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?
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.)