$F$ is a diffeomorphism $\Leftrightarrow$ $0$ is not in the image of this function

Question

Let $f,g,h:\mathbb R\to \mathbb R$ be differentiable functions and define $F(x,y)=(f(x)\cdot h(y), g(y))$. Suppose $f$ and $g$ are diffeomorphisms from $\mathbb R$ to $\mathbb R$. I would like to prove the following equivalence:

$$F\ \text{is a diffeomorphism} \Leftrightarrow 0\notin h(\mathbb R)$$

Notice the determinant of the Jacobian matrix is $Jf(x,y)=h(y)f'(x)g'(y)$ for every $x,y\in \mathbb R $ and by inverse function theorem on $\mathbb R$ (in particular, $f$ and $g$ are differentiable functions) we get the following equivalence:

$$Jf(x,y)\neq 0\ \text{for every $x,y\in \mathbb R$}\Leftrightarrow 0\notin h(\mathbb R)$$

How can I use this fact to prove the former equivalence above?

Answer 1

You can't, at least not directly; knowing the Jacobian is zero only proves that $F$ is locally invertible.

While you could use some method to use local invertibility to prove global invertibility, such as muttering something about $\mathbb{R}^2$ being simply connected, that would be a tremendous amount of overkill — in this case you can simply write down the inverse of $F$.