Let $f:A\subseteq \Bbb R^n\to\Bbb R^n$ be differentiable and $Df(x)$ denote the Jacobian of $f$ at $x$.
If, $$Df(x)\neq 0,\;\;\;\forall x\in \Bbb R^n-------(*)$$
Then by Inverse Function Theorem we can conclude that, $\forall x \in \Bbb R^n \;\;\exists $ a neighborhood of $x$ in which $f$ is invertible, also $f$ is injective therein.
My question is- Does $(*)\implies$ $f$ is injective on $A$?
In general how do we check whether $f$ is injective or surjective on $A$ ? Are we supposed to simply use the definitions of injectivity and surjectivity or can we make use of the values of Jacobian to conclude something?
Counterexamples are trivial if $A $ is not connected.
Let $n=2$, $$A=\{(x,y):\ xy\ne0\}, $$ and $f (x,y)=(x^2,y^2)$. The Jacobian is $$Df (x,y)=\det\begin{bmatrix}2x&0\\0&2y\end{bmatrix}=4xy. $$ The conditions on $A $ guarantee that $Df\ne0$; and $f (1,1)=f (-1,1) $.
When $A $ is connected, the result is true for $n=1$.
For $n\geq2$, the situation is likely more subtle.