# CW complex structure of the projective space $\mathbb{RP}^n$

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?

#### Solutions Collecting From Web of "CW complex structure of the projective space $\mathbb{RP}^n$"

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

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 $RP^n$.