Univalent holomorphic function on an annulus

$f$ is holomorphic function on open unit disk $B(O,1)$. There exists a positive number $r$ such that $f$ is univalent on the annulus $r<|z|<1$. Is it sufficient to conclude $f$ to be univalent on $B(O,1)$?

• If it is not univalent in the hole of the annulus, the image (by the function minus a value it takes more than once) of a loop in the annulus would have to wind more than once around the origin without intersecting itself. – Peyton Jul 30 '17 at 13:01

Suppose that $f|_{B(0,r)}$ is not injective. Then, for some $a\in\mathbb C$, the equation $f(z)=a$ has two distinct solutions in $B(0,r)$. Now, let $\gamma\colon[0,2\pi]\longrightarrow\mathbb C$ be the loop defined by $\gamma(t)=r'e^{it}$, for some $r'\in(r,1)$ chosen such that the equation $f(z)=a$ has no roots with absolute value equal to $r'$. Then, by the argument principle and since $f$ has no poles, $\frac1{2\pi i}\int_\gamma\frac{f'(z)}{f(z)-a}\mathrm dz$ is a natural number greater than $1$. On the other hand, if $\Gamma=f\circ\gamma$, then$$\frac1{2\pi i}\int_\gamma\frac{f'(z)}{f(z)-a}\mathrm dz=\frac1{2\pi i}\int_\Gamma\frac{\mathrm dz}{z-a}=\operatorname{Ind}_\Gamma(a).$$So, the winding number of $\Gamma$ with respect to $a$ is greater than $2$. So, $\Gamma$ loops around $a$ twice (at least). But then it must intersect itself, and therefore the restriction of $f$ to the image of $\gamma$ cannot be injective.