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Let $\mathbb{R}P^m$ be the $m$-dimensional real projective space and let $\mathbb{R}P^m\setminus\{*\}$ be the punctured space. I observe:

  1. $\mathbb{R}P^2\setminus\{*\}$ is homeomorphic to a (open) Mobius strip, that is, the total space of a non-trivial line bundle over $\mathbb{R}P^1$.

  2. $\mathbb{R}P^1\setminus\{*\}$ is homeomorphic to $\mathbb{R}$, which can be regarded as $\mathbb{R}P^0\times\mathbb{R}$.

Could I generalize (1) and (2) to the following?

(i). If $m$ is odd, then $\mathbb{R}P^{m+1}\setminus\{*\}$ is homeomorphic to the total space of a non-trivial line bundle over $\mathbb{R}P^{m}$.

(ii). If $m$ is even, then $\mathbb{R}P^{m+1}\setminus\{*\}$ is homeomorphic to $\mathbb{R}P^{m}\times\mathbb{R}$.

Question: Whether are (i) and (ii) true or false? Are there any related references?

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    $\begingroup$ You might want to read about Thom spaces. $\endgroup$
    – Todd Trimble
    Commented May 23, 2016 at 11:35
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    $\begingroup$ Thom spaces? All we do is remove two antipodal points from the round sphere and take the quotient by the antipodal map. The projection to the equator is an $\mathbb R$-bundle. It is nontrivial in the quotient: try to smoothly push $RP^m$ off itself. Then there would be $\mathbb Z_2$-invariant push off upstairs, and visibly it does not exist. Alternatively, you could note that the antipodal map of $S^n$ changes orientation iff $n$ is even, so $RP^{m+1}\setminus\{*\}$ and $RP^m$ cannot be both orientable, hence the normal bundle is non-orientable, and in particular non-trivial. $\endgroup$
    – Igor Belegradek
    Commented May 23, 2016 at 11:58
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    $\begingroup$ @IgorBelegradek The OP was asking about related references, and so I'm just mentioning that the Thom space of the tautological line bundle over $\mathbb{RP}^m$ is $\mathbb{RP}^{m+1}$. $\endgroup$
    – Todd Trimble
    Commented May 23, 2016 at 13:31
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    $\begingroup$ Alternatively one could say that the line bundle over ${\mathbb {RP}}^{n-1}$ is nontrivial because its restriction over ${\mathbb {RP}}^1$ is nontrivial. $\endgroup$
    – Allen Hatcher
    Commented May 23, 2016 at 17:52

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(i) is true and (ii) is false. In fact, (i) is true for all $m\ge1$. More precisely, dropping a coordinate yields a map from $\mathbb RP^{m+1}\setminus\{*\}\to\mathbb RP^m$ which is a locally trivial bundle with fiber $\mathbb R$. This bundle is not trivial for $m\ge1$ because, e.g., $\mathbb R P^m$ is orientable if and only if $m=0$ or $m$ is odd.

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