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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.
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.