Homotopy equivalence of $\Bbb{R}^2\setminus\ \bigg\{[0,1] \times \{0\}\bigg\}$ and $\Bbb{R^n}\setminus\{0\}$ I am currently working on an exercise problem, in which I have proove that $ \mathbb{R}^2\backslash [0,1] \times\{0\}$ and $\mathbb{R}^n\backslash \{0\}$ are homotopy equivalent.
My initial idea was to break it down to showing that $ \mathbb{R}^2\backslash \{0\}$ is homotopy equivalent to $\mathbb{R}^n\backslash \{0\}$, but it didn't work out, since I couldn't find a continuous function that is a homotopy equivalence between those two spaces. So is that actually possible (and if yes how), or should I use a different approach to prove the statement.
 A: For $n=2$,
You have $\Bbb{R^{2}}\setminus([0,1]\times{\{0\}})$ is homotopy equivalent to $S^{1}$.
To see this take the circle of radius $2$ and use the (strong)deformation retract $H(x,t)= \frac{x}{1-t+\frac{1}{2}t\cdot ||x||}$ . Then this is a deformation retract from the given space to the circle centered at origin and of radius $2$ .
And $\Bbb{R}^{2}\setminus{\{0\}}$ also deformation retracts to the unit circle using a similar deformation retraction(which is homotopic to the circle of radius $2$) . (Take $H(x,t)=\frac{x}{1-t+t\cdot ||x||}$) .
But if $n \geq 3$ then $\Bbb{R}^{n}\setminus\{0\}$ is homotopy equivalent to $S^{n-1}$ which has trivial fundamental group and hence they cannot be homotopy equivalent due to them having different fundamental groups.
A: I don't give a full answer, but it is possible to show that $\mathbb{R}^{n}\setminus\{0\}$ is homotopy equivalent with $\mathbb{S}^{n-1}$ (why?).
And maybe you are able to show that $\mathbb{R}^2\setminus ([0,1]\times\{0\})$ is homotopy equivalent to $\mathbb{R}^2\setminus\{0\}$.
Can you conclude things from the previous two statements?
