication, because according to the definition of
multiplication, there should be an inverse.
For real or rational numbers, maybe.
But also think about Integers, or matrices.
[ 1 2 ] * [ 3 ] = [ 11
?! I found out what a group is:
?! A group is a monoid each of whose elements is
?! invertible.
?!
OK.
?! Only I still find it weird that join is called a
?! multiplication, because according to the definition of
?! multiplication, there should be an inverse. I think,
No, it ain't.
If you ta
--- "Iavor S. Diatchki" <[EMAIL PROTECTED]> wrote:
> hi ron,
>
> here are the relations between the two formulations
> of monads:
> (using haskell notation)
>
> map f m = m >>= (return . f)
> join m = m >>= id
>
> m >>= f = join (fmap f m)
>
> there are quite a few general co
Iavor S. Diatchki wrote:
Ron de Bruijn wrote:
I am pretty sure, that >>= is to monads what * is to
for example natural numbers, but I don't know what the
inverse of >>= is. And I can't really find it anywhere
on the web(papers, websites, not a single sole does
mention it.
this is not quie correc
hi ron,
here are the relations between the two formulations of monads:
(using haskell notation)
map f m = m >>= (return . f)
join m = m >>= id
m >>= f = join (fmap f m)
there are quite a few general concepts that you need
to understand in what sense monads are monoids, but
to unde
I am pretty sure, that >>= is to monads what * is to
for example natural numbers, but I don't know what the
inverse of >>= is. And I can't really find it anywhere
on the web(papers, websites, not a single sole does
mention it.
It should have type, at least that's what I think:
inv::M a->M b
I sa
On 2004 June 07 Monday 15:19, Ron de Bruijn wrote:
> newtype S a = State -> (a,State) -- functor T to map
> objects
> mapS::(a-> b) -> (S a -> S b) -- functor T to map
> morphisms
> unitS :: a -> S a --\eta
> joinS::S(S a)-> S a -- \mu
>
> This is a complete monad using a direct mapping from
> Ca
Hello,
The last 3 or 4 days I have been studying a lot of
Category Theory so that I would be able to explain the
concept of a monad to some people at the university in
a "learn to present something"-course.
newtype S a = State -> (a,State) -- functor T to map
objects
mapS::(a-> b) -> (S a -> S
13 matches
Mail list logo
