Writing $f(x,y)$ as $\Phi(g(x) + h(y))$ Could you prove or disprove the following statement? 

Let 
  $f\colon[0,1]^2\rightarrow \mathbb R$ 
  be a continuous function.  Then
  there are continuous functions 
  $g,\ h\colon [0,1]\rightarrow \mathbb R$ and
  $\Phi\colon \mathbb R \to \mathbb R$
  such that  $$ f(x,y) = \Phi(g(x) + h(y)).$$

(This problem popped up in my mind while I was thinking about this related one on MO.  I couldn't find an easy proof or a disproof. This version is much weaker than the one asked at MO, since $g$ and $h$ do depend on $f$ here.)
 A: The statement is not true.
To make the counterexample simpler, I will take $f:[-1,1]^2 \rightarrow \mathbb{R}$.  Let f(x,y)=xy.  So we have 
---0+++
---0+++
0000000
+++0---
+++0---

Consider the image of xy=0 in g(x)+h(y).  Since it's a connected, compact set, the image is connected and compact, so an interval [m,n], with $\Phi([m,n])=0$.  We may assume that $\Phi(n+\epsilon)>0$ and $\Phi(m-\epsilon)<0$.  Let A=g(1), a=g(-1), B=h(1), b=h(-1).
Then we have $A+B>n, \; a+b>n, \;  A+b<m, \; a+B<m$, which leads to a contradiction.
A: Suppose $f$ is such that $f(x,0)$ and $f(0,y)$ are monotonic in $x$ and $y$ respectively over $[0,1]$. Since $f(x,0) = \Phi(g(x) + h(0))$, it must be the case that $g$ is monotonic over $[0,1]$. Similarly, $h$ is monotonic over $[0,1]$. So $g(x) + h(y)$ is the sum of monotonic functions. It's not hard to show that for any $(x,y)$ in the interior of $[0,1]^2$, there must be a point on the boundary with the same value of $g(x) + h(y)$, and therefore the same value of $\Phi(g(x) + h(y))$. Since it's easy to construct a continuous $f$ which satisfies the first assumption yet takes a larger range of values in the interior than it does on the boundary (for any $f$ that doesn't, try adding $M x(1-x)y(1-y)$ for large enough $M$), this contradicts the conjecture.
