Why is the return type of Deref::deref itself a reference?

I was reading the docs for Rust's Deref trait:

pub trait Deref {
    type Target: ?Sized;
    fn deref(&self) -> &Self::Target;
}

The type signature for the deref function seems counter-intuitive to me; why is the return type a reference? If references implement this trait so they can be dereferenced, what effect would this have at all?

The only explanation that I can come up with is that references don't implement Deref, but are considered "primitively dereferenceable". However, how would a polymorphic function which would work for any dereferenceable type, including both Deref<T> and &T, be written then?

that references don't implement Deref

You can see all the types that implement Deref, and &T is in that list:

impl<'a, T> Deref for &'a T where T: ?Sized

The non-obvious thing is that there is syntactical sugar being applied when you use the * operator with something that implements Deref. Check out this small example:

use std::ops::Deref;

fn main() {
    let s: String = "hello".into();
    let _: () = Deref::deref(&s);
    let _: () = *s;
}

error[E0308]: mismatched types
 --> src/main.rs:5:17
  |
5 |     let _: () = Deref::deref(&s);
  |                 ^^^^^^^^^^^^^^^^ expected (), found &str
  |
  = note: expected type `()`
             found type `&str`

error[E0308]: mismatched types
 --> src/main.rs:6:17
  |
6 |     let _: () = *s;
  |                 ^^ expected (), found str
  |
  = note: expected type `()`
             found type `str`

The explicit call to deref returns a &str, but the operator * returns a str. It's more like you are calling *Deref::deref(&s), ignoring the implied infinite recursion.

Xirdus is correct in saying

If deref returned a value, it would either be useless because it would always move out, or have semantics that drastically differ from every other function

Although "useless" is a bit strong; it would still be useful for types that implement Copy.