For the sake of completeness, let us start with the definition of an abstract manifold.

Let $$X$$ be a topological space. A chart (or coordinate chart) is an open set $$U\subseteq X$$ together with a homeomorphism $$\phi: U \to \mathbb{R}^n$$. An atlas is a collection of charts $${(U_{\alpha},\phi_{\alpha})}$$ with the condition $$X = \cup_{\alpha\in A} U_{\alpha}$$. A smooth atlas is an atlas such that whenever $$U_{\alpha} \cap U_{\beta} \neq \emptyset$$, the composition $$\phi_{\beta} \circ \phi_{\alpha}^{-1} : \phi_{\alpha}(U_{\alpha} \cap U_{\beta}) \to \phi_{\beta}(U_{\alpha}\cap U_{\beta})$$ i.e. every transition map is smooth. Two smooth atlases are equivalent if their union is also a smooth atlas. This gives a natural equivalence class of smoothly equivalent atlases which we call a smooth structure. More specifically, define an equivalence relation $$\sim$$ where $$P \sim Q$$ if $$P, Q$$ are smooth atlases that are equivalent. The equivalence class generated by this relation is called a smooth structure.

A topological manifold $$X$$ is a Hausdorff, second countable topological space $$X$$ for which one can find an atlas. A smooth manifold $$X$$ is a topological manifold with a distinguished smooth structure.

Before we talk about the real projective space $$\mathbb{R}P^n$$, we talk about the $$n = 2$$ case.

### Definition (Real Projective Plane)

Let $$\mathbb{R}P^2$$ denote the set of lines in $$\mathbb{R}^3$$ that go through the origin. We say that $$\mathbb{R}P^2$$ is the real projective plane.

To show this set is an abstract manifold, we first have to define a topology on it, which we do using the quotient topology. To do this consider $$\mathbb{R}^3$$ and the equivalence relation $$\sim$$ on it where $$x\sim y$$ if and only if $$x = \lambda y$$ for $$\lambda\in\mathbb{R},\lambda\neq 0.$$ Now, consider the mapping $$p(x\in \mathbb{R}^3-{0}) = [x]$$. Call $$Y$$ the set mapped to by $$p$$ and say that a set $$V$$ in $$Y$$ is open if and only if $$p^{-1}(V)$$ is open in $$\mathbb{R}^3-0$$. Then, $$p$$ is clearly a quotient map so that $$Y$$ is given the quotient topology. We can clearly identify $$\mathbb{R}P^2$$ with $$Y$$ by the mapping $$f: Y \to \mathbb{R}P^2$$ where $$f(x) = L(x)$$ where $$L(x)$$ is the line that goes through the points $$x$$ and $$0$$. Via, this identification, we can say a set $$V$$ in $$\mathbb{R}P^2$$ is open if and only if $$f^{-1}(V)$$ is open. Since $$Y$$ and $$\mathbb{R}P^2$$ are equivalent, we could have also defined $$\mathbb{R}P^2$$ to simply be the topology given by quotient $$(\mathbb{R}-0)/\sim$$.

For each line, identify two antipodal points each magnitude 1 from the origin. This defines the set $$S^2$$. A simple figure is provided below.

### How does this fit in with my previous understanding of coordinates?

You may have thought coordinates as something which was implicitly connected to the space you were working in (e.g. $$\mathbb{R}, \mathbb{R}^2, …$$) in the way that giving a coordinate was the same as pointing out a unique place on the space. This is in fact, still correct. In this case, let $$X = \mathbb{R}^n$$ and if our neighbourhood $$U \subset X$$ is of dimension $$k$$, then we let it be the identity mapping.