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

[Brian Hulley]

Johan Gršnqvist wrote:
> 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.

But f /= h so by the above reasoning you get a proof by contradiction that 
the figure is not a category.

Brian.
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