Why is there a homeomorphism from ($\{ (x,y) \in R^2 | xy = 1 \}$) (a hyperbola) to two lines. In class it was stated that this set $\{ (x,y) \in R^2 | xy = 1 \}$ (a hyperbola) is homeomorphic with two real lines.
Informally it was stated that there exists a projection of the hyperbola on two real lines and this gives us the homeomorphism, but how would this be done?
I don't mean to obtain the homeomorphism explicitly but at-least understand how is this projection constructed. 
If you have a link to a figure please share it.
 A: The line $x+y=2$ is tangent to the branch of the hyperbola in the first quadrant, and the line $x+y=-2$ is tangent to the branch in the third quadrant, as in the (rather crude) picture below:

Project the upper branch of the hyperbola at $45^\circ$ down and to the left onto the upper tangent line, and project the lower branch of the hyperbola at $45^\circ$ up and to the right onto the lower tangent line, as indicated by the blue arrows.
A: One intuitive way to do this would be to start with the orthogonal projection $\pi:\mathbb{R}^2 \to L$, where $L$ is the line $\{(x,y) \mid x+y=0\} \subset \mathbb{R}^2$.
Clearly, $L$ is homeomorphic to $\mathbb{R}$.  Also, when restricted to either component of $H=\{(x,y) \mid xy=1\}$, $\pi$ has a continuous inverse (drop a perpendicular), and so it is a homeomorphism.

In fact, this approach can give us a nice explicit description of a homeomorphism mapping one branch of $H$ to $\mathbb{R}$, such as $(x,y) \mapsto x-y$.
A: An alternate approach is to just consider the vertical projection onto the $x$-axis, i.e. the map $(x,y)\mapsto y$.  You can show that this a homeomorphism from $\{(x,y)\in\mathbb{R}^2:xy=1\}$ to the subset $(-\infty,0)\cup(0,\infty)\subset\mathbb{R}$.  This is perhaps more accurately described as two (open) rays rather than two lines, but you can also show that $(0,\infty)$ and $(-\infty,0)$ are both themselves homeomorphic to $\mathbb{R}$ (see, for instance, this question), so you can also think of it as two lines.
