Is $\{(x_1,x_2) \in \mathbb{R}^2 -\{(1,0)\} : x_1>-\frac{1}{2}\}$ homotopy equivalent to $\mathbb{R}^2-\{(0,0)\}$?

Question

Is $\{(x_1,x_2) \in \mathbb{R}^2 -\{(1,0)\} : x_1>-\frac{1}{2}\}$ homotopy equivalent to $\mathbb{R}^2-\{(0,0)\}$?

Actually I am trying to find the cohomology of $\mathbb{R}^2 -\{(1,0)\}$ by Mayer-Vietoris sequence and I wanted to split $\mathbb{R}^2 -\{(1,0)\}$ into $U = \{(x_1,x_2) \in \mathbb{R}^2 -\{(1,0)\} : x_1>-\frac{1}{2}\}$ and $V = \{(x_1,x_2) \in \mathbb{R}^2 -\{(1,0)\} : x_1<\frac{1}{2}\}$.

Answer 1

Not only are they homotopy equivalent, they are actually homeomorphic. I'll give you some hints:

• Hint 1 : $\Bbb R^2$ and $X = \{(x_1, x_2) \in \Bbb R^2: x_1 > -\frac{1}{2} \}$ are homeomorphic.

Let $f : X \to \Bbb R^2$ be such a homeomorphism.

• Hint 2 : Let $g : \Bbb R^2 \to \Bbb R^2$ be the translation by $- f(1,0)$. Then, $(g|_{\Bbb R^2 - f(1, 0)}) \circ (f|_{X - (1, 0)})$ is the required homeomorphism.