Basic Application of Implicit Function Theorem I am studying an old exam for a course in real analysis and came across this problem (not a homework problem).
Let
$f(x, y, z) = (xyz, (z \cos x) + y - 1)$
and observe that $f(2\pi, 1, 0) = (0, 0)$. Only one of the following is true, circle and carefully
justify the correct statement:
(a) There exists an open set $U \subset \mathbb R^1$ containing $2\pi$ and a $C^1$ function
$g = (g_1,g_2) : U \to \mathbb R^2$
such that $f(x, g_1 (x), g_2(x)) = (0, 0)$ for all $x \in U$.
(b) There exists an open set $U \subset \mathbb R^2$ containing $(1, 0)$ and a $C^1$ function
g : $U \to \mathbb R^1$
such that $f(g(y, z), y, z) = (0, 0)$ for all $(y, z) \in U$.
The answer is marked as (b) but it seems that it should be (a) because the implicit function theorem says that given a function $f: \mathbb R ^{m+n} \to \mathbb R^m$ and some conditions on the derivative it is possible to find a function from $ g(x) \mathbb R^n \to \mathbb R^m$ s.t. $(g(x), x) = 0$ for all $x \in \mathbb R^n$. The rest of the problem (verifying that these conditions hold) I am not worried about. Am I just confused about what the implicit function theorem states/how to use it? Or is the answer (a)?
 A: I actually agree with you. Discussion:
The implicit function theorem basically says that if you have $m$ independent scalar equations in $n+m$ variables, then you can (locally) solve for $m$ of the variables in terms of $n$ of the variables. (The technical details of the result are about what "independent" and "locally" really mean.) In the process, the number of independent variables drops by the number of scalar constraints. One way to remember this is to note that when the system is linear, the result is exactly the same as the rank-nullity theorem in the case of a full rank matrix. 
In your case you start with $3$ independent variables and impose $2$ scalar constraints, so you should have $3-2=1$ independent variable at the end.
The one possible problem is that the Jacobian with respect to your chosen "$y$" variables, which in the first statement are $y$ and $z$, might not be invertible at the point in question. But it looks to me like it is; I find that it is $\begin{bmatrix} 0 & 2 \pi \\ 1 & 0 \end{bmatrix}.$
