no point where assumptions of Inverse Function Theorem hold?

Question

Let $\varphi: \mathbb{R}^3 \to \mathbb{R}$, $\psi: \mathbb{R}^3 \to \mathbb{R}$ are continuously differentiable. Define $f: \mathbb{R}^3 \to \mathbb{R}^3$ by$$f({\bf x}) = (\varphi({\bf x}), \psi({\bf x}), 1 + \varphi({\bf x})\psi({\bf x}) + \varphi({\bf x})^3).$$ Why is there no point ${\bf x} \in \mathbb{R}^3$ where the conditions of the Inverse Function Theorem holds for $f$? Can we assert that the image of $f$ is a $2$-manifold, and thus cannnot contain any open balls in $\mathbb{R}^3$?

Answer 1

The image does not actually need to be a $2$-manifold, depending on the specific nature of the functions. Also, the approach you suggest relies on the fact $2$-manifold cannot also be a $3$-manifold. It is of course true that a manifold has a unique dimension, but showing this, in the most general setting, is not a triviality. It is not too hard to show the uniqueness of dimension for differentiable manifolds, however for merely continuous manifolds a decent amount of algebraic topology is required. So I will show the statement with a bit more of an elementary approach in two different ways, one analytic and one geometric.

Analytic Approach

Define $g: \mathbb{R}^2 \to \mathbb{R}$ by $g(x, y) = x^2 + xy + 1$, so that $f({\bf x}) = (\varphi({\bf x}), \psi({\bf x}), g(\varphi({\bf x}), \psi({\bf x})))$. Out of laziness we suppress the ${\bf x}$ and use subscripts to denote partial derivatives, for instance $\psi_2 = {{\partial \psi}\over{\partial x_i}}({\bf x})$. The Jacobian of $f$ is, using the chain rule to calculate the bottom row,$$\begin{pmatrix} \varphi_1 & \varphi_2 & \varphi_3 \\ \psi_1 & \psi_2 & \psi_3 \\ g_1\varphi_1 + g_2\psi_1 & g_1\varphi_2 + g_2\psi_2 & g_1\varphi_3 + g_2\psi_3 \end{pmatrix}.$$Observe that the last row is the first row multiplied by $g_1$ plus the second row multiplied by $g_2$. Hence at any point the rows are linearly dependent, and thus the determinant is equal to $0$. So the main assumption of the Inverse Function Theorem—the invertibility of the Jacobian—is not satisfied anywhere.

Geometric Approach

Above we showed that the assumptions of the Inverse Function Theorem cannot hold; here we show explicitly the conclusion cannot either. One of the conclusions of the Inverse Function Theorem is that if $U$ is a sufficiently small open ball around a point ${\bf x}$, then $f(U)$ is an open ball around $f({\bf x})$. We show that $\text{Im}\,f$ can contain no open balls. Let ${\bf x} \in \text{Im}\,f$, such that ${\bf x} = (x, y, z) = (x, y, g(x, y))$. Observe that for a given $(x, y) \in \mathbb{R}^2$ there is at most one $z$ such that $(x, y, z) \in \text{Im}\,f$ because we must have $z = g(x, y)$. So for any $\epsilon > 0$, the point $(x, y, z + \epsilon/2) \in B_\epsilon({\bf x})$ but it is not contained in $\text{Im}\,f$. So $f$ can contain no open balls, and the conclusion of the Inverse Function Theorem does not hold.