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In this problem, we consider $S^2$ with the coordinate charts $(U, \phi)$, and $(U', \phi')$, where:

$$U = \{(x,y,z) \in S^2 : z < 0\} \quad \phi(x,y,z) = (x,y), \ \mathrm{and}$$ $$U'\ = \{(x,y,z) \in S^2 : x > 0\} \quad \phi'(x,y,z) = (y,z)$$

My question concerns explicitly writing out the sets $\phi(U \cap U')$, and $\phi'(U \cap U')$. My attempt for writing out $\phi(U \cap U')$ is:

$$\phi(U \cap U') = \{(x,y) \in \mathbb{R^2} : x = \sqrt{1 - y^2 - z^2}\}$$

This incorporates the fact that $x > 0$ and $(x,y,z) \in S^2$. However, since $\phi$ drops the $z$ term, it feels inappropriate the include a $z$ in the definition of $\phi(U \cap U')$, and this makes it difficult to invert $\phi$ when computing the transition function. I am confused as to how to write the set explicitly that makes it simple to compute the transition function.

Solarflare0
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I think that the confusion lies in that you are assuming the image of $\phi(U\cap U^{\prime})$ has to line in $S^{2}$, when really it lies in $\mathbb{R}^{2}$. To elaborate, I'll write $U_{x,+} = \{(x,y,z)\in S^{2} : x > 0 \}$ and $\phi_{x,+}(x,y,z) = (y,z)$ for its coordinate chart. But since $\phi_{x,+}:U_{x,+} \rightarrow \phi_{x,+}(U_{x,+}) = V_{x,+}$, say its inverse $\phi_{x,+}^{-1}:V_{x,+} \rightarrow U_{x,+}$ has to have its image lie within $U_{x,+} \subset S^{2}$, so we take

$$ \phi_{x,+}^{-1}(y,z) = \bigg(\sqrt{1 - y^{2} - z^{2}}, y, z\bigg) \in U_{x,+} \subset S^{2}. $$

Very similar coordinate functions for the other coordinate neighbourhoods have similar properties. As to your question, I would say that the information regarding $z$ is encoded more subtly in the form of the inverse coordinate functions (note that whilst these choices of coordinate functions are usually many-to-one if the domain was $\mathbb{R}^{3}$, the fact that it is $S^{2}$ then forces them to be injective.).

EDIT - Further elaboration: I'm basing this on an example in Do Carmo's 'Riemannian Geometry', on page 21.

The coordinate charts can be seen as orthogonal projections from $S^{2}$ onto some subset of $\mathbb{R}^{2}$, e.g. $$ \phi_{x,+}(x,y,z) = (0,y,z) = (y,z) \in \mathbb{R}^{2}. $$ In the ambient $\mathbb{R}^{3}$ however, say after the inclusion $(y,z) \mapsto (y,z,u)$ there would indeed be two points in $\mathbb{R}^{3}$ such that $y^{2} + z^{2} + u^{2} = 1$, and indeed making a choice of sign will send us back to a unique point in $S^{2}$, so the inverse chart function would be e.g. $$ \phi_{z,-}^{-1}(u,v) = (u,v,-\sqrt{1-u^{2}-v^{2}}) \in U_{z,-}. $$ Combining these, say on $\phi_{x,+}(U_{x,+}\cap U_{z,-})\subset \mathbb{R}^{2}$, then: $$ \phi_{z,-}\circ \phi_{x,+}^{-1}(u,v) = \phi_{z,-}(\sqrt{1 - u^{2} - v^{2}},u,v) = (\sqrt{1 - u^{2} - v^{2}},u)\in \phi_{z,-}(U_{x,+}\cap U_{z,-}) $$

BenCWBrown
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  • Hi, thanks for answering! I understand more clearly how to take the inverse function now, but I am still confused regarding the set $\phi(U \cap U')$. In particular, must not the coordinate $(x,y) \in \mathbb{R^2}$ still have to satisfy the equation of the sphere (since you are just projecting into $\mathbb{R^2}$). How do I determine the constrants (if there are any), on the elements in $\mathbb{R^2}$ that lie in $\phi(U \cap U')$? – Solarflare0 Feb 24 '19 at 20:51
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    I've updated my answer to hopefully elaborate further. Essentially the constraints are still there just in the ambient larger space $\mathbb{R}^{3}$, in the sense that the inclusion $\phi(U \cap U^{\prime})\hookrightarrow \mathbb{R}^{3}$ would lie in $S^{2}$. – BenCWBrown Feb 25 '19 at 19:22