If the image of an injective immersion is a topological manifold, must it be an embedded smooth submanifold?

Question

Suppose $M,N$ are smooth manifolds (Hausdorff, second-countable). Let $ F:M\to N $ be an injective smooth immersion. Suppose that with the subspace topology inherited from $ N $, the image $ F(M) $ is a topological manifold (Hausdorff, second-countable). Then is $ F(M) $ necessarily an embedded smooth submanifold of $ N $?

Answer 1

If you allow your manifolds to have boundary, then no: consider $M=[0,1)$, $N=\mathbb{R}^2$, and let $F$ be an immersion whose image is a topological circle which has a non-smooth corner at $F(0)$ (with $F(t)$ approaching $F(0)$ as $t\to 1$).

If you do not allow boundaries, then yes, by invariance of domain. Invariance of domain implies that any continuous bijection between topological manifolds without boundary is an open map and thus a homeomorphism.* In particular, $F:M\to F(M)$ must be a homeomorphism so $F:M\to N$ is a smooth embedding.

*Note that if $f:X\to Y$ is a continuous bijection between (second-countable) nonempty topological manifolds then $X$ and $Y$ must have the same dimension, and so invariance of domain applies. This follows from the fact that $\mathbb{R}^n$ does not embed in $\mathbb{R}^m$ if $n>m$, which is another consequence of invariance of domain (restrict such an embedding to an $m$-dimensional subspace of $\mathbb{R}^n$ and apply invariance of domain to get a contradiction). This immediately implies that $\dim X\leq \dim Y$. On the other hand, $f$ is an embedding when restricted to each compact subset of $X$, and the images of these compact subsets must have empty interior if $\dim X<\dim Y$. Since $X$ is $\sigma$-compact this implies $f$ cannot be surjective by the Baire category theorem.