$\mathbb{R}^2\setminus (\mathbb{R}\times \{0\})$ is homotopic to $\mathbb{R}\setminus {0}$

Question

How to show that $\mathbb{R}^2\setminus (\mathbb{R}\times \{0\})$ is homotopic to $\mathbb{R}\setminus {0}$?

Pictorialy it is almost clear. But unable to prove it.

Answer 1

The space $\Bbb R^2\setminus (\Bbb R\times \{0\})$ deformation retracts to its subspace $(\{0\}\times\Bbb R)\setminus\{(0,0)\}$ which is homeomorphic to $\Bbb R\setminus\{0\}$.

Answer 2

Let $A=\mathbb{R^2}-(\mathbb R \times \{0\})$, $B=\mathbb R-\{0\}$, $f:A \rightarrow B$, $f(x,y)=y$, $g:B \rightarrow A$, $g(y)=(0,y)$. Then $f \circ g=id_B$ and let $h=g \circ f$, take homotopy $h_t(x,y)=(xt,y)$.

Answer 3

Note that $\Bbb R^2\setminus (\Bbb R\times \{0\}) = \Bbb R\times (\Bbb R\setminus\{0\})$