In Lee's Introduction to Topological Manifold on page 134, it states that the set $\mathbb{B}^2\setminus \{(x,0): x\in[0,1)\}$ is an open 2-cell. However, I'm not quite sure as to how to define a homeomorphism from $\mathbb{B}^2$ onto the set. ($\mathbb{B}^2$ is defined to be the open disc in $\mathbb{R}^2$).
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There is probably a much easier way to do it but here we go:
Let $X=\{z\in\mathbb{C}\mid z\notin[0,1]\subset\mathbb{R}\}$. By applying the square-root, i.e. "halfing the angle", we get a biholomorphic map from $X$ to the upper half disk, i.e. $E=\{z\in\mathbb{C}:\left\rvert z\right\rvert<1,\operatorname{Im} z>0\}$. Now look at the Möbius-transform $T:E\to\mathbb{C}$, $T(z)=\frac{z+1}{-z+1}$. If I didn't mess up the calculations this should map $E$ to the first quadrant. Now by squaring we get a biholomorphism with the upper half complex plane $\mathbb{H}$. Now define a similar Möbius-transform that gives the isomorphism between $\mathbb{H}$ and the open disc, e.g. the Cayley-transform.
Composing all those maps we get a biholomorphic map from $X$ to the open unit disc.
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