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