To prove f(0) = 0 and hence f is a bijection let $ f:\mathbf{D}\rightarrow \mathbf{D}$ be a holomorphic function. And suppose that it is Bijective on ${D}\setminus \left \{ 0 \right \}\rightarrow \mathbf{D}\setminus \left \{ 0 \right \} $. Can we conclude that
f(0) = 0?  
I can conclude that  $\begin{vmatrix}f(0)\end{vmatrix} < 1$ because other wise if  $\begin{vmatrix}f(0)\end{vmatrix} = 1$ then by maximum modulus principle f(z) is constant but that cannot happen because the function is a bijection. but from this is it possible to deduce that f(0)=0 ?
I'm referring the correct answer of conformal self maps on punctured disk 
 A: Let $f(0) = a$ with $0<|a|<1.$ As $f:\mathbb D /\{0\} \mapsto \mathbb D /\{0\} ,$ So there exist a point $b\in \mathbb D /\{0\}$ so that $f(b)= a.$ Consider the function $g: \mathbb D \mapsto \mathbb D$ defined by $g(z)= f(z)- a.$ The function $g$ has a zero at $b$ and $0.$ Let order of the zero of $g$ at $b$ and $0$ be $m_1, m_2$ respectively. Note that $m_1,m_2 \geq 1.$
So there exist $\epsilon_1,\epsilon_2 >0$ and a $\delta>0$ so that each point $w$ in $B(0,\delta),$  the equation $g(z)=w$ has exactly $m_1$ simple root in $B(b,\epsilon_1)$ and $m_2$ simple root in $B(o,\epsilon_2).$ Note that $\delta$ can be taken so small so that corresponding $\epsilon_1,\epsilon_2$ will have the property that $B(o,\epsilon_2) \cap B(b,\epsilon_1) = \emptyset. $ In that case bijectivity of $f$ in $\mathbb D/\{0\}$ will be contradicted.
Hence $f(0)$ must be $0.$
A: I'm Referring the book Function Theory of One Complex Variable
By Robert Everist Greene, Steven George Krantz.. 
Supppose that   
$f(0) \neq 0 $ then  $ \exists z` \in  \mathbb{D} \setminus  \begin{Bmatrix}0\end{Bmatrix}  $ such that $f(0) = z`$. Also note that  $ \exists z_{1} \in  \mathbb{D} \setminus  \begin{Bmatrix}0\end{Bmatrix}  $ such that $f(z_{1}) = z`$.  
Now by using Theorem 5.2.2 we have $ \delta_{0},  \varepsilon _{0}, \delta_{1},  \varepsilon _{1} $ such that $ \forall z \in D(z`,\varepsilon _{0}) \setminus \{z` \}  $ has exactly one pre image on $ D(0,\delta_{0}) $. 
and similarly $ \forall z \in D(z`,\varepsilon _{1}) \setminus \{z` \}  $ has exactly one pre image on $ D(0,\delta_{1}) $.    
now by considering the intersection of two open balls $D(z`,\varepsilon _{0}) \setminus \{z` \}$ and $D(z`,\varepsilon _{1}) \setminus \{z` \}$ which is nonempty we can contradict with the fact that the function is one to one on the punctured disc. 
