Continuous surjective map Does there exist a continuous surjective map from $\{(x, y)\in \mathbb{R^2} | x^2-y^2=1\}$ to $\mathbb{R}$?
Any help would be greatly appreciated. I was trying to work with taking one branch of the hyperbola and restricting the function to it. But I cannot understand how to proceed much further.
 A: One possibility is to project to the $y$-axis, i.e. define $\pi_y : \mathbb{R}^2 \to \mathbb{R}$ by
$$ \pi_y((x,y)) := y. $$
This map will be continuous with respect to the usual topologies on $\mathbb{R}^2$ and $\mathbb{R}$ (note that the preimage of an open set $U$ is a rectangle of the form $\mathbb{R}\times U$, which is open in $\mathbb{R}^2$), which means that the restriction of $\pi_y$ to the hyperbola will be continuous (with respect to the subspace topology).
To see that the map is surjective, note that if $y\in\mathbb{R}$, then the point
$$ \left( \sqrt{1+y^2}, y \right) $$
is in the hyperbola, as
$$ \left(\sqrt{1+y^2}\right)^2 - y^2= (1+y^2) - y^2 = 1. $$
But then $\pi_y$, restricted to the hyperbola, is a continuous surjection.
A: Yes! Ascribe points of hyperbola with $x>0$ to positive numbers and points with $x<0$ to negative ones. Also map $(1,0)$ and $(-1,0)$ to 0.
A: Define $f: \mathbb R^2 \setminus (\{0\}\times \mathbb R)$ by
$$f(x,y) = \begin{cases} x&& x>0\\ x+2 && x<0 \end{cases}$$
Then $f$ is continuous on $\mathbb R^2 \setminus (\{0\}\times \mathbb R).$ Since the hyperbola, let's call it $H,$ is contained in the domain of $f,$ $f$ is continuous on $H.$ Check that $f$ sends the right branch of $H$ onto $[1,\infty),$ and sends the left branch of $H$ onto $(-\infty,1].$ Hence $f|_H$ is a continuous map of $H$ onto $\mathbb R.$
