$\Bbb{RP}^n$: Saturated Open Set

Question

In the above entry of $\mathbb{RP}^n$ which is from Lee's Introduction To Smooth Manifolds, it says that $\tilde{U_i}$ is a saturated open set under the projection $\pi$. But I don't understand why that is true. Note that a set $U$ being saturated under a quotient map is such that $\pi^{-1}(\pi(U)) = U$.

Answer 1

Suppose $\pi(y) = \pi(x)$ for some $x \in \widetilde U_i$. That is to say, $y \in \pi^{-1}(\pi(\widetilde U_i))$. If $x = (x_1, \cdots, x_{n+1})\in \widetilde U_i$, then $x_i \neq 0$. To say that $\pi(y) = \pi(x)$ means that $y$ and $x$ span the same line. This is only true if $y$ and $x$ are scalar multiples of one another: $y = cx$ for some $c \neq 0$.

If $x = (x_1, \cdots, x_{n+1})\in \widetilde U_i$, then $x_i \neq 0$. Therefore $y_i = cx_i \neq 0$. So $y \in \widetilde U_i$. So we have $\pi^{-1}(\pi(\widetilde U_i)) \subset \widetilde U_i$, as desired; the two sets are equal.

Answer 2

Let $H_i \subset \mathbb{R}^{n+1}$ denote the hyperplane $x_i = 0$ which is closed in $\mathbb{R}^{n+1}$. Hence $H_i' = H_i \setminus \{ 0\}$ is closed in $\mathbb{R}^{n+1}\setminus \{ 0\}$ so that $\tilde{U}_i = (\mathbb{R}^{n+1}\setminus \{ 0\}) \setminus H'_i$ is open.

Let $x \notin \tilde{U}_i$ and $y \in \tilde{U}_i$, i.e. $x \in H'_i$ and $y \notin H'_i$.

The line $\pi(x)$ spanned by $x$ is contained in $H_i$ and the line $\pi(y)$ spanned by $y$ is not contained in $H_i$, hence $\pi(x) \ne \pi(y)$. This means that $\pi(x) \in U_i = \pi(\tilde{U}_i)$ implies $x \in \tilde{U}_i$. Hence $\tilde{U}_i$ is saturated.