# Finding the Inverse of a Function on the Sphere

I am attempting to find the inverse of the function: $$f:S^2 \to S^2$$, defined as:

$$f(x,y,z)=(x\cos(z)+y\sin(z), x\sin(z)-y\cos(z),z)$$

My approach so far has been to use linearity and express the function as a sum of scaled basis vectors:

$$f(x,y,z)=x(\cos(z),\ sin(z),0)+y(\sin(z), -\cos(z),0)+z(0,0,1)$$

However, I'm unsure how to proceed from this point. Any guidance or assistance would be greatly appreciated! Thanks.

Your transformation can be written under a matrix form :

$$\pmatrix{x'\\y'\\z'}=\pmatrix{\cos(z) & \sin(z) & 0\\ \sin(z) & -\cos(z) & 0\\0&0&1}\pmatrix{x\\y\\z}$$

Its inverse transformation is therefore given by the inverse matrix (the inverse being obtained by inverting the diagonal blocks) :

$$\pmatrix{x\\y\\z}=\pmatrix{\cos(z) & \sin(z) & 0\\ \sin(z) & -\cos(z) & 0\\0&0&1}\pmatrix{x'\\y'\\z'}$$

• ... which are in fact the same formulas ! The transformation is its own inverse. Dec 9, 2023 at 13:22
• Geometric-geographical explanation : this transformation sends a point onto its opposite point on the same meridian line. Dec 9, 2023 at 15:20

If we write $$x = \cos(\theta)$$, $$y = \sin(\theta)$$, then

$$f(x,y,z)=(\cos(z+\theta), \sin(z-\theta),z)$$

which is an involution (is equal to its own inverse).