Suppose we have a holomorphic injective function $f:B_1(0)\setminus\{x\}\to B_1(0)\setminus\{x\}$ where $B_1(0)$ is the unit ball and $x\in B_1(0)$. Now since $f$ is bounded in a neighborhood of $x$ we may extend it to a function $g$ holomorphic to all $B_1(0)$
I want to prove that $g$ is still injective. My idea was that $B_1(0)\setminus\{x\}$ is a dense subset of $B_1(0)$ so we can find a sequence $x_n$ inside $B_1(0)\setminus \{x\}$ converging to $x$. Then $g'(x)=g'(lim x_n)=lim g'(x_n)$ where all $g'(x_n)\not=0$ but this is not enough to conclude.
You need to use the Open Mapping Theorem.
Assume that $\,g:B(0,1)\to B(0,1)\!\smallsetminus\!\{x\}\,$ is not $\,1\!-\!1$. Hence, there exists a $\,y\in B(0,1)\!\smallsetminus\!\{x\}$, such that $\,g(x)=g(y)$.
Open Mapping Theorem implies that if $g$ holomorphic and non-constant then for every $U$ open, so is $g[U]$. Let $0<\delta<|x-y|/3$, then $\,w=g(x)=g(y)\in g[B(x,\delta)]\cap g[B(y,\delta)]$. But as $g[B(x,\delta)]\cap g[B(y,\delta)]$ is open, then there exists an $r>0$, such that $$ B(w,r)\subset g[B(x,\delta)]\cap g[B(y,\delta)]. $$ Hence, every value in $B(w,r)$ is taken at least twice by $f$.