Topology of punctured plane vs topology of circle?

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SUMMARY

The topology of R^n minus the origin is not homeomorphic to the (n-1)-dimensional sphere, despite their similarities. The discussion highlights the concept of homotopy equivalence and deformation retractions, which indicate that while these spaces share certain topological properties, their cutsets differ significantly. The fundamental groups and homology groups are isomorphic due to the functoriality of induced maps, but the infinite-to-one nature of the standard map from R^2\{0} provides further distinction between the two topologies.

PREREQUISITES
  • Understanding of basic topology concepts, including homotopy and homeomorphism.
  • Familiarity with fundamental groups and homology groups in algebraic topology.
  • Knowledge of deformation retractions and their implications in topology.
  • Experience with R^n and its properties, particularly in relation to punctured spaces.
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  • Research the concept of homotopy equivalence in greater detail.
  • Explore the implications of deformation retractions in various topological spaces.
  • Study the properties of fundamental groups and homology groups in algebraic topology.
  • Examine the differences between punctured spaces and their corresponding spheres.
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Mathematicians, particularly those specializing in topology, algebraic topology students, and anyone interested in the properties of punctured spaces and their relationships to spheres.

nonequilibrium
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So how does the topology of R^n minus the origin relate to that of the (n-1)-dimensional sphere?

I would think the topology of the former is equivalent to that of an (n-1)-dimensional sphere with finite thickness, and open edges. But I suppose that is as close as one can get to the (n-1)-dimensional sphere itself? Hm, they seem pretty similar, but not the same... Is there a way in which they are approximately the same topology-wise?
 
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Check out homotopy equivalence and deformation retractions.
 
Yes; a homotopy induces an isomorphism of fundamental groups and homology groups:

by functoriality of induced maps , if fg=IdX and gf=IdY , then

(fg)* =f*g* =IdX* , and

same idea for (gf)*. Then the induced maps are isomorphisms.

Still, note that the cutsets of the two are different, so that they are not homeomorphic

(tho maybe the fact that the standard map from R^2\{0| is infinite-to-one may give a clue

of this..
 
Hey, thank you both. Micromass' suggestion made me look up stuff that indirectly made me understand Bacle's post (and more). The homotopy equivalence of spaces is what I was looking for :)
 
one has the same topology as the product of the other with the real line.
 

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