1
$\begingroup$

Consider a triangle, $T$, in $\mathbb{R}^3$ with vertices $(0,0,1), (0,1,0)$, and $(1,0,0)$. Let $S$ denote the sphere centered at the origin with radius 1 and let $S_1$ denote the portion of the sphere in the same quadrant at the triangle. We can define a map $f: T \rightarrow S_1$ by $$f(x) = \frac{x}{|x|}$$ which is one-to-one and onto.

The map is a bijection, so the inverse should exist. Maybe I am losing my mind but, what does the explicit formula for $f^{-1}: S_1 \rightarrow T$ look like?

$\endgroup$
  • $\begingroup$ By "look like", do you mean that you want an explicit formula for $f^{-1}$? $\endgroup$ – Travis Willse Sep 4 '15 at 20:31
  • 2
    $\begingroup$ $(x,y,z) \mapsto \left(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z}\right)$. $\endgroup$ – achille hui Sep 4 '15 at 20:31
  • $\begingroup$ Yep, that's what I meant. Thanks for your answer. I can mark it as an answer if you post it. $\endgroup$ – user1515302 Sep 4 '15 at 20:35
1
$\begingroup$

Notice under the direct mapping, any point $p$ on $T$ get mapped to a point $f(p)$ on $S$ which is a scalar multiple of $p$. So for any point $q = (x,y,z) \in S$, the inverse mapping $f^{-1}$ will send $q$ to a point which is a scalar multiple of $q$. This means there exists a function $\lambda : S \to \mathbb{R}$ such that

$$q = (x,y,z)\quad\mapsto\quad f^{-1}(q) = ( \lambda(q)x, \lambda(q)y, \lambda(q)z)$$

It is clear the triangle $T$ lies on the plane $x + y + z = 1$, this means $\lambda(q)$ satisfy the constraint:

$$\lambda(q)(x+y+z) = 1 \quad\iff\quad \lambda(q) = \frac{1}{x+y+z}$$

As a result, we have

$$f^{-1}(x,y,z) = \left(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z}\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.