Why holomorphic functions on the unit disk which has star-shaped range are injective? I am recently reading a book named "Univalent  Functions: A primer", written by Derek K. Thomas.
Here are some definitions in the book. Here, $\mathbb{D}$ denotes the unit open disk in the complex plane.


*

*$f: \mathbb{D} \rightarrow \mathbb{C}$ is in the family $\mathcal{A}$ if $f$ is analytic, $f(0)=0$, $f'(0)=1$.

*$f \in \mathcal{A}$ is called "starlike" if the range of $f$ is star-shaped with respect to the origin. That is, if $z \in f(\mathbb{D})$ and $t \in [0,1]$, then $tz \in f(\mathbb{D})$. 


The author claims that starlike functions are injective. For proof of this fact, he quotes the book "Univalent Functions" written by Peter L. Duren.
But  Duren's book defines "starlike functions" as an injective map satisfying the conditions above, so that the reasoning seems to be circular.
Can someone provide me a proof (or a precise reference) for the fact that all starlike functions are injective?
 A: The statement is actually false:
Pick an automorphism of the unit disk $\beta$ not mapping $0$ to $0$. 
Then for all complex numbers $a,b$, $f(z)=a\beta(z)^2+b$ has a star-shaped range and is not injective. 
For suitable $a$, $b$, we can ensure $f(0)=0$, $f’(0)=1$. 
A: Actually, one can characterize all functions $f$ holomorphic on the unit disc $D$ with $f(0)=0, f'(0)=1, f(D)=U$ starlike (at $0$) as the set of all compositions $f(z)=\frac{1}{B'(0)}h(B(z))$ where $h$ is a normalized ($h(0)=0, h'(0)=1$) univalent starlike function on $D$ and $B$ ranges through the set of holomorphic functions on $D$ that satisfy $B(0)=0, B'(0)>0, B(D)=D$.
Note that for example, finite Blaschke products (of order $n \ge 1$) that have a simple zero at the origin are examples of $B$ above after possibly a rotation (multiplication by $e^{i\theta}$) to make $B'(0) >0$ as $B(0)=0, B'(0) \ne 0$, and $B$ is $n$ to $1$ so surjective on the unit disc, while the only univalent such $B$ is the identity because the only Riemann map from $D$ to itself that satisfies $B(0)=0, B'(0) >0$ is the identity (by the unicity of the Riemann map). 
Note also that by Schwarz lemma for all such $B$, $B'(0) \le 1$ with equality precisely iff $B(z)=z$, while any $f(z)=\frac{1}{B'(0)}h(B(z))$ as above clearly has a starlike image and is normalized as required, so we only need to prove the converse, so given $f$, find $h,B$ as required.
Let $g(0)=0, g'(0)>0$ be the unique Riemann map from $D$ to $U$ and let $B(z)=g^{-1}(f(z))$ so $f(z)=g(B(z))$. Clearly $B(0)=0, B'(0)>0, B(D)=D$ and $g'(0) =\frac{1}{B'(0)} \ge 1$ with equality iff $B(z)=z, g=f$, so $f$ was univalent to start with. 
The function $h(z)=B'(0)g(z)$ is normalized univalent starlike and obviously $f(z)=\frac{1}{B'(0)}h(B(z))$. 
By the unicity of the Riemann map, $f$ univalent iff $f=g$ iff $B(z)=z$ iff the conformal radius of $U$ at $0$ is $1$. 
