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I was wondering if the following two ways of defining topology on $P^n(\mathbb{R})$ are the same and why?

  1. 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?).
  2. 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!

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    Identify $P^n(\mathbb{R})$ with a quotient space of $S^n$. This space will be compact. Then look for continuous bijections to the two models you gave and show that these two models are Hausdorff. Apply the fact that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. I strongly suggest that you do this yourself in detail...2011-12-25
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    @t.b.: Thanks! I am really not sure, so may I ask what the two models/spaces that I gave are?2011-12-25
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    You gave two (a priori) different topologies on $P^n(\mathbb{R})$, I described a third one. The two different topologies give you two a priori different spaces but they turn out to be the same. I outlined one way how to prove that this is the case.2011-12-25
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    The tradition map from Euclidean space to projective space doesn't include $0$ in the domain.2011-12-25
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    @ThomasAndrews: Thanks for pointing it out!2012-01-01
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    @t.b.: Sorry, I thought about your hint for a while and then couldn't help but diverged to something else. (1) In your hint, I wonder why it is "a continuous bijection from a compact space to a Hausdorff space is a homeomorphism"? Can I find it from some book like the topology book by Munkres? (2) In the Wiki's construction of topology on $P^n(\mathbb{R})$, "a subset of $U_i$ is open iff its image under the above isomorphism is an open subset of $\mathbb{R}^n$". Does it mean the minimal topology on $U_i$ so that the isomorphism to $\mathbb{R}^n$ is continuous? Thanks!2012-01-01
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    (1) yes, that's in every topology book. See also [here](http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2006;task=show_msg;msg=2558.0001). (2) It's the [quotient topology](http://en.wikipedia.org/wiki/Quotient_space) induced from the map $\mathbb{R}^{n+1} \smallsetminus \{0\} \to P^n(\mathbb{R})$, that is to say the *finest* (largest) topology on $P^n(\mathbb{R})$ making the projection continuous.2012-01-01
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    @t.b.: Thanks! About (2), I was referring to Wiki's construction as in part 2 of my post, and I think you were thinking about part 1 of my post?2012-01-01

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