π‘Ÿπœƒ:𝑆2→𝑆2π‘Ÿπœƒ(π‘₯,𝑦,𝑧)=(π‘₯cosπœƒβˆ’π‘¦sinπœƒ,π‘₯sinπœƒ+𝑦cosπœƒ,𝑧)

is a diffeomorphism.

  • It’s obviously that for π‘₯,𝑦,𝑧 component functions are differentiable, hence π‘Ÿπœƒ is smooth

  • Also, notice 𝑧 didn’t change so we can ignore it.

Because π‘Ÿπœƒ is a rotation by z-axis, we check that π‘Ÿβˆ’πœƒβˆ˜π‘Ÿπœƒ(π‘₯,𝑦,𝑧)=(π‘₯,𝑦,𝑧)

(π‘₯cosπœƒβˆ’π‘¦sinπœƒ)cos(βˆ’πœƒ)βˆ’(π‘₯sinπœƒ+𝑦cosπœƒ)sin(βˆ’πœƒ)=(π‘₯cosπœƒβˆ’π‘¦sinπœƒ)cos(πœƒ)βˆ’(π‘₯sinπœƒ+𝑦cosπœƒ)sin(βˆ’πœƒ)=(π‘₯cosπœƒβˆ’π‘¦sinπœƒ)cos(πœƒ)+(π‘₯sinπœƒ+𝑦cosπœƒ)sin(πœƒ)=π‘₯(sin2πœƒ+cos2πœƒ)=π‘₯

and

(π‘₯cosπœƒβˆ’π‘¦sinπœƒ)sin(βˆ’πœƒ)+(π‘₯sinπœƒ+𝑦cosπœƒ)cos(βˆ’πœƒ)=(βˆ’π‘₯cosπœƒ+𝑦sinπœƒ)sin(πœƒ)+(π‘₯sinπœƒ+𝑦cosπœƒ)cos(πœƒ)=𝑦(sin2πœƒ+cos2πœƒ)=𝑦

hence the inverse map π‘Ÿβˆ’1πœƒ=π‘Ÿβˆ’πœƒ, and π‘Ÿβˆ’πœƒ is also smooth, π‘Ÿπœƒ is a diffeomorphism.