Why does `\x -> f x x` = `join f`?

Question

I found out recently that the pointfree for \x -> f x x is join f, and wanted to understand why. I started here:

join :: Monad m => m (m a) -> m a

then got stumped because I'm not familiar with the "function monad". Could someone help me with the type derivation that explains the equality?

Answer 1

This is a pretty simple algebraic transformation of the Monad ((->) r) on the type level. Watch as we specialize and simplify the type of join.

join :: Monad m => m        (m        a)  -> m        a
join ::            ((->) r) (((->) r) a)  -> ((->) r) a  -- Specializing
join ::            (r ->    (r ->     a)) -> (r ->    a) -- Infix
join ::            (r ->     r ->     a)  -> r  ->    a  -- Simplifying

Answer 2

If we substitute x -> for m in the type of join, we get (x -> x -> a) -> x -> a. If we apply f to that (which I assume has the type x -> x -> a for some x and a), we get x -> a, which is also the type of \x -> f x x.