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I am having problems with the following question: what is the relation between the rank of the Jacobian matrix of f and the dimension of the image of f? (f being continuosly differentiable),is there a theorem about it? Thank you

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What is the domain and range of $f$? –  Jesse Madnick Jun 29 '11 at 1:04

2 Answers 2

Look up the rank theorem in Spivak's Calculus on Manifolds or Flemings Functions of Several Variables.

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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.

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