We want to show $\mathbb R^2 \setminus \left\{0\right\}$ and $\mathbb R^2$ are homeomorphic (Goal), but I already proved that $\mathbb R^2 \setminus \left\{0\right\}$ and $S^1$ are homeomorphic. And I already proved that $S^1$ and $B^2$ are not homeomorphic.
So I thought it only need to show that $B^2$ and $\mathbb R^2$ are homeomorphic. And I tried to construct a deformation retraction $H$ of $\mathbb R^2$ onto $B^2$, defined by $$ H(x, t)= \begin{cases} x & \;\text{ if }\; |x|\leq 1,\\ x \cdot (t/|x| + 1-t) & \; \text{ if }\; |x|>1, \end{cases}$$ (where $x$ is in $\mathbb R^2$) so that points of outer circle to be continuously collapse to the unit circle.
Is it okay to do that?
Thank you.
It is not true that ${\mathbb{R}^2 - (0,0)}$ and ${\mathbb{R}^2}$ are homeomorphic. In fact, they don't even have the same homotopy type, since ${\mathbb{R}^2 - (0,0)\simeq S^1}$, we have $$ \pi_1(\mathbb{R}^2 - (0,0)) = \pi_1(S^1) = \mathbb{Z},\text{ but }\pi_1(\mathbb{R}^2) \equiv 0 $$