0

When $\operatorname{dim}X = \operatorname{dim} Y$, show that immersions $f: X \rightarrow Y$ are the same as local diffeomorphism.

If $\operatorname{dim}X = \operatorname{dim} Y$, then $\operatorname{dim}T(X) = \operatorname{dim} T(Y)$. Hence, injectivity of $df$ implies bijectivity. However, the tangent plane of a space is locally isomorphic to the space. Hence, $f$ is a local bijection.

But I don't find anything to prove smoothness.

Another attempt: I try to show that $df$ never equals zero. Stuck immediately.

Thank you for your very much patience.. :=)

1LiterTears
  • 4,572
  • 1
    Do you know the inverse function theorem? – Najib Idrissi Jul 02 '13 at 16:06
  • @nik, yes I thought about it, but it really confuses me. Inverse function theorem requires $df_x \neq 0$, but the injectivity of $df$ discuss situation when $df_x = 0$. So I was so confused that I assume $df_x \neq 0$ is just a sufficient condition.. – 1LiterTears Jul 02 '13 at 16:09
  • 1
    An injective map between two finite dimensional vector spaces of the same dimension is an isomorphism by the rank - nullity theorem. –  Jul 03 '13 at 06:20
  • 1
    @Jellyfish: The inverse function theorem requires $df_x$ to be an isomorphism, not merely $\neq 0$. An immersion is when all the $df_x$ are injective (so situations where $df_x \cdot \xi = 0$ for a tangent vector $\xi$); not when $df$ is injective itself. – Najib Idrissi Jul 03 '13 at 06:23

1 Answers1

1

This is a direct application of the inverse function theorem:

Let $x \in X$. Since $f$ is an immersion, $df_x : T_xX \to T_{f(x)}Y$ is injective. Since the two spaces have the same dimension $\mathrm{dim}(X) = \mathrm{dim}(Y)$, $df_x$ is actually an isomorphism. By the inverse function theorem, this implies that there is a neighborhood $U$ of $x$ such that $f_{|U} : U \to f(U)$ is a diffeomorphism. This is the definition of a local diffeomorphism: every point has a neighborhood like that.

Najib Idrissi
  • 54,185