How to show $P^n$ (real projective space of dimension n) is triangulable? That is, how to show there exists a triangulation of $P^n$? By triangulation, I mean a simplicial complex $K$ and a homeomorphism from $|K|$ (the polyhedron of K) to $P^n$.
I'm thinking of $P^n$ as the closed ball $B^n$ with antipodal points of the boundary identified. Let $q:B^n\rightarrow P^n$ be the quotient map that identifies antipodal points of the boundary. I think if I have a triangulation $K$ of $B^n$ (i.e. if K is a simplicial complex homeomorphic to $B^n$) and I take the union of the image of each simplex of K under $q$, then I should get a simplicial complex which is homeomorphic to $P^n$. In particular if $S$ is a simplex in $K$, I'm thinking that $q(S)$ is a simplex. Is this correct? So for example, the unit simplex $\Delta^n$ is homeomorphic to $B^n$. Does that mean that $q(\Delta^n)$ is a simplex (or simplicial complex)?
I'm pretty confused about triangulations, so my question may reveal particular misunderstandings. If so, please point them out.