Smooth non-surjective interpolation I am looking for a smooth interpolation of the function, inside the annulus confined by radii $R>0$ and $2R$,
$$f(z)=\begin{cases} &z\text{ for }\vert z \vert >2R\\&0 \text{ for }\vert z \vert \le R\end{cases}$$ such that $f:\mathbb C\to \mathbb C$ is not onto.
It would be easy to directly interpolate down to $0$ by keeping the argument (of the complex number) constant on each ray, but this way, the interpolation would indeed be onto.
 A: There is no such map which is even continuous, much less smooth. Also, your assumption that $f(z) = 0$ for $|z|\leq R$ is not needed to obtain the contradiction.
Without loss of generality we can set $R=\frac{1}{2}$.
So we are trying to prove that it is impossible for a continuous function $f: \mathbb{C} \to \mathbb{C}$ to satisfy $f(z) = z$ for $|z|>1$ and be non-surjective.
Assume to the contrary that $p$ is not in the image of $f$.
$p$ must be inside the disk of radius $1$.
Let $P: D(0,1) - \{p\} \to C(0,1)$ be given by radial projection from $p$.
Let $S: \mathbb{C} \to \mathbb{C}$ be defined by
$$
S(z) = 
\begin{cases}
z \textrm{ if $|z|\leq 1$}\\
\frac{z}{|z|} \textrm{ if $z >1$}
\end{cases}
$$
Then $g = P \circ S \circ f: D(0,1) \to C(0,1)$ is a continuous map from the disk to its boundary circle which restricts to the identity on the boundary.  This is impossible:  this is one of the first applications you will see in any textbook on algebraic topology of the fundamental group.
Considering the inclusion of $\iota: C(0,1) \to D(0,1)$ we would have maps $ \pi_1(C(0,1)) \stackrel{\pi_1(\iota)}\longrightarrow \pi_1(D(0,1)) \stackrel{\pi_1(g)}\longrightarrow \pi_1(C(0,1))$.  But $\pi_1(C(0,1)) \cong \mathbb{Z}$ and $\pi_1(D(0,1)) \cong 0$.  So we cannot have $ \pi_1(g) \circ \pi_1(\iota) = \textrm{Id}_{\pi_1(C(0,1))}$ when $\pi_1(\iota)$ is the zero map.  Here we are using the fact that the fundamental group is a functor:  since $g \circ \iota = \textrm{Id}_{C(0,1)}$ we must have $\pi_1(g) \circ \pi_1(\iota) = \textrm{Id}_{\pi_1(C(0,1))}$
