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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.

rogerl
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Monolite
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3 Answers3

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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:

hyperbola with tangent lines

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.

mercio
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Brian M. Scott
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  • @Slade Thank you for your answers! If I wanted to explicitly write the orthogonal projection how would I do this? Could I have a reference or a link on how to project a function? – Monolite Nov 25 '15 at 14:22
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    @Monolite: It’s probably easier here to compute the inverse of the projection. Start at a point $\langle x,2-x\rangle$ on the upper line, for instance. If you move northeast perpendicular to the line, you go through points of the form $\langle a+x,a+2-x\rangle$. You hit the hyperbola when $(a+x)(a+2-x)=1$, which can be rearranged to $a^2+2a-(x-1)^2=0$. Since $a\ge 0$, this gives us $a=-1+\sqrt{1+(x-1)^2}$, so $\langle x,2-x\rangle$ corresponds to $$\langle x-1+\sqrt{1+(x-1)^2}, 1+\sqrt{1+(x-1)^2}-x\rangle$$ on the hyperbola. You can do the same sort of thing for the other line. – Brian M. Scott Nov 25 '15 at 21:21
  • Thank you so much Brian that all made sense! Just to see if I got it: if I wanted to project my point in the direction of $y = x+1$ instead of $y=x$ I would obtain the points $\langle a+1+x,a+2-x\rangle$ right? Now I am left with the doubt on how I write the homeomorphism explicitly though since I can't do this procedure for infinite points. – Monolite Nov 25 '15 at 21:49
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    @Monolite: You’re very welcome. What I wrote there actually gives you a formula for a homeomorphism from the upper line to the upper branch of the hyperbola: $$h(\langle x,2-x\rangle)=\left\langle x-1+\sqrt{1+(x-1)^2}, 1+\sqrt{1+(x-1)^2}-x\right\rangle;.$$ – Brian M. Scott Nov 25 '15 at 21:55
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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 $(x,y) \mapsto x-y$.

Andrew Dudzik
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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.

Community
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Eric Wofsey
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