How can I prove that a quotient of $\mathbb{R}^2$ is homeomorphic to $\mathbb{R}^2$? Given X=$\mathbb{R}^2$, $A=\overline{B((0,0),1)}$ and $Y=X/A$, I want to prove that $X$ is homeomorphic to $Y$.
I'm pretty convinced that the function:
$f:X\rightarrow Y\ ;\ f(0)=A,\ f(x)=\frac{||x||+1}{||x||}x\ \forall x \in X,\ x\neq 0 $
is a homeomorphism with inverse
$g:Y\rightarrow X\ ;\ g(A)=0,\ g({x})=\frac{||x||-1}{||x||}x\ \forall x \in X,\ ||x||>1$
The fact that $g$ and $f$ are inverses is easily verifiable. I also managed to prove that $f$ is open by checking that the images of open balls (with radius $\delta<||x||$ if centered in $x \neq 0$) is open.
The only thing that is left to prove and that I have no idea how to approach is the continuity of $f$ (or the openness of $g$).
If anyone could help me or at least lend me a hint they would have my eternal gratitude!
 A: Your approach is fine, but I think you did not really use the fact that $Y$ is endowed with the quotient topology.
Consider the closed subsets $A = \{x \in \mathbb R^2 \mid \lVert x \rVert \le 1 \}$ and $C = \{x \in \mathbb R^2 \mid \lVert x \rVert \ge 1 \}$ of $\mathbb R^2$. Define
$$\phi : \mathbb R^2 \to \mathbb R^2, \phi(x) = \begin{cases} 0 & x \in A \\ \frac{||x||-1}{||x||}x & x \in C \end{cases}$$
This map is well-defined because both parts of the definition agree on $A \cap C = \{x \in \mathbb R^2 \mid \lVert x \rVert = 1 \}$. It is continuous because it is continuous on the closed sets $A, C$. Since $\phi(x) = 0$ for $x \in A$, the universal property of the quotient gives us an induced continuous
$$g : \mathbb R^2/A \to \mathbb R^2$$
such that $g \circ p = \phi$, where $p : \mathbb R^2 \to \mathbb R^2/A$ is the quotient map. Clearly $\phi(p^{-1}(M)) = g(p(p^{-1}(M))) = g(M)$ for $M \subset \mathbb R^2/A$.
Let us prove that $\phi$ is a closed map. So let $D \subset \mathbb R^2$ be closed and $(x_n)$ be a sequence in $\phi(D)$ which converges to some $\xi \in \mathbb  R^2$. We have to show that $\xi \in \phi(D)$. Let $y_n \in D$ such that $\phi(y_n) = x_n$. If $x_n \ne 0$, then $y_n = \frac{||x_n||+1}{||x_n||}x_n$. If $x_n = 0$, then $y_n  \in A$. Obviously $(y_n)$ is a bounded sequence, thus by picking a subsequence we get $y_{n_k} \to \eta \in \mathbb R^2$ (and of course $x_{n_k} \to \xi$). Since $D$ is closed, we have $\eta  \in D$ and by continuity $x_{n_k} = \phi(y_{n_k}) \to \phi(\eta) \in \phi(D)$. Hence $\xi = \phi(\eta) \in \phi(D)$.
You have already shown that $g$ is a bijection. To prove that is a homeomorphism it suffices to verify that it is a closed map. So let $E \subset \mathbb R^2/A$ be closed. Then $p^{-1}(E)$ is closed in $\mathbb R^2$ and $g(E) = \phi(p^{-1}(E))$ is closed in $\mathbb R^2$.
