MHB Homeomorphism between a cylinder and a plane?

AI Thread Summary
A homeomorphism exists between a punctured plane and a cylinder, demonstrated through a specific mapping function. The mapping transforms polar coordinates from the punctured plane to cylindrical coordinates, effectively linking the two spaces. However, the entire plane is not homeomorphic to the punctured plane due to the difference in their topological properties, specifically regarding simple connectivity. This distinction highlights the importance of understanding the implications of topological characteristics in homeomorphism discussions. The conversation emphasizes the nuanced relationships between different geometric structures in topology.
Fernando Revilla
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I quote a question from Yahoo! Answers

Because there is a homeomorphism between a cylinder and a plane?

I have given a link to the topic there so the OP can see my response.
 
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We can construct an explicit homeomorphism between the punctured plane (for example $\mathbb{R}^2\setminus\{(0,0)\}$) and a cylinder (for example $C=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=1\}$) given by:
$$f:\mathbb{R}^2\setminus\{(0,0)\}\to C\;,\quad f(r\cos\theta,r\sin\theta)=(\cos \theta,\sin\theta,\ln r)\;(r>0)$$
But $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^2\setminus\{(0,0)\}$, because simply connected is a toplogical property.
 
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