Let $S$ := {$(x, y) ∈ \mathbb R^2: x^2 + y^2 = 1$} be the unit circle. Let $p = (0, -1)$ ∈ $S$.
Define a map $F$ from $\mathbb R$ × {$0$} to $S$ as follows: Given $(t, 0)$, let $F((t, 0)) = (x, y)$ ∈ $S$ be the intersection of $S$ with the line through $p$ and $(t, 0)$.
Derive a formula for $F$.
Show that $t$ → $F((t, 0))$ is a bijective map from $\mathbb R$ to $S$ \ {$p$}, and also derive a formula for its
inverse.
If we express $S$ in polar coordinates:
$$S:=\{(r,\varphi):r=1,\varphi∈(-\frac\pi{2},\frac{3\pi}{2})\}$$
then, from simple trigonometric relations, the inverse function $\Phi$ is:
$$\Phi(\varphi)=\tan\frac{\frac\pi2-\varphi}2$$
and its inverse $F$ is:
$$F(t)=\frac\pi{2}-2\arctan t$$
After changing coordinates back to cartesian, $F$ will look like:
$$F(t)=(\cos(\frac\pi{2}-2\arctan t),\sin(\frac\pi{2}-2\arctan t))$$
which can be further simplified to:
$$F(t)=(\frac{2 t}{t^2 + 1},\frac{1 – t^2}{t^2 + 1})$$
Is this the formula of $F$? Is it correct?
And I know that to show that $\phi$ : $\mathbb R$ → $S$ \ {$p$} is a bijective map, I have to show that $\phi$ is injective and surjective. But how do I do that? Can someone please help me show that $\phi$ is bijective?
and how can I derive a formula for its inverse?
Best Answer
This is a variant of stereographic projection. Working with polar coordinates unnecessarily complicates finding $F$.
The line through $p = (0,-1)$ and $(t,0)$ has the equation $$l(s) = (t,0) + s((0,-1) - (t,0)) = ((1-s)t,-s),\quad s \in \mathbb R .$$ We have to determine $s$ such that $l(s) \in S \setminus \{p\}$ which means $(1-s)^2t^2 + s^2 = 1$ and $l(s) \ne p$. As solutions of the quadratic equation we get $$s = \frac{t^2 \pm 1}{t^2 +1 } .$$ The plus-sign gives $s=1$ and $l(s) = l(1) = p$ which is excluded, thus the intersection point is $$F(t) = l\left(\frac{t^2 - 1}{t^2 +1 }\right) = \left(\frac{2t}{t^2 +1 },\frac{1-t^2}{t^2 +1} \right)$$ which is the same as your solution.
To get the inverse $G$ of $F$, we consider the line through $p$ and $(x,y) \in S \setminus \{p\}$ which has the equation $$L(s) = (x,y) + s((0,-1) - (x,y)) = ((1-s)x,y-(1+y)s),\quad s \in \mathbb R .$$ Then $G(x,y)$ is the intersection point of this line with $\mathbb R \times \{0\}$ which is determined by the equation $y-(1+y)s= 0$. This gives $s = \dfrac{y}{1+y}$ and thus $$G(x,y) = \left(\dfrac{x}{1+y},0 \right) .$$ It is geometrically clear that $F, G$ are inverse to each other, but it can also be shown formally by verifying that $G(F(t,0)) = (t,0)$ and $F(G(x,y)) = (x,y)$. In fact, this is a routine exercise.
Edited on request:
It seems that some clarification is needed concerning the concept of a bijection. Let $f : A \to B$ be function between sets $A, B$. Then the follwing are equivalent:
(1) $f$ is bijective (i.e. injective and surjective).
(2) $f$ has an inverse $g : B \to A$ (this means that $g \circ f = id_A$ and $f \circ g = id_B$).
This is a well-known elementary result. Let us nevertheless prove it.
(1) $\Rightarrow$ (2) : Since $f$ is surjective, each $b \in B$ has the form $b = f(a)$ with $a \in A$. Since $f$ is injective, we see that there exists exactly one $a \in A$ with this property. Writing $a = f^{-1}(b)$, we get a function $f^{-1} : B \to A$. By construction it has the property $f^{-1}(f(a)) = a$, i.e. $f^{-1} \circ f = id_A$, and $f(f^{-1}(b)) = b$, i.e. $f \circ f^{-1} = id_B$. Thus $f^{-1}$ is an inverse of $f$.
(2) $\Rightarrow$ (1) : $f$ is surjective because for each $b \in B$ we have $f(g(b)) = b$. $f$ is injective because if $f(a) = f(a')$, then $a = g(f(a)) = g(f(a')) = a'$.
Note that an inverse of $f$, if it exists, is unique and will usually be denoted by $f^{-1}$. In fact, if $g,g'$ are inverses of $f$, then $g = g \circ id_B = g \circ (f \circ g') = (g \circ f) \circ g' = id_A \circ g' = g'$.
In my answer we constructed two maps $F : \mathbb R \times \{0\} \to S$ and $G : S \to \mathbb R \times \{0\}$ such that $G \circ F = id$ and $F \circ G = id$. Now the above theorem applies.