Let $M$ be a topological $n$-manifold and $\varphi \colon U\to \mathbb{R}^n$ a chart of $M$. Assume that $\varphi(U)=B(q, \varepsilon)$ is an open ball in $\mathbb{R}^n$ and that the map $f\colon B(q, \varepsilon) \to \mathbb{R}^n$ is an homeomorphism between $B(q, \varepsilon)$ and $\mathbb{R}^n$. Let $s\colon \mathbb{S}^n - \{ P \} \to \mathbb{R}^n$ ($P=(0, \ldots, 0, 1)$) be the stereographic projection. Define a function $g\colon M\to \mathbb{S}^n$ by \begin{equation*} g(x)=\begin{cases} s^{-1} \circ f \circ \varphi(x) & \text{if } x\in U, \\ P & \text{if } x\in M - U. \end{cases} \end{equation*}
My question is: Can we show that the map $g$ is continuous ? Which properties must be satisfied by $M$ in order to prove that $g$ is continuous ?
Thanks in advance for all your answers !!! Cheers...
The maps $\varphi: U \to \Bbb{R}^n$ and $f: B(q, \varepsilon) \to \Bbb{R}^n$ are clearly continuous. Therefore $f \circ \varphi: U \to \Bbb{R}^n$ is continuous. Also, $(f \circ \varphi)(U) = \Bbb{R}^n$, so $f \circ \varphi$ is surjective.
The stereographic projection $s: \Bbb{S}^n - \{P\} \to \Bbb{R}^n$ is a homeomorphism so $s^{-1}: \Bbb{R}^n \to \Bbb{S}^n - \{P\}$ is continuous.
Now $g|_U(x) = s^{-1} \circ f \circ \varphi(x): U \to \Bbb{S}^n - \{P\}$ is continuous.
Now remember that we can think the stereographic projection also as a mapping $\bar{s}: \Bbb{S}^n \to \Bbb{\bar{R}}^n$, where $\Bbb{\bar{R}}^n$ is the one point compactification of $\Bbb{R^n}$, meaning that we add a "point at infinity" to it and define $\bar{s}(P) = \infty$ and $\bar{s}(x) = s(x)$ otherwise. Now $\bar{s}$ is still a homeomorphism, so setting $g|_{M-U}(x) = P$ makes $g: M \to \Bbb{S}^n$ a continuous mapping.
Of course, you can also check the inverse image of an open set. If $V \subset \Bbb{S}^n$ is open, we have \begin{align} g^{-1}(V) &= (\bar{s}^{-1} \circ f \circ \varphi)^{-1}(V) \\ &= \varphi^{-1} f^{-1} \bar{s}(V)\,. \end{align} The set $\bar{s}(V)$ is open because $\bar{s}$ is a homeomorphism so it is an open map. Then, because $\varphi$ and $f$ are continuous, the set $\varphi^{-1} f^{-1} \bar{s}(V)$ is open.
Intuitively, $g$ is a quotient map from $M$ to $\Bbb{S}^n$ which identifies the set $M-U$ with a single point $P$. Topologically, $U$ is an open ball $B^n$ and so is $\Bbb{S}^n - \{P\}$. Then we take everything outside this open ball and think of it as a single point. This is like gluing the whole boundary of $B^n$ together, making it a sphere $S^n$. The cases $n=1,2$ are easy to visualize and you should try to build your intuition with these special cases.