[Haskell-cafe] Re: Article review: Category Theory

[=?x-mac-roman?Q?Johan_Gr=9Anqvist?=]

Ulf Norell skrev:
> 
> On Jan 16, 2007, at 7:22 PM, David House wrote:
> 
> 
> In the section on the category laws you say that the identity morphism 
> should satisfy
> 
>   f . idA = idB . f
> 
> This is not strong enough. You need
> 
>   f . idA = f = idB . f
> 

(I do not know category theory, but try to learn from the 
tutorial/article/introduction.)

Given this, and looking at the figure accompanying exercise 2. Can I not 
then show that f=h from:

f.g = idA (f.g is A -> A and idA is the only morphism A -> A, closedness 
gives the equality)

h.g = idA (same argument)
g.f = g.h = idB (same argument)

thus (using the laws for id and associativity)

f =  idA . f  =  (h . g) . f = h . (g . f)  = h . idB  = h

Thus in the figure f=h must hold, nad one arrow can be removed from the 
  graph.



Regards

johan