Yitzchak Gale wrote:
I wrote:
...it was recently claimed on this list that tuples
are not products in that category.
I've not been convinced yet.
I'm going to try convince you :) The crucial problem of Haskell's
product is that (_|_,_|_) ≠ _|_ but that the two projections
fst :: (A,B) -> A
snd :: (A,B) -> B
cannot distinguish between both values. But if (,) were a categorial
product, fst and snd would completely determine it. We would have
the universal property that for every
f :: C -> A
g :: C -> B
there is a _unique_ morphism
f &&& g :: C -> (A,B)
subject to
f = fst . (f &&& g)
g = snd . (f &&& g)
In other words, there is a unique function
(&&&) :: forall c . (c -> A) -> (c -> B) -> (c -> (A,B))
f &&& g = \c -> (f c, g c)
In the particular case of C=(A,B), f=fst and g=snd , the identity
function is such a morphism which means
fst &&& snd = id
due to uniqueness. But
id _|_ ≠ id (_|_,_|_)
while clearly
(fst &&& snd) _|_ = (fst &&& snd) (_|_,_|_)
Derek Elkins wrote:
Also, there is a Haskell-specific problem at the very get-go.
The most "obvious" choice for the categorical composition operator
assuming the "obvious" choice for the arrows and objects does not work...
...This can
easily be fixed by making the categorical (.) strict in both arguments
and there is no formal problem with it being different from Haskell's
(.), but it certainly is not intuitively appealing.
Note that the problem with (.) is "seq's fault" (pun intended :)
Otherwise, it would be impossible to distinguish _|_ from its
eta-expansion \x._|_ .
Regards,
apfelmus
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