Question about the one point compactification $\mathbb{R}^2 \cup \infty$ of $S^2$ Question about the one point compactification $\mathbb{R}^2 \cup \infty$ of $S^2$.
Given $a,b \in S^2$, can somebody give me the explicit homeomorphism $\gamma$ from $S^2$ to $\mathbb{R}^2 \cup \infty$ such that $\gamma(a) = 0$ and $\gamma(b) = \infty$? For some reason in my brain I thought this was only possible if $a$ and $b$ are antipodal points, that seems like something that might be true, right? ha. Anyway apparently it's not according to the proof of Lemma 61.2 in Munkres topology, which takes arbitrary $a$ and $b$ and then maps one to $0$ and the other to $\infty$.
thanks in advance :-)
 A: Stereographic projection:   the north pole $P$ corresponds to the point at infinity, the South pole to zero.
In Cartesian coordinates it's $(x,y,z)\to (\frac y{1-x},\frac z{1-x})$.
As for mapping two arbitrary points on the sphere to the origin and infinity, see the comment by @Cheerful Parsnip.
A: Well not really explicit at all, but Naturally we can consider $S^2$ to be $\text{bd}(B_{\mathbb{R}^3}[(0,0,1/2),1/2])$. If we take $a=0$ and $b=(0,0,1),$ then we can find a homeomorphism by stereographic projecting the points of $S^2$ onto the homeomorphic copy of $\mathbb{R}^2$ that is $\{(x,y,0)|x,y\in\mathbb{R}\}.$ The only thing that's not obvious is where do we map $(0,0,1)?$ That's easy though, infinity. And neighborhoods $\epsilon$ of $(0,0,1)$ will get mapped to complements of closed discs in $\mathbb{R}^2$. Complements of closed disc are neighborhoods of $\infty$ as we define an open set containing $\infty$ to be the complement of a compact set in $\mathbb{R}^2$
Now you did mention $a,b$ being arbitrary distinct points, but that's not so hard to fix. You can find a homeomorphism of $S^2\to S^2$ sending $a\to 0$ and $b\to(0,0,1)$ using the disc lemma.
A: I will use the following fact: if $\equiv$ denotes 'being homeomorphic' and $Z^*$ is the one point compactification of a space $Z$, then $X \equiv Y$ imples $X^* \equiv Y^*$. The idea is to take a homeomorphism $f : X \to Y$ and define $f^* :X^* \to Y^*$ via $f^*(x) = f(x)$ when $x \in X$ and $f^*(\infty_X) = \infty_Y$. After checking that this map is continuous, you just have to observe that $(f^*)^{-1} = (f^{-1})^*$.
Now, for any $k \geq 1$, the map
$$
g(x_1, \dots, x_{k+1}) = \frac{1}{1-x_{k+1}}(x_1, \dots, x_k)
$$
from $X_k := S^k \setminus \{(0,\dots,0,1)\}$ to $\mathbb{R}^k$ is a homeomorphism. Visually, one takes the line that joins $N_k := (0,\dots,1)$ with $x \in S^k$, and $g(x)$ is the intersection of the line with $\{x_{k+1} = 0\} \subset \mathbb{R}^{k+1}$. Hence, this induces a homeomorphism of the one-point compactifications, $g^* : (X_k)^* \to (\mathbb{R}^k)^*$. You can check that $S^k \equiv (X_k)^*$ via $f(x) = x$ when $x \neq N_k$, and $f(N_k) := \infty_{X_k}$. The composition then gives an explicit (since $g^*$ depends only of $g$ which we know) homeomorphism between $S^k$ and $\mathbb{R}^k$, 
$$
\begin{align}
\Phi : & S^k \rightarrow (\mathbb{R}^k)^* \\
&x\mapsto \cases{\frac{1}{1-x_{k+1}}(x_1, \dots, x_k) \quad \ \text{ if $x \neq N_k$} \\ \infty_{\mathbb{R}^k} \ \text{if $x = N_k$}} 
\end{align}
$$
It is clear from here that $\Phi(N_k) = \infty$ and $\Phi(0,0,\dots,-1) = 0$. Thus, it will be enough to find a homeomorphism $S^2 \to S^2$ such that $a \mapsto N_k$ and $b \mapsto (0,0,\dots,-1)$ and post-compose it with $\Phi$. Can you take it from here?
A: If you know complex function theory, working on the Riemann sphere $\mathbb{C} \cup \{\infty\}$, just take a Moebius transform sending $a$ to $b$ and $b$ to $\infty$. 
