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

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This was asked several times on math.se but none of them was answered.

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.

Edit: Since I can't comment I'm writing here: @JackLee by $x$ I mean the identification of $[x]$ in $[0,1]/∼$