How to determine the independence of multiple scalar functions Let $x,y\in R$ be two arbitrary scalar numbers. $f_1(x,y),...,f_n(x,y)$ are n scalar functions of $x,y$. Note $x,y$ only have two degree of freedom (DOF), so I think the n functions $f_1(x,y),...,f_n(x,y)$ at most have two DOF. That probably means the vector $[ f_1(x,y),...,f_n(x,y) ]^T$ can not be an arbitrary vector in $R^n$ space. My question is: how to determine whether two functions are independent? And how to determine which functions are redundant among the n functions?
 A: if you want $f_1,f_2,f_3$ st $\{(f_1(x,y),f_2(x,y),f_2(x,y)) : x,y\in\mathbb{R}\}=\mathbb{R}^3$ then probably not.  each equation $f_i(x,y)=a_i$ is basically a curve in the plane and the intersection of three curves may not exist.
you could (artificially) make 3 functions that do satisfy this however.  establish a bijection $B=(B_1(x,y),B_2(x,y),B_3(x,y)):\mathbb{R}^2\to\mathbb{R}^3$, then let $f_i=B_i$ so that for any point in $\mathbb{R}^3$ you can look up the corresponding $x,y$.  however i doubt these functions will be continuous.
A: Thanks everyone. I come back again. I think I didn't express myself well in the question. Here is my own answer to this question. What I use is 'Diffeomorphism'.
Definition: A function $\phi: R^n \mapsto R^n$ is called diffeomorphism if it is smooth and if its inverse $\phi^{-1}$ exists and is smooth.
Given a function, we can use the singularity of its Jacobian matrix to determine whether the function is diffeomorphism. Here is an example.
$$\left[ \begin{array}{ccc} z_1 \\\ z_2\end{array} \right]
=\phi (x)
=\left[ \begin{array}{ccc} 2x_1+5x_1x_2^2 \\\ 3\sin x_2\end{array} \right]$$
Then the Jacobian of $\phi (x)$ is
$$J=\left[ \begin{array}{ccc} 2+5x_2^2 & 10x_1x_2 \\\ 0 & 3\cos x_2 \end{array} \right]$$
It is apparent that the Jacobian is singular if and only if $\det (J)=3(2+5x_2^2)\cos x_2=0$. That is $x_2=k\pi+\pi/2$. Hence the inverse function $\phi ^{-1}$ exists only when $x_2\neq k\pi+\pi/2$.
This might be the right way to understand the question.
