[Haskell-cafe] Confusion on the third monad law when using lambda abstractions

[Hans van Thiel]

On Thu, 2009-06-18 at 08:34 -0500, Jake McArthur wrote:
[snip]
> So, `(=<<)` is just like `($)` except for the information carried along 
> by the monad.
> 
> Anyway, the "obvious" thing to do is to drop the `x` from both sides of 
> the definition for `bar`. To do that with `foo` earlier, we had to 
> substitute `($)` with `(.)`. What we are looking for is an equivalent 
> operator for monads:
> 
>      (.)   ::            (b      c) -> (a ->   b) -> (a ->   c)
Just to show I'm paying attention, there's an arrow missing, right?
       (.)   ::            (b  ->  c) -> (a ->   b) -> (a ->   c)

Many thanks, also to the others who've replied. I've wondered about
(=<<) usage for a long time too, and this is all very illuminating. I'll
work this through and put it in my monad tutorial, if I may (without
implicating you guys in any way, of course, unless you insist...)

Regards,

Hans van Thiel
[snip]