Does a bijective submersion imply diffeomorphism

Question

Submersion, by definition is, $df_x: T_x(X)\rightarrow T_y(Y)$ is surjective and if it is also bijective, then intuitively it is a diffeomorphism for me as $df_x^{-1}:T_y(Y)\rightarrow T_x(X)$ is well defined. Am I correct? Thanks and appreciate a hint.

Answer 1

1. Yes. Injective (smooth) submersions are (smooth) embeddings. Therefore, bijective submersions are surjective embeddings, also known as diffeomorphisms.

2. If $f$ is a bijective submersion, then we don't have right away that $df_{x}$ has an inverse. The given bijection is $f$ not $df_{x}$. Of course however, diffeomorphisms $g$ are both submersions and immersions, and thus $dg_{x}$ is invertible (i.e. both a submersion and immersion).

Answer 2

Your claim follows from the Global Rank theorem which says that if $F:M \rightarrow N$ is a smooth map of constant rank then the following holds true:

1. If $F$ is injective then $F$ is an immersion.
2. If $F$ is surjective then $F$ is a submersion.
3. If $F$ is bijective then $F$ is a diffeomorphism.

Now in particular (3) follows from (1) and (2) in a straightforward way as follows:

It is a known result that if $F$ is both an immersion and submersion then $F$ is a Local Diffeomeorphism. So your claim follows from the fact that every bijective local diffeomorphism is a diffeomorphism and submersions are by definition smooth maps of constant rank.

For the detailed proof of all the above stated facts please check out Theorem 4.14 (Global Rank Theorem), Pg-83, Introduction to Smooth Manifolds, John M. Lee, 2nd Edition.