Can we determine the injectivity of a map $\mathbb{R}^n \rightarrow \mathbb{R}^n$ on subset of $\mathbb{R}^n$ by looking at the Jacobian? Specifically, I'm looking for an analogue of the following theorem in the case of real functions:

Let $f: A \subset \mathbb{R} \rightarrow \mathbb{R}$ be monotone on $A$, then $f$ is injective on $A$.

My question is, given a differentiable function $F : \mathbb{R}^n \rightarrow \mathbb{R}^n$, is there a similar theorem we could state in terms of the sign of the Jacobian, and possibly some other simple topological restrictions on the set $A$?
The inverse function theorem doesn't work for me, as I am wondering about the injectivity on a specific set $A$, not just on some neighborhood of a point.
 A: Suppose that $A\subseteq \mathbb R^n$ is convex. Let $F$ be once continuously differentiable, denote by $F'(x)$ the Jacobian of $F$ at $x\in A$, and let $I$ denote the $n\times n$ identity matrix. Suppose that we have
$$
\sup_{x\in A}\lVert F'(x)-I\rVert<1.
$$
It follows from the fundamental theorem of calculus that
\begin{align}
F(x)-F(y) &=\int_0^1 \left[\left.\frac{d}{dt}\right|_{t=s} F(tx+(1-t)y)\right] ds\\
&=\left[\int_0^1 F'(sx+(1-s)y)ds\right] (x-y) \\
&=\left[I+ \int_0^1 \left[F'(sx+(1-s)y)-I\right]ds\right] (x-y).
\end{align}
Invoking the triangle inequality for integrals, we find
$$
\lVert\int_0^1 \left[F'(sx+(1-s)y)-I\right]ds\rVert \leqslant \sup_{x\in A}\lVert F'(x)-I\rVert.
$$
Then an application of the triangle inequality for the Euclidian norm ensures that
\begin{align}
\lvert F(x)-F(y) \rvert &=\lvert (x-y)+ \left[\int_0^1 \left[F'(sx+(1-s)y)-I\right]ds\right](x-y)\rvert\\
&\geqslant \left[1-\sup_{x\in A}\lVert F'(x)-I\rVert\right]\lvert x-y\rvert.
\end{align}
Since we assume $\sup_{x\in A}\lVert F'(x)-I\rVert<1$, this suffices to conclude that $F$ is injective.
A: A sufficient condition, somewhat weaker than that proposed by 
user161825, is that the Jacobian $F'$ is continuous and $F' + (F')^T$ is positive definite everywhere on the convex set $A$.  This implies that for $y \ne x$, 
$$ (y - x)^T (F(y) - F(x)) = \int_0^1 (y-x)^T F'(x + t(y-x)) (y-x) \; dt > 0$$
A: In dimensions $2$ and higher, local injectivity does not imply global injecivity. A nice counterexample, expressed with a complex variable $z=x+iy$, is
$$f(z) = e^z
$$
which is expressible in real variables as
$$f(x,y) = (e^x \cos(y), e^x \sin(y))
$$
This is infinite-to-one over each point except the origin.
A: *

*One more cool theorem along these lines which deserves to be better known, it is due to V.Zorich:


Definition. A map $f: R^n\to R^n$ is said to have bounded distortion (or is quasiregular) if the quantity
$$
M_f(x):= ||Df(x) (Df(x))^{T}|| J_f^{-2n}(x)
$$
is uniformly bounded, where $J_f(x)$ is the Jacobian determinant of $f$ at $x$. (For simplicity, $f$ is $C^1$, although one needs only $f\in W^{1,n}_{loc}$.) Geometrically speaking, $f$ maps infinitesimally small spheres to infinitesimally small ellipsoids of uniformly bounded eccentricity. 
Theorem (V.Zorich [1]). Suppose that $n\ge 3$ and $f: R^n\to R^n$ has bounded distortion and is a local homeomorphism. (In the smooth setting just assume that $J_f(x)\ne 0$ for all $x$.) Then $f: R^n\to R^n$ is a homeomorphism. 
Note that this theorem fails for ${\mathbb R}^2={\mathbb C}$, where one can take $f(z)=e^z$ as a counter example. The proof is quite nontrivial, see: 
[1] V. Zorich, The theorem of M.A. Lavrent’ev on quasiconformal mappings in space. Math. Sb. 74 (1967)  p. 417–433.  


*The following is the (notoriously difficult) Jacobian Conjecture which is somewhat related to your question:


Conjecture. Suppose that $f: {\mathbb C}^n\to {\mathbb C}^n$ is a polynomial mapping whose complex Jacobian is a nonzero constant. Then $f$ is bijective (and its inverse is again a polynomial mapping).
Note that (somewhat surprisingly) versions of this conjecture for holomorphic maps and for real polynomial maps (with variable nonvanishing Jacobian) are false.   
