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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 :-)

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.

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 ${(x,y,0)|x,yinmathbb{R}}.$ 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 $S^2to S^2$ sending $ato 0$ and $bto(0,0,1)$ using the disc lemma.

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 $mathbb{R}^k$,

$$
begin{align}
Phi : & S^k rightarrow (mathbb{R}^k)^* \
&xmapsto cases{frac{1}{1-x_{k+1}}(x_1, dots, x_k) quad text{ if $x neq N_k$} \ infty_{mathbb{R}^k} text{if $x = N_k$}}
end{align}
$$

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?

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 $infty$.