I am trying to show that $\{(x,y)\in\mathbb{R}^2:x>0\}$ is homeomorphic to the disk $\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}$. I know that the disk is homeomorphic to the whole plane. A homeomorphism from $\mathbb{R}^2$ to the disk could be $f(x,y)=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})$. I know that being homeomorphic is an equivalence relation. So I was looking for a homeomorphism between the half plane and the whole plane, but I couldn't find one. Any idea?
Asked
Active
Viewed 2,343 times
8
-
4You can try to find a homeomorphism from $(0,\infty)$ to $\mathbb R$ first. – May 17 '15 at 22:11
-
I think you can do this regarding $\mathbb R^2$ as $\mathbb C$ and then just finding a Mobius tranformation whose image of the imaginary axis is the unit circle and maps the open right half-plane onto the open unit diisk. – Matematleta Dec 16 '23 at 16:15
3 Answers
9
Hint: $x \mapsto e^x$ is a homeomorphism between $\mathbb{R}$ and $(0, \infty)$
Rob Arthan
- 48,577
-
So $f:\text{half plane}\to \mathbb{R}^2$ defined as $f(x,y)=(\ln(x),y)$ would an isomorphism, then I compose it with the one I found? – 3x89g2 May 17 '15 at 22:41
-
2
Use the Riemann Mapping theorem, since the right half is simply-connected. A biholomorphic function is a homeomorphism. You may be able to prove this last by invariane of domain.
Gary.
- 2,432
0
What about the map $f : \mathscr H \longrightarrow \mathbb D^2$ defined by $$(x,y) \mapsto \left (\frac {2y} {1 + x^2 + y^2}, \frac {x^2 + y^2 -1} {1 + x^2 + y^2} \right ).$$ The inverse of which is given by $$(x,y) \mapsto \left (\frac {\sqrt {1 - x^2 - y^2}} {1 - y}, \frac {x} {1 - y} \right ).$$
Hint $:$ Stereographic Projection.
Akiro Kurosawa
- 485
- 11