Construct an explicit isomorphism

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    Explicit Isomorphism
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SUMMARY

The discussion focuses on constructing an explicit isomorphism between the bundle ##E = \{([x], v) : [x] ∈ \Bbb{R}P^1, v ∈ [x]\}## and the Mobius bundle ##T = [0, 1] × R/ ∼##, where ##(0, t) ∼ (1, −t)##. The user verifies that ##\Bbb{R}P^1## is homeomorphic to ##\Bbb{S}^1## and proposes the mapping ##([x],v) \to (x,(1-x)v+xe^v)##. The user questions the naturalness of this mapping and considers using a pullback of the bundle via the homeomorphism for a more intuitive approach.

PREREQUISITES
  • Understanding of $\Bbb{R}P^1$ and its properties
  • Familiarity with Mobius bundles and their structure
  • Knowledge of homeomorphisms and their applications in topology
  • Basic proficiency in constructing mappings in differential geometry
NEXT STEPS
  • Research the properties of $\Bbb{R}P^1$ and its relation to $\Bbb{S}^1$
  • Study the construction and applications of Mobius bundles
  • Explore the concept of pullbacks in the context of fiber bundles
  • Investigate explicit isomorphisms in differential geometry
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Mathematicians, particularly those specializing in topology and differential geometry, as well as students seeking to understand the intricacies of bundle isomorphisms and their applications.

bedi
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$\Bbb{R}P^1$ bundle isomorphic to the Mobius bundle

I'm trying to construct an explicit isomorphism from ##E = \{([x], v) : [x] ∈ \Bbb{R}P^1, v ∈ [x]\}## to ##T = [0, 1] × R/ ∼## where ##(0, t) ∼ (1, −t)##. I verified that ##\Bbb{R}P^1## is homeomorphic to ##\Bbb{S}^1## which is homeomorphic to ##[0,1]/∼## where ##0∼1##. So this is the map I have in my mind: ##([x],v)\to (x,(1-x)v+xe^v)##. Does that work? It doesn't look very natural.
 
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How about pulling back the bundle using the homeomorphism?
 

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