Proving that the family $\{U_i,\phi_i\}_{i=1}^{n+1}$ is an atlas of the projective space.

Question

We have the binary equivalence relation $$xRy \iff \exists\lambda\in\mathbb{R}:x=\lambda y$$ so the quotient space $$(\mathbb{R}^{n+1}\setminus\{0\})/{R}$$ is the projective space.

Then, we define $V_i=\{x\in\mathbb{R}^{n+1}-\{0\}:x_i\neq 0\}$, $U_i=\pi(V_i)$, where $\pi$ is the application that maps a point $x \in \mathbb{R}^{n+1}\setminus\{0\}$ to its equivalence class $[x]_R$ and

$$\phi_i:U_i\longrightarrow \mathbb{R}^n$$ as $$\phi_i([x]_R)=\dfrac{1}{x_i}(x_1,...,x_{i-1},x_{i+1},...,x_{n+1})$$

I have to prove that the family $\{U_i,\phi_i\}_{i=1}^{n+1}$ is an atlas for the projective space, i.e, with the definition I am using, I have to prove that

1. $\bigcup\limits_{i=1}^{n+1} U_i =\mathbb{R}P^n$ and $\phi_i$ is injective $\forall i=1,...,n+1$

2. For $i\neq j,\quad \phi_i(U_i\cap U_j)\quad$ and $\quad \phi_j(U_i\cap U_j) \quad$ are open sets, and $\phi_i\circ\phi_j^{-1}$ is differentiable of class $C^k$.

I have already done the first part but I am having trouble proving that the sets mentioned in 2 are open. I have checked the answer provided and I don't understand the procedure followed: first, we prove that $\phi_i(U_i)=\mathbb{R}^n$ and then we take $[x]_R\in U_i\cap U_j$ and we prove that $$\phi_i([x])=\phi_i([\frac{1}{x_i}x])\in \mathbb{R}^{j-2}\times\mathbb{R}-\{0\}\times\mathbb{R}^{n-j}$$

I get the first step but I am not sure why this is not enough to prove that the intersection is an open set. If $\phi_i(U_i)$ is an open set and $\phi_i(U_j)$ is also an open set, does this not mean that $\phi_i(U_i\cap U_j)$ is also an open set? Why do we need the last step too?

Answer 1

There are lot of things which have to be checked.

The key is to observe that the quotient map $\pi: \mathbb R^{n+1} \setminus \{0\} \to \mathbb RP^n$ is an open map. Note that all open maps are quotient maps, but the converse fails. See for example When is a quotient map open?

To see that $\pi$ is open, let $V \subset \mathbb R^{n+1} \setminus \{0\}$ be open. Then $\pi^{-1}(\pi(V)) = \{ \lambda x \mid \mathbb R^{n+1} \setminus \{0\}, x \in V \} = \bigcup_{\mathbb R^{n+1} \setminus \{0\}} \lambda V$, where $\lambda V = \{ \lambda x \mid x \in V \}$. Since multiplication with $\lambda \ne 0$ gives a self-homeomorphism on $\mathbb R^{n+1} \setminus \{0\}$ (its inverse being multiplication with $1/\lambda$), all $\lambda V$ are open and so is their union. We conclude that $\pi(V)$ is open in $\mathbb RP^n$.

Since $V_i$ is open in $\mathbb R^{n+1} \setminus \{0\}$, the restriction $\pi_i : V_i \to U_i$ of $\pi$ is an open map, hence a quotient map.

We conclude that $\phi_i$ is continuous (this follows from the obvious continuity of $\phi_i \circ \pi_i : V_i \to \mathbb R^n$).

The map $f_i : \mathbb R^n \to U_i, f_i(y_1,\ldots,y_n) = (y_1,\ldots,y_{i-1},1,y_i,\ldots, y_n)$ is clearly continuous. Thus $\psi_i = \pi_i \circ f_i : \mathbb R^n \to V_i$ is continuous. It is easy to check that $\psi_i \circ \phi_i = id$ and $\phi_i \circ \psi_i = id$, and this shows that $\phi_i$ is a homeomorphism.

Since $\bigcup_{i=1}^{n+1}U_i = \mathbb RP^n$, the homeomorphisms $\phi_i$ form an atlas for $ \mathbb RP^n$. As in any atlas, the intersections $U_i \cap U_j$ are open and mapped by $\phi_i, \phi_j$ homeomorphically to open subsets $\phi_i(U_1 \cap U_j)$, $\phi_j(U_1 \cap U_j)$ of $\mathbb R^n$.

It remains to show that the transition functions $\phi_{ji} = \phi_i \circ \phi_j^{-1} : \phi_j(U_1 \cap U_j) \to \phi_i(U_1 \cap U_j)$ are smooth (i.e. of class $C^\infty$). This is trivial for $i = j$. Let us consider the case $i < j$ (the case $j < i$ is treated similarly). We have $$\phi_i \circ \phi_j^{-1} = \phi_i \circ \psi_j = \phi_i \circ \pi_j \circ f_j$$ and therefore $$\phi_{ji}(y_1,\ldots,y_n) = \phi_i([y_1,\ldots,y_{j-1},1,y_j,\ldots, y_n]) \\= (\frac{y_1}{y_i}, \ldots, \frac{y_{i-1}}{y_i}, \frac{y_{i+1}}{y_i}, \ldots, \frac{y_{j-1}}{y_i}, \frac{1}{y_i},\frac{y_j}{y_i}, \ldots, \frac{y_n}{y_i}) .$$

All $n$ component functions are clearly smooth, thus $\phi_{ji}$ is smooth.

Although it is irrelevant, let us finally determine $U_i \cap U_j$. Clearly $\pi^{-1}(U_i \cap U_j) = \pi^{-1}(U_i) \cap \pi^{-1}(U_j) = V_i \cap V_j$, thus $U_i \cap U_j = \pi(V_i \cap V_j)$.