Suppose $m,n>0$, $U$ an open subset of $\mathbb{R}^{m+n}$ and let $f: U \to \mathbb{R}^n$ be continuously differentiable. Is it possible for $f$ to be injective?
My thinking is that continuous differentiability+injectivity suggests that this is sort of getting at a converse to the Inverse Function Theorem. We know the Jacobian is not invertible at any point in $U$, and the question is asking if this implies that $f$ cannot be invertible in any neighborhood contained in $U$.
Your reasoning using invariant of domain is correct, but overkill for the question you are asking. After all, you did not use differentiability of $f$ at all.
A different approach would be the following: Let $k \leq n$ be the maximal rank of all differentials $Df(x)$ for $x \in U$. It is then not hard to show that the set $$V := \{x\in U\mid \text{rank}(Df(x)) = n\} = \{x\in U\mid \text{rank}(Df(x)) \geq n\}$$
is open (use that for $x\in V$, there is a $k\times k$ submatrix of $Df(x)$ which is invertible).
By assumption, $f$ is injective on $U$ and hence on $V$. But the constant rank theorem (see e.g. Constant rank theorem) shows that $f$ is (on $V$, up to diffeomorphisms) locally of the form $(x_1,\dots ,x_{n+m})\mapsto (x_1,\dots , x_k)$, which is not injective.
