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I'm trying to understand the CW complex structure of the projective space $\mathbb{RP}^n$, but some things are unclear. I understand we start by identifying $\mathbb{RP}^n$ with $S^n/R$ where $R$ is the equivalence relation identifying antipodal points on the sphere. This is fine. But then $S^n/R$ is identified to $D^n/R$ with R this time restricted to the border $S^{n-1}$ of $D^n$. Here is my problem: can anybody provide an explicit map of this identification? And secondly, how can we identify this last space to the adjoint space of $\mathbb{RP}^{n-1}$ and $D^n$, in other words, how does $\mathbb{RP}^{n-1}$ becomes a (n-1) skeleton of the CW-complex from here?

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Since you were using $R$ for two different things, I took the liberty to denote real projective space by $\mathbb{RP}^n$ –  M Turgeon Sep 12 '12 at 15:32

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The natural inclusion of the hemisphere $D^n \to S^n$ respects the relation $R$ as described. So, it induces a map $D^n/R \to S^n/R$, which is a homeomorphism. Now, consider the inclusion of the boundary $S^{n-1} \to D^n$, and see that this too respects the relation $R$, thus inducing an inclusion map $S^{n-1}/R = \mathbb R P^{n-1} \to D^n/R \cong \mathbb RP^n$. It should now be easy to see that we obtain $D^n/R \cong \mathbb RP^n$ from $\mathbb R P^{n-1}$ by attaching $D^n$ along the quotient map $S^{n-1} \to S^{n-1}/R = \mathbb RP^{n-1}$.

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It's a bit better but I'm still unsure. So I guess you mean that the map $D^n \rightarrow S^n \rightarrow S^n/R$ is the map respecting the relation and therefore induces $D^n/R \rightarrow S^n/R$? But I still can't find the bijection. Also, I can see that we can create a n-cellular extension with $D^n$, $RP^{n-1}$ and $S^{n-1}$, but again don't we need to prove that the resulting space is actually homeomorphic to $RP^n$? –  Tom Sep 12 '12 at 16:08
    
Write out the coordinates and see what it means to be injective/surjective for that map. Once that is done, we have already proved that the "n-cellular extension" namely $D^n/R$ is homeomorphic to $\mathbb R P^n$. –  Justin Young Sep 13 '12 at 7:12

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