Jacobian in 3-Space Given three functions $U,V,W$, show that if the Jacobian determinant $\frac{\partial U, V, W}{\partial x, y, z}$ is equal to zero implies a functional relation $F(u,v,w) = 0$. This was a problem in a textbook. 
This seems like a contradiction to the implicit function theorem which states that $D_{f}(a,b,c)$ being invertable implies that there exists ${\{(x, g(x) : g(x) = y}\} =\{ {(x,y) : f(x,y) = 0\}}$. In this case the Jacobian is given as irreversible, so what is going on? 
 A: If the determinant of the Jacobian is zero, it must have an eigenvector with eigenvalue zero. We can use this to find a vector perpendicular to all the tangent vectors to the manifold $\{(U(x,y,z),V(x,y,z),W(x,y,z)):x,y,z \in \mathbb{R}\}$.
In particular, change the notation to make the summations easier to write: let the functions be $u_i(x)$, $x=(x_1,\dotsc,x_d)$. Any tangent vector to the manifold $\{(u_1(x),\dotsc,u_d(x)):x \in \mathbb{R}^d\}$ is given by
$$ V' = JV, $$
where $V = (V_1,\dotsc,V_n) \in \mathbb{R}^d$ and
$$ J_{ij} = \frac{\partial u_i}{\partial x_j} $$
is the Jacobian matrix. We have $\det{J}=0$, so there is a nonzero vector $N$ so that $NJ=0$. But then for any $V$ we have $0=(NJ)V = N \cdot (JV) $, by linearity, so $N$ is normal to any vector tangent to the manifold. Hence the manifold cannot have dimension $d$. (We can't determine what its dimension is unless we can find either the number of zero eigenvalues/eigenvectors or the number of linearly independent tangent vectors.)

Now, the implicit function theorem is talking about a function $f(x,u)=0$ (with arguments in both the input and output, i.e. domain and codomain). Given a function of this form with a nonzero Jacobian (which is a linear map on the product or direct sum of the tangent spaces of the domain and codomain). This is not what you have with $F$: $F$ is a real-valued function on the output variables only. 
What would correspond most explicitly to the implicit function theorem's initial function would be something like
$$ 0 = f(x,u) = u-U(x). $$
This is the case of the inverse function theorem, which applies here to say that it is not possible to write $x=V(u)$ for some function $V$, even locally. (This has the simple interpretation that some of the $n$ dimensions of the domain are removed: on the tangent space, this just looks like the result that a nontrivial kernel implies noninjectivity in linear algebra, because the Jacobian determinant being zero also implies the existence of a nontrivial set of right-eigenvectors. At each point there is at least one direction one can move in locally in the domain without affecting the position of the output point in the codomain.)
