Continuous map from the ring on the unit circle 
Is there a surjective continuous map from the ring $r<x^2+y^2<1\,(0<r<1)$ on the unit circle $x^2+y^2<1$ ?

It seems NO, but how can it be done ?

Edit: what if we add the injectivity to the map ?

 A: Yes. Let me denote $R = \sqrt{r}$, so the ring is defined by $R < |p| < 1$ where $p=(x,y)$. Divide the ring into an inner subring
$$R < |p| \le (R+1)/2
$$
and an outer subring 
$$(R+1)/2 \le |p| < 1
$$
Map the entire inner subring to the origin. Map the outer subring by pulling its inner boundary inward to the origin: for each angle $\theta$, the radial segment of the outer ring of angle $\theta$ connecting the circle of radius $(R+1)/2$ and the circle of radius $1$ is stretched by a constant factor over the radial segment of angle $\theta$ connecting the origin to the circle of radius $1$. 
A: For $\alpha\in(r,(r+1)/2]~$, the function $f(x)=(x-\alpha)^2$ satisfies $f((r,1))=[0,1)$. Then the function $g(x,y)=f(x^2+y^2)$ sends the ring onto the open unit disk.
A: I find visual building of a map from an open ring to an open circle.
In the three dimension space imagine an open ring as a lateral surface of the cylinder. Now let pull the central line of the cylinder to the point. We have two cones with a common vertex and open boundaries. Finally, project (two-fold covering) these two cones to an open circle (two boundaries of cones map to the boundary of the circle).
