There is a holomorphic and bijective function of $\mathbb{D}$ in $\mathbb{D}\setminus(0,1]$? where $\mathbb{D}$ denotes the open unit disk in the complex plane.
Idea: I really need to prove that there is a holomorphic and bijective function of $R={\{z\in\mathbb{C}: Re(z)>0 , Im(z)>0}\}$ in $\mathbb{D}\setminus(0,1]$ But just find a function with such properties of $\mathbb{D}$ in $\mathbb{D}\setminus(0,1]$, and then make the composition with the transformation of cayley and the function $f(z)=z^2$.
Thanks for the help!!
No, there isn't a homeomorphism, let alone a biholomorphism.
Suppose there exists a homeomorphism $f\colon \mathbb{D}\backslash (0,1] \rightarrow \mathbb{D}$. We may remove $0$ and $f(0)$ ans still get a homeomorphism $f\colon \mathbb{D}\backslash [0,1] \rightarrow \mathbb{D}\backslash \{f(0)\}$.
Since $\mathbb{D}\backslash [0,1]$ is a simply connected open subset of the plane and, on the other hand, $\mathbb{D}\backslash \{f(0)\}$ is not simply connected, we have a contradiction.
Handwavy heuristic approach:
Let $p=f(0)\in \mathbb{D}$ and take an arbitrary small disc $D$ with $p\in D\subset \mathbb{D}$. Since $f$ is a homeomorphism, $f^{-1}(\bar D)$ must contain $0$ and its boundary must be a closed simple curve enclosing $p$ totally inside $\mathbb{D}\backslash (0,1]$ which is impossible.
Added: To be precise, a holomorphic bijective map is a biholomorphism as a consequence of the Open Mapping Theorem
Added 2: Just a courious fact. It follows directly from the Open Mapping Theorem that any holomorphic map $f\colon \mathbb{D} \rightarrow \mathbb{D}\backslash (0,1]$ whose image contains $0$ is constant.