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The most common definitions of real projective spaces are:

  • $\mathbb{R} \mathbb{P} ^n = (\mathbb{R}^{n+1} - 0)/ \sim$, where $x,y \in \mathbb{R}^{n+1}-0$ satisfies $x \sim y$ iff $x = \lambda y$ for such a $\lambda \in \mathbb{R}^*$;

  • $\mathbb{R} \mathbb{P} ^n = S^n / \sim_1$, where $\sim_1$ is the classical antipodal equivalence.

Where can I find a complete and rigorous proof that the two definitions are equivalent, in the sense that the two structures are diffeomorphic as differentiable manifolds? I clearly see the point by an intuitive point of view, but I need a rigorous proof.

Edit: Yes, the answers recevied are ok. Furthermore, I have found a complete and detailed proof on the book "Introduction to Global Analysis", by Donald W. Kahn [pag. 40, Proposition 2.1]

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    $\begingroup$ In the first definition, if you take only $\lambda >0$, the resulting space is $S^n$. $\endgroup$
    – Arthur
    Nov 28, 2014 at 11:05

1 Answer 1

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Hint: consider $$i: S^n\longrightarrow\Bbb R^{n+1}$$ the inclusion and $$p: \Bbb R^{n+1}\longrightarrow\Bbb{RP}^n$$ the quotient (first definition) map. The composition $$p\circ i: S^n\longrightarrow\Bbb{RP}^n$$ is smooth (why?). What is $S^n/(p\circ i)$ (points with the same image are identified)?

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  • $\begingroup$ Thank you Martìn, but please may I can ask you a clarification? For instance, why $p \circ i$ is smooth? I think it could be that $p$ is smooth, and so their composition...but I am getting confuse in understanding why the projection is smooth... $\endgroup$
    – Biagio
    Dec 13, 2014 at 16:04
  • $\begingroup$ @Biagio, consider near (say) $(0,\cdots,0,1)$ the function $\Bbb R^{n+1}\rightarrow\Bbb R^n$ $(x_1/x_{n+1},\cdots,x_n/x_{n+1})$ and use it to construct a chart of $\Bbb{RP}^n$ (first definition). Study the differentiability of $p$ using the chart. Cheat sheet: homepage.math.uiowa.edu/~idarcy/COURSES/133/3_2rpns.pdf $\endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 13, 2014 at 17:49
  • $\begingroup$ thank you for the comment. Now the point is the following: let $p$ be a chart $\phi^{-1}$ as stated in the link, actually a smooth and injective map $p: \Bbb R^{n+1}\longrightarrow\Bbb{RP}^n$. Consequently, because $i$ is smooth and injective too, so is their composition. As a result we have the map $p \circ i: S^n \longrightarrow \Bbb{RP}^n$ smooth and injective. I suppose, but I am not so sure, that the quotient $p\circ i: S^n/(p\circ i) \longrightarrow\Bbb{RP}^n$ becomes smooth, bijective and with inverse smooth...is it true? How can I prove it? $\endgroup$
    – Biagio
    Dec 18, 2014 at 15:25
  • $\begingroup$ $p\circ i$ isn't injective ($p$ neither), but is surjective and locally injective. The quotient map is bijective and... See also www2.math.uu.se/~tkragh/1MA259-2013/Solution-3.5.pdf. $\endgroup$
    – Martín-Blas Pérez Pinilla
    Dec 18, 2014 at 17:28

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