Showing diffeomorphism between $S^1 \subset \mathbb{R}^2$ and $\mathbb{RP}^1$ I am trying to construct a diffeomorphism between $S^1 = \{x^2 + y^2 = 1; x,y \in \mathbb{R}\}$ with subspace topology and $\mathbb{R P}^1 = \{[x,y]: x,y \in \mathbb{R}; x \vee y \not = 0  \}$ with quotient topology and I am a little stuck.
I have shown that both are smooth manifolds, and I used stereographic projection for $S^1$, but now I am runing into trouble when I give the homeomorphism between $S^1$ and $\mathbb{RP}^1$ as the map that takes a line in $\mathbb{RP}^1$ to the point in $S^1$ that you get when letting the parallel line go through the respective pole used in the stereographic projection. 
If I use the south and north poles I get a potential homeomorphism, but I cannot capture the horizontal line in my image, but when I pick say north and east then my map is not well defined as I get different lines for the same point in $S^1$. Can somebody give me a hint how to make this construction work, or is it better to move to a different representation of $S^1$ ?
 A: I am not sure that I follow all parts of the question.  In particular, it is unclear to me what ``a line in $\mathbb{RP}^{1}$'' is, but this might be something that is clear to others.  However, I can say that I do not like the idea of viewing both $S^{1}$ and $\mathbb{RP}^{1}$ in the same copy $\mathbb{R}^{2}$ as (I think) that it could lead to confusion since $S^{1}$ is a double cover of $\mathbb{RP}^{1}$ with the quotient map.  This may or may not be where your confusion comes in.  
Anyway, here is how I see the construction (I can't promise that I preserved the orientations of each, but you can fiddle with that if needed).  Start with $S^{1} \subset \mathbb{R}^{2}_{(x, y)}$ and view $\mathbb{RP}^{2}$ as coming form a quotient of $\mathbb{R}^{2}_{(u, v)}$, where I use the subscripts $(x, y)$ and $(u, v)$ to denote coordinates. (Note:  This is admittedly a superficial labeling, but it provides a clearer picture (at least for me)).  
I will walk through the entire of the construction both geometrically and in coordinates, and I apologize if the response is too wordy.  The intuition should be  be clear however: cut a $S^{1}$ and compress it appropriately so that the ends will be close to the same point in $\mathbb{RP}^1$.  With the hindsight of having a working construction, there are easier ways to see what is happening.
Now, define a map $\phi : S^{1}\backslash N \to \mathbb{R}$ via stereographic projection from the North Pole.  For a point $P(x, y) \in S^{1}\backslash N$ we thus have $\phi(P) = \frac{-x}{y -1}$.  This identifies $S^{1}\backslash N$ with $\mathbb{R}_{t}$, and now we simply need to map this copy of the real line to cover one hemisphere of $S^{1} \subset \mathbb{R}^{2}$ and then send the North Pole to the appropriate equivalence class.
Define $\gamma : \mathbb{R}_{t} \to \mathbb{R}^{2}_{(u,v)}$ by 
$$\gamma(t) = \left( \frac{t}{\sqrt{t^2 + 1}}, \frac{1}{\sqrt{t^2 + 1}}\right),$$
and observe that this covers all of the (strict) northern hemisphere of $S^{1} \subseteq \mathbb{R}^{2}_{(u, v)}$. This map is described geometrically below. 
Your desired map from $S^{1} \subseteq \mathbb{R}^{2}_{(x, y)} \to \mathbb{RP}^{1}$ is then given by
$$
\Phi(P) =
\begin{cases}
[\gamma \circ \phi (P(x,y))] = \left[\frac{-x}{\sqrt{x^2 + (y-1)^2}}, \frac{1-y}{\sqrt{x^2 + (y-1)^2}}\right], & \text{if } P(x, y) \ne N \text{ } \\
[1,0], & \text{if  } P(x, y) = N
\end{cases}.
$$
We have come this far (and I don't particularly like coordinates for maps like these), so I will briefly summarize the construction for the more geometric minded and point to a similar one that I observed after the fact.


*

*Stereographic projection from $S^{1}\backslash N$ to $\mathbb{R}_{t}$.

*Identify $\mathbb{R}_{t}$ with the line $u = 1$ in $\mathbb{R}^{2}_{(u, v)}$.

*Define a map from the line $u = 1$ to the northern hemisphere of $S^{1} \subseteq \mathbb{R}^{2}_{(u, v)}$ by looking at the intersection of the position vector $\langle u, 1 \rangle$ with the northern hemisphere. (2. and 3. define the map $\gamma$).

*Take equivalence classes and send $N$ to $[1,0]$.


After the fact it occurred to me that you could just as easily do the following (without superficial labels on the  coordinates):


*

*Stereographically project from $S^{1}\backslash (South Pole)$ to the line $y = 1$

*Repeat the map described in 3. above.

*Take Equivalence classes and send the  South Pole to [1,0].

A: Consider the stereographic atlas $\{(U_\alpha, \varphi_\alpha), (U_\beta, \varphi_\beta)\}$  where $U_\alpha=\mathbb S^1-\{(0, 1)\}$ and $U_\beta=\mathbb S^1-\{(0, -1)\}$. Here $\varphi_\alpha$ is given by $$(x, y)\mapsto u=\frac{x}{1-y},$$ whereas $\varphi_\beta$ is given by $$(x, y)\mapsto v=\frac{x}{1+y}.$$ I don't know if I have given the right names for the south and north projection, but you decide later..Now consider the atlas $\{(U_1, \varphi_1), (U_2, \varphi_2)\}$ for $\mathbb R\mathbb P^1$ given by $\varphi_1:U_1\rightarrow \mathbb R-\{0\}$, $$[(x, y)]\mapsto \frac{y}{x}$$ and $\varphi_2:U_2\rightarrow \mathbb R-\{0\}$ is given by, $$[(x, y)]\mapsto \frac{x}{y}.$$ Define $\psi:\mathbb R\mathbb P^1\rightarrow \mathbb S^1$ setting $$\psi(u)=\left\{\begin{array}{ccc}\varphi_\alpha^{-1}\circ \varphi_1&\textrm{if}&u\in U_1\\ \varphi_\beta^{-1}\circ \varphi_2&\textrm{if}&u\in U_2\end{array}\right..$$ In this case $\psi$ is defined in $U_1\cup U_2=\mathbb R\mathbb P^1$, but for it to be well defined you must check that in $U_1\cap U_2$, $$\varphi_\alpha^{-1}\circ \varphi_1=\varphi_\beta^{-1}\circ \varphi_2.$$ This will be your diffeomorphism..
A: The easiest explicit map I know is:
$$(\cos(\theta), \sin(\theta))\mapsto [\cos(\theta/2):\sin(\theta/2)]$$
Note that although $\cos(\theta/2)$ and $\sin(\theta/2)$ depend on $\theta$ and not just $\sin(\theta)$ and $\cos(\theta)$, the map is well-defined so long as we use the same value of $\theta$ when computing coordinates in $S^1$.That is,  $$\cos(\frac{\theta+2\pi}{2})=-\cos(\theta/2) \text{ and } \sin(\frac{\theta+2\pi}{2})=-\sin(\theta/2)$$ So the choice of $\theta$ modulo $2\pi$ does not affect $[\cos(\theta/2):\sin(\theta/2)]$, since $[x,y]=[-x,-y]$.
