# Proving projective space is a topological manifold

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

• 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. – Lee Mosher Dec 10 '18 at 15:24
• Oh sorry I forgot that line, already edited it. – Johanna Dec 10 '18 at 15:30

## 1 Answer

$$U_i:=\{[x_0,\dots , x_n] : x_i\neq 0\}$$ 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 $$f_0([2,2,0 \dots, 0]=(2,0,\dots, 0)$$

• 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 – Johanna Dec 10 '18 at 15:38
• You’re right sorry – Federico Fallucca Dec 10 '18 at 15:46