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

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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?