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I'm trying to prove that $\mathbb{P}^n(\mathbb{R})$ is a topological manifold of dimension n.

So I can define $f_i: U_i \rightarrow \mathbb{R}$ such that $f_i([x_0:x_1: \dots : x_n])= (x_0, \dots x_{i-1}, x_{i+1}, \dots x_n)$ where $U_i=\{[x_0:\dots : x_n] : x_i=1\}$. then I only need to prove this is an homeomorphism. Open sets on $\mathbb{P}^n(\mathbb{R})$ are $\pi(A)$ where $A$ is an open set of $\mathbb{R}^{n+1}$. For $f_0$:

Is open because if $A=A_0 \times ... \times A_n$ $f_0(\pi(A))=f_0( .\cup_{x_0\in A_0} [1:\frac{A_1}{x_0}: \dots : \frac{A_n}{x_0}]) = \cup_{x_0 \in A_0} \frac{A_1}{x_0}\times \dots \times \frac{A_n}{x_0}$ which is open.

But I'm not sure how to see this is continuous. I would appreciate any help for doing this Thanks

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  • $\begingroup$ You wrote "I only need to prove that this is an homeomorphism". A formula does not define a function, and hence does not define a homeomorphism. You must also specify the domain and range. Notice how this is done in the answer of @FedericoFallucca. $\endgroup$ – Lee Mosher Dec 10 '18 at 15:24
  • $\begingroup$ Oh sorry I forgot that line, already edited it. $\endgroup$ – Johanna Dec 10 '18 at 15:30
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$U_i:=\{[x_0,\dots , x_n] : x_i\neq 0\}$ is an open set of $\mathbb{P}^n$ and you can define

$\phi_i: U_i\to \mathbb{R}^n$ such that

$\phi_i([x_0,\dots , x_n] ):=(\frac{x_0}{x_i},\dots , \frac{x_n}{x_i})$

That it is an omeomorphism but $\cup_{i=0}^n U_i=\mathbb{P}^n$ and for every $i\neq j$

$\phi_j\circ \phi_i^{-1}(x_1,\dots , x_n)= (\frac{x_1}{x_j},\dots ,\frac{1}{x_j},\dots , \frac{x_n}{x_j}) $

and in $\phi_i(U_i\cap U_j)=\{(x_1,\dots , x_{n}) : x_j\neq 0\}$ Is infinitely differentiable

Your map is not well defined.. for example $f_0([1,1,\dots , 0])=(1,0, \dots ,0)$ but $[1,1,0,\dots , 0]=[2,2,0 \dots, 0]$so $f_0([2,2,0 \dots, 0]=(2,0,\dots, 0)$

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    $\begingroup$ But I'm using the representative of the class that has a 1 on the i-th coordinate, so $[2,2,0...0] $ is in fact represented by $[1,1,0...0]$. I don't see why I can't do this $\endgroup$ – Johanna Dec 10 '18 at 15:38
  • $\begingroup$ You’re right sorry $\endgroup$ – Federico Fallucca Dec 10 '18 at 15:46

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