Submersion map from $S^2$ to $S^2$ is a diffeomorphism

Question

If $f:S^2\mapsto S^2$ be a submersion. Show that $f$ is a diffeomorphism.

I got that $f$ is surjective using $S^2$ is connected and compact.(Theorem : If $X$ is compact and $Y$ is connected, then every submersion $f:X→Y$ is surjective)

Answer 1

To say that $f:S^2\to S^2$ is a submersion means that at any $s\in S$ the tangent map $T_sf:T_sS^2\to T_sS^2$ is surjective, hence bijective since source and target of that linear map have same dimension (namely 2).
We deduce that $f$ is a local diffeomorphism.
But a local diffeomorphism (or even local homeomorphism) with compact source is a covering space, so that $f$ is actually a differentiable covering map.
And now for the coup de grâce: since $S^2$ is simply connected a connected covering map must be one sheeted, so that $f$ is a bijective local diffeomorphism.
Its inverse $f^{-1}$ is also differentiable (by some avatar of the implicit function theorem), and finally we can proudly assert:
Yes, $f$ is a diffeomorphism.