TITLE: Jacobian matrix rank and dimension of the image
QUESTION [8 upvotes]: I am having problems with the following question: 

What is the relation between the rank of the Jacobian matrix of $f$ (which is continuously differentiable) and the dimension of the image of $f$?

Is there a theorem about it?

REPLY [3 votes]: Let $U\subset \mathbb R^n$ be open and consider a continuously differentiable map
$f:U\to \mathbb R^m$. Since  you speak  of the rank of $df$, I assume that you mean that for all $x\in U$ the rank of $d_x(f)$ is some fixed integer $k$ (the map $f \;$  is then called a subimmersion : a  generalization of both immersions and  submersions).
The constant rank theorem then says that locally near every fixed $a\in U$ the morphism $f \;$ has in suitable coordinates the form 
$$ (x_1,x_2,\ldots,x_n) \mapsto (x_1,x_2,\ldots,x_k,0,0,\ldots,0)$$
So in a sense the image has dimension $k$. For example, if $k=m$ you have a good old submersion and the image of $f\; $ is an open subset  $f(U)\subset \mathbb R^m$, which has of course dimension $m$.   
End of story? Not so fast! As I said the displayed form for $f\;$ is only true locally and unfortunately the image of $f \;$ is not a submanifold of $\mathbb R^m$ so that it is not trivial to make precise the idea that the image is of dimension $k$. The simplest example of that difficulty is the notorious injective immersion $\mathbb R \to  \mathbb R^2$ whose image is a figure eight, which has a node and thus is not a submanifold of $\mathbb R^2$, although it certainly has dimension one in every conceivable interpretation of dimension.