# Is the inverse function theorem for manifolds applicable here?

Let $$\mathbb{M}$$ be some $$m$$-dimensional manifold with boundary and let $$\varphi \mathpunct{:} \mathbb{R}^m \to \mathbb{M}$$ be some smooth function. If the Jacobian $$D\varphi(x)$$ has rank $$m$$ in some point $$x$$, can I make any statements about local invertability and smoothness of the inverse? What about if $$\varphi(x)$$ is in the interior of $$\mathbb{M}$$? What if I have some open set $$U \subset \mathbb{R}^m$$ such that $$\varphi$$ restricted to $$U$$ is injective and $$D\varphi$$ has rank $$m$$ on all of $$U$$?

This feels like the inverse function theorem but the conditions are not quite correct since $$\mathbb{M}$$ has a boundary and I have information on the rank of the Jacobian instead of the tangent map. Can I still do anything here?

In fact, if $$D\varphi(x)$$ has maximal rank, then $$\varphi(x)$$ cannot lie in the boundary of $$M$$.
Consider $$d\!M$$ the double of $$M$$, which consists in two copies of $$M$$ glued together along their common boundary. It is a $$m$$-dimensional manifold without boundary, and since $$M\subset d\!M$$, then $$\varphi\colon \Bbb R^m \to d\!M$$ is well defined. We still have that $$\mathrm{Im}(\varphi)\subset M$$.
Let $$x\in \Bbb R^n$$ be such that $$D\varphi(x)$$ has maximal rank. By the inverse function Theorem (for manifolds without boundary), $$\varphi$$ induces a diffeomorphism between an open neighbourhood $$U$$ of $$x$$ in $$\Bbb R^n$$ and an open neighbourhood $$\varphi(U)$$ of $$\varphi(x)$$ in $$d\!M$$. Recall that $$\varphi(U)\subset \mathrm{Im}(\varphi)\subset M$$. Then, $$\varphi$$ induces a diffeomorphism between an open neighbourhood of $$x$$ in $$\Bbb R^n$$ and an open neighbourhood of $$\varphi(x)$$ in $$M$$.
To conclude, note that no open neighbourhood of any boundary point of a manifold can be diffeomorphic (even homeomorphic) to an open subset of $$\Bbb R^n$$. Thus, $$\varphi(x)$$ did not lie in $$\partial M$$.
Remark: instead of working with the double, one would have preferred to glue a collar neighbourhood $$\partial M \times [0,1)$$ to $$M$$ along $$\partial M$$ (with $$x \sim (x,0)$$). The rest of the proof remains the same.