Re: [Haskell-cafe] Questions about the Functor class and it's use in "Data types à la carte"

[Dominic Steinitz]

>> Do you have a counter-example of (.) not being function composition in
>> the categorical sense?
> 
> Let bot be the function defined by
> 
> bot :: alpha -> beta
> bot = bot
> 
> By definition,
> 
> (.) = \ f -> \ g -> \ x -> f (g x)
> 
> Then
> 
>    bot . id
> = ((\ f -> \ g -> \ x -> f (g x)) bot) id
> = (\ g -> \ x -> bot (g x)) id
> = \ x -> bot (g x)

I didn't follow the reduction here. Shouldn't id replace g everywhere?

This would give

= \x -> bot x

and by eta reduction

= bot

> 
> which /= bot since (seq bot () = bot) but (seq (\ x -> M) () = ())  
> regardless of what expression we substitute for M.
> 

Why is seq introduced?