Scala: Type inference and subtypes/higher-kinded-types

Question

I've been playing around with Scalaz to get a little bit of the haskell feeling into scala. To understand how things work in scala I started implementing various algebraic structures myself and came across a behavior that has been mentioned by the Scalaz folks.

Here is my example code which implements a functor:

trait Functor[M[_]] {
  def fmap[A, B](a: M[A], b: A => B): M[B]
}

sealed abstract class Foo[+A]
case class Bar[A]() extends Foo[A]
case class Baz[A]() extends Foo[A]

object Functor {

  implicit val optionFunctor: Functor[Option] = new Functor[Option]{
    def fmap[A, B](a: Option[A], b: A => B): Option[B] = a match {
      case Some(x) => Some(b(x))
      case None => None
    }   
  }

  implicit val fooFunctor: Functor[Foo] = new Functor[Foo] {
    def fmap[A, B](a: Foo[A], b: A => B): Foo[B] = a match {
      case Bar() => Bar()
      case Baz() => Baz()
    }   
  }
}

object Main {
  import Functor._

  def some[A](a: A): Option[A] = Some(a)
  def none[A]: Option[A] = None

    def fmap[M[_], A, B](a: M[A])(b: A => B)(implicit f: Functor[M]): M[B] =
      f.fmap(a, b)

  def main(args: Array[String]): Unit = { 
    println(fmap (some(1))(_ + 1)) 
    println(fmap (none)((_: Int) + 1)) 
    println(fmap (Bar(): Foo[Int])((_: Int) + 1))                                                    
  }
}

I wrote a functor instance for Option and a bogus sumtype Foo. The problem is that scala cannot infer the implicit parameter without an explicit type annotation or a wrapper method

def some[A](a: A): Option[A] = Some(a)

println(fmap (Bar(): Foo[Int])((_: Int) + 1))

Scala infers types like Functor[Bar] and Functor[Some] without those workarounds.

Why is that? Could anyone please point me to the section in the language spec that defines this behavior?

Regards, raichoo