I was wondering if the following two ways of defining topology on $P^n(\mathbb{R})$ are the same and why?
- Since $P^n(\mathbb{R})$ is the quotient space of $\mathbb{R}^{n+1}$, define the topology on $P^n(\mathbb{R})$ to be the quotient topology ( i.e. the maximal topology that can make the quotient map $q: \mathbb{R}^{n+1} - \{0\} \to P^n(\mathbb{R})$ continuous, if I understand correctly or do I?).
From Wikipedia:
Consider the following subsets of $P^n(\mathbb{R})$: $ U_i = \{[x_0:\cdots: x_n], x_i \neq 0\}, i=0, \dots,n. $ Their union is the whole projective space. Furthermore, $U_i$ is in bijection with $\mathbb{R}^n$ via the following maps: $ [x_0:\cdots: x_n] \mapsto \left (\frac{x_0}{x_i}, \dots, \widehat{\frac{x_i}{x_i}}, \dots, \frac{x_n}{x_i} \right ) $ $ [y_0:\cdots: y_{i-1}: 1: y_{i+1}: \cdots y_n] \leftarrow \left (y_0, \dots, \widehat{y_i}, \dots y_n \right ) $ (the hat means that the $i$-th entry is missing).
Then define a topology on projective space by declaring that these maps shall be homeomorphisms, that is, a subset of $U_i$ is open iff its image under the above isomorphism is an open subset (in the usual sense) of $\mathbb{R}^n$. An arbitrary subset $A$ of $P^n(\mathbb{R})$ is open if all intersections $A ∩ U_i$ are open. This defines a topological space.
Thanks and regards!