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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?

glS
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Tom
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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

2 Answers2

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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 $S^{n-1} \to S^{n-1}/R = \mathbb RP^{n-1}$.

Justin Young
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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
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    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 07:12
  • By the "natural inclusion of the hemisphere $D^n\to S^n$" you mean the map $(x_1,...,x_n)\mapsto(x_1,...,x_n,\sqrt{1-x_1^2-x_2^2-...-x_n^2})$ or the map $tu\mapsto (\sin(t\pi/2)u,\cos(t\pi/2))$ where $t\in [0,1]$ and $u\in S^{n-1}$?. – Zero Aug 29 '15 at 03:56
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    The first map, I suppose, I am identifying $D^n \subset S^n$ as the $x\in S^{n}$ such that $x_{n+1} \ge 0$. – Justin Young Aug 29 '15 at 11:39
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First see the CW-structure of $S^n$ as 2 0-cells,2 1-cells, ... , 2 n-cells and attaching maps are natural.Then see that $Z/2$ act on $S^n$ by antipodal action and this $Z/2$ action flip each 2 $i$-cells.Note $RP^n = S^n / Z/2$ This gives the CW complex structure of $RP^n$.

Math
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