Stereographic projection of $S^1$ onto $\mathbb R$

Can you tell me if this is correct:

I define $f_{(0,1)}: S^{1+} \setminus \{(0,1)\} \to \mathbb R$ as $f((x,y)) =\frac{x}{1-y}$ where $S^{1+} = \{ (x,y) \in S^1 \mid y \geq 0 \}$. Since I removed the north pole this map is continuous and it is easy to verify that $f$ is also differentiable in each variable.

Hence $S^1$ is a differentiable manifold with the atlas $\{ (S^{1+} \setminus \{(0,1)\}, f_{(0,1)}), (S^{1-} \setminus \{(0,-1)\}, f_{(0,-1)}) \}$.

Another question I have is: is this also called stereographic projection? I google-searched the term but all results seemed to be only about $S^2 \to \mathbb R^2$.

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Why do you take $y \geq 0$ making the set look like two half-open intervals? This makes your charts meet only at $(\pm1,0)$, so that's no good... Think about the transition maps. – t.b. Sep 11 '12 at 14:27
@t.b. Why is that no good? Then they are an open cover of $S^1$, what am I missing? – Rudy the Reindeer Sep 11 '12 at 14:29
Student said whatever I have to say about this. – t.b. Sep 11 '12 at 14:32

1. Your first map looks like a restriction of the stereographic projection. Note that you don't need to restrict to $S^{1+}$, you really only need to exclude $\{(0, 1)\}$, which you already did. Also note that $S^{1+}$ is closed inside $S^1$, so it's not a good choice of domain; you should pick an open set (like the whole $S^1\backslash\{(0, 1)\}$).
2. If you are formally checking that $S^1$ is a manifold, you shouldn't say "$f$ is differentiable". After all, this doesn't make sense until you prove that $S^1$ is a manifold. What you have to check is the differentiability of the transition maps, namely, $f_{(0, -1)} \circ f_{(0, 1)}^{-1} : \mathbb{R} \rightarrow \mathbb{R}$.
3. About the nomenclature: yes, this is called the stereographic projection (at least when defined on the whole $S^1\backslash\{(0, 1)\}$), and you can do this in all dimensions, that is, for all spheres $S^n$, by removing one point. The $S^2$ case is more common, I guess.
So I guess if I restrict then my transition map is the boring map whose image consists of $-1$ and $1$? Which would not be a transition map because I want it to be surjective? (although it does not mention surjectivity on Wikipedia) – Rudy the Reindeer Sep 11 '12 at 14:35
@bananalyst you want the domains of $f_{(0, 1)}$ and $f_{(0, -1)}$ to cover your manifold and to be open subsets of $S^1$. The way you defined, they're not open, so that's no good. If you restrict then yes, the transition map is something discrete like that. – student Sep 11 '12 at 14:38
Ah, ok, thanks. So if I change the restriction to $S^{1+} = \{ (x,y) \in S^1 \mid y > -\varepsilon \}$ for some $\varepsilon > 0$ then it becomes a valid chart? – Rudy the Reindeer Sep 11 '12 at 14:41
@bananalyst it does. Just be careful to also exclude $\{(0, 1)\}$. – student Sep 11 '12 at 14:43
: D Thank you! ${}$ – Rudy the Reindeer Sep 11 '12 at 14:45