Homeomorphism between a cylinder and a plane?

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SUMMARY

The discussion addresses the homeomorphism between a cylinder and a punctured plane, specifically $\mathbb{R}^2\setminus\{(0,0)\}$ and the cylinder defined as $C=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=1\}$. An explicit homeomorphism is constructed using the function $$f:\mathbb{R}^2\setminus\{(0,0)\}\to C\;,\quad f(r\cos\theta,r\sin\theta)=(\cos \theta,\sin\theta,\ln r)\;(r>0)$$. The discussion concludes that while the punctured plane is homeomorphic to the cylinder, the full plane $\mathbb{R}^2$ is not homeomorphic to the punctured plane due to the property of simple connectivity.

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  • Understanding of homeomorphism in topology
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  • Knowledge of cylindrical coordinates
  • Basic concepts of simple connectivity
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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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