Is there a continuous surjection from $\Bbb{R}^3\setminus S^2$ to $\Bbb{R}^2\setminus\{(0,0)\}$

Question

In the above question $S^2$ denotes the unit sphere in $\Bbb{R}^3$ represented by $x^2+y^2+z^2=1$. I know that $\Bbb{R}^3\setminus S^2$ is disconnected while $\Bbb{R}^2\setminus\{(0,0)\}$ is connected. So there cannot be a continuous surjection from $\Bbb{R}^2\setminus\{(0,0)\}$ to $\Bbb{R}^3\setminus S^2$. But what about the reverse implication?

In general can the continuous image of a disconnected set be connected?

Thanks in advance.

Answer 1

The image of a connected set by a continuous map is connected. Let $f:X\rightarrow Y$ be continuous and $X$ connected, suppose that there are two disjoint open subsets $U,V$ such that $f(X)\subset U\cup V$, $X\subset f^{-1}(U)\cup f^{-1}(V)$ and $f^{-1}(U)\cap f^{-1}(V)$ is empty. This implies that $f^{-1}(U)$ or $f^{-1}(V)$ is empty since $X$ is connected. We deduce that $U$ or $V$ is empty and henceforth $f(X)$ is connected.

Answer 2

There is such a map :

project $\Bbb R^3 \setminus S^2$ onto $\Bbb R^2$ (projection on the first two coordinates, for instance);

apply the exponential map from $\Bbb R^2$ onto $\Bbb R^2\setminus\{(0,0)\}$.

Both are continuous surjections, so their composition still is continuous