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I just found that the projective space $\mathbb{R}P^{n}$ (the space of all lines throught the orige on $\mathbb{R^{n+1}}$) is homeomorphic to the disc obtained by identifying the antipodal points of $S^{n-1}$.

How can I show this spaces are diffeomorphics?

I Would like any answer or reference about it.

Thanks in advanced.

Rodrigues
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  • The $n=1$ case is already on MSE: https://math.stackexchange.com/questions/2081895/real-projective-space-mathbbrp1-is-diffeomorphic-to-s1 . – Eric Towers Sep 07 '19 at 21:12

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First map $D^n \mapsto \mathbb R^{n+1} - \{\mathcal O\}$ (where $\mathcal O$ denotes the origin) by the function $$f(x_1,...,x_n) = \left((x_1,...,x_n,\sqrt{1-x_1^2-...-x_n^2}\right) $$ Then compose with the standard quotient map $$q : \mathbb R^{n+1} - \{\mathcal O\} \mapsto \mathbb RP^n $$ taking each point $P \in \mathbb R^{n+1} - \{\mathcal O\}$ to the element of $\mathbb RP^n$ represented by the line $\overline{\mathcal O P}$.

Do a little work to convince yourself that this composition $$f \circ q : D^n \to \mathbb R P^n $$ is also a quotient map (you can apply a theorem of topology that any continuous, surjective function from a compact space to a Hausdorff space is a quotient map), and that two distinct points $A \ne B \in D^n$ have the same image $f(A)=f(B)$ if and only if $A,B$ are antipodal points on $S^{n-1}$. And then you're done.

Added: Quotient spaces do not ordinarily have smooth structures defined on them, so that part of the question does quite really make sense. Sometimes one can observe that the quotient map is a smooth map, as it is in this case, so at least the smooth structures on the domain and range are consistent with each other

In this case, one can say a little bit more. Suppose $P,-P \in S^{n-1}$ is a pair of antipodal points. Choose a smooth manifold-with-boundary coordinate chart near $P$, which has the form $$f : [0,1) \times \mathring{D}^{n-1} \to U_P $$ for some neighborhood $U_P$ of $P$, and assume that $U_P \cap -U_P = \emptyset$. One obtains a smooth manifold-with-boundary chart $$g : [0,-1) \times \mathring{D}^{n-1} \to -U_P $$ by defining $g(t,x) = -f(-t,x)$. One can then put these together to define a smooth manifold coordinate chart near the point of $\mathbb RP^n$ represented by $\{P,-P\}$, namely the function $$h : (-1,+1) \times \mathring{D}^{n-1} \to q(U_P \cup - U_P) \subset \mathbb R P^n $$ defined by $$h(t,x) = \begin{cases} q \circ f(t,x) &\quad\text{if $t \in [0,-1)$} \\ q \circ g(t,x) &\quad\text{if $t \in (-1,0]$} \end{cases} $$ In this manner, using a smooth atlas on $D^n$ I can derive a smooth atlas for $\mathbb RP^n$.

However, I do not like saying that the quotient space is diffeomorphic to $\mathbb RP^n$, unless I first define exactly what I mean by a general concept of a smooth structure on a quotient space. This definition would have strict preconditions which do not hold for quotient spaces in general.

Lee Mosher
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  • Why $D^{n}/ \sim$ is diffeomorphic to $\mathbb{R}P^{n}$? You show that $q \circ f$ induces a homeomorphism $\bar{f}:D^{n}/\sim \rightarrow \mathbb{R}P^{n}$, don't you? thanks for your answer. – Rodrigues Sep 07 '19 at 23:31
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    I've added some additional comments regarding the issue of "diffeomorphism". – Lee Mosher Sep 08 '19 at 13:50