Regularity of maps between embedded manifolds

Question

Let $M\subseteq \mathbb{R}^m$, $N\subseteq \mathbb{R}^n$ be topological/smooth/real analytic embedded manifold. Let $f: M \rightarrow \mathbb{R}^n$ be a continuous/smooth/real analytic map such that $Im(f)\subseteq N$. Does this imply that the function $f: M \rightarrow N$ is continuous/smooth/real analytic? How about the case of complex manifolds?

Answer 1

Yes, in all cases.

In the topological category, it just follows from the fact that $N$ has the subspace topology. (This is the characteristic property of the subspace topology.)

In the other categories, it follows from the constant-rank theorem in the appropriate category, which in turn follows from the inverse function theorem in the same category. (That is, if $U,V$ are open subsets of $\mathbb R^n$ or $\mathbb C^n$ and $\Phi\colon U\to V$ is a smooth/real-analytic/holomorphic map with nonsingular derivative at a point $p\in U$, then it has a smooth/real-analytic/holomorphic inverse in a neighborhood of $\Phi(p)$.)