Does the rank of the Jacobian contain information about invertibility. Let $m\geq n$ and consider  $f: \mathbb{R}^n \to \mathbb{R}^m: x\to f(x)$  continuous and differentiable. Does 
the rank of the Jacobian of $f$ (at a point $x_0$) contain any information of whether it has an (local) inverse? 
I don't know, but have managed to google me to the multivariable inverse function theorem: Wikipedia, but this seems to assume that $n = m$ for some reason, but I need it to hold also if $m>n$. Is it still true? A source would be plenty.
 A: Derivatives can only tell you about local invertibility; global invertibility is a totally different matter.  This answer focuses on local inverse, and on the case when the Jacobian has rank $ n$.
The Jacobian matrix at $\mathbf x_0$   has more rows than columns. By assumption, its columns are linearly independent. You can add extra columns to this matrix so it gets into square shape, and becomes invertible. (A linearly independent set of vectors can be extended to a basis.) 
Adding columns $v_{n+1},\dots,v_m$ means  introducing  extra dimensions $x_{n+1},\dots,x_m$ to the domain, and extending $f$ to the map 
$$
F(x_1,\dots,x_n,x_{n+1},\dots,x_m) = 
f(x_1,\dots,x_n) + \sum_{i=n+1}^{m}  x_i  v_{i}   
$$
where  $v_{n+1},\dots,v_m$ are constant vectors in $\mathbb R^m$. You can check that the Jacobian of $F$ consists of the Jacobian of $f$ with those extra columns $v_{n+1},\dots,v_m$. 
Applying the inverse function theorem to $F$, we get a smooth inverse map $F^{-1}$. It goes from some neighborhood $U$ of $f(\mathbf x_0)$ in $\mathbb R^{m}$ into $\mathbb R^{m}$. We can disregard the added dimensions now, projecting back to $\mathbb R^n$. The result is a smooth map $g$ from $U$ to $\mathbb R^n$. It satisfies $g(  f(x)) = x$. 
The composition $f\circ g$ is also an interesting map: it is a smooth projection from $U$ onto the image of a neighborhood of $\mathbf x_0$ under $f$. The above construction also shows that this image is a smooth $n$-dimensional manifold.  
