Show the real projective plane is an n-dimensional topological manifold 
Define $S^n:=\{x\in \mathbb{R}^{n+1}:\|x\|=1\}$
We define the real projective space of dimension n as $\mathbb{R}P^n:=S^n/(x\sim -x)$.
Show that $\mathbb{R}P^n$ is an n-dimensional topological manifold.

I have shown that $\mathbb{R}P^n$ is a second countable Hausdorff space, but I am not sure how to show that for each $x\in \mathbb{R}P^n$, there is an open neighborhood $U\subset \mathbb{R}P^n$ containing $x$, such that there exists a homeomorphism $f:U\to V$, where $V\subset \mathbb{R}^n$ is open.
I know there are plenty of resources online regarding this topic, but I am not sure I fully understand them. What would such sets $U$ and their homeomorphisms be, and furthermore, how would we prove such functions $f:U\to V$ are homeomorphisms?
I am quite new to this topic, so I apologize in advance for the lack of work. Any help would be much appreciated. I have proved, however, that $S^n$ is an $n4-dimensional manifold, if that is of any use.
 A: There's a few ways you could try and proceed. Here's the approach I find the most enlightening.
Hint. Try and establish a homeomorphism between $\mathbb{RP}^n$ and
$$(\mathbb{R}^{n+1} \setminus \{0\})/ (\sim'),$$
where $\sim'$ is the equivalence relation generated by the relation $(x_0,\ldots,x_n) \sim (\lambda x_0,\ldots\lambda x_n)$ for any $\lambda \in \mathbb{R} \setminus \{0\}$. We commonly write $[x_0 : \cdots : x_n]$ to denote the equivalence class of $(x_0,\ldots,x_n)$ in this set. After you've done that, consider the open set
$$U_i = \big\{ [x_0 : \cdots : x_n] \in \mathbb{RP}^n : x_i \neq 0\big\}.$$
Show that each of these opens are homeomorphic to $\mathbb{R}^n$, and consider their transition functions.
Note. Personally I think it's important to make sure to picture what happens. Regarding projective space, the maps and homeomorphisms may seem arbitrary on first sight, but there's some geometric intuition behind it all. Try and picture the case $n = 2$, where $\mathbb{RP}^n$ is just the usual sphere $S^2$, except that opposite points are being identified.
Projective spaces are geometrically weird on first sight. To train your geometric intuition, maybe consider trying the following informal exercises. They are not relevant for your question, but they might aid in your mental picture.

*

*Convince yourself that the above homeomorphism tells you that $\mathbb{RP}^n$ is, in a sense, the set of straight lines through the origin in $\mathbb{RP}^{n+1}$, endowed with a topology that essentially encodes how close to each other two given lines are.


*Let $p$ be a person living on $S^2$. Imagine $p$ has a rope. Say $p$ walks in a straight direction, leaving a trail of the rope behind him, until he finds himself back at the origin where we started. He now ties one end of the rope, which he left behind, to the other end of the rope, which he took with him. Note that he can collect back all of the rope, without having to break the knot that he made.
Now suppose $p$ lives on $\mathbb{RP}^2$ instead, and he does the exact same thing. Try to convince yourself that he can no longer collect the rope without breaking breaking open the knot!

*

*Now say that $p$ goes around the surface of $\mathbb{RP}^2$ again, except not once, but twice, leaving a trail of the rope behind him all this time. After going around twice he ties the ends of the rope. Convince yourself that this time, he can collect the rope.

A: You may look at $\mathbb{R}P^n:=S^n/(x\sim -x)$ which is the same as $S^n/\Gamma$ where $\Gamma$ is just discrete 2 element group acting on $S^n$ by doing nothing or reflecting the points via origin. Given that you may use the first answer form this topic: Under what conditions the quotient space of a manifold is a manifold?
