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.. :=)
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.