In this paper [1], the following vector-to-vector function is introduced:
$$f(\mathbf{z}) = \mathbf{z} + \mathbf{u} h(\underbrace{\mathbf{w}^T \mathbf{z}}_{\text{dot product}} + b)$$
where $\mathbf{z}, \mathbf{w}, \mathbf{u} \in \mathbb{R}^D$, $b\in\mathbb{R}$, and $h$ is the element-wise tanh function (a strictly increasing function). The paper calls this the planar flow transformation, though the name itself is unimportant for the context of this question.
The paper claims that $f$ is invertible as long as $\mathbf{w} ^T \mathbf{u} \geq -1$, but I disagree (see comment at the end).
Here's a summary of their constructive proof for finding $\mathbf{z}$, given that we know $\mathbf{w}, \mathbf{u}, b$ and $\mathbf{y}=f(\mathbf{z})$:
(1) Decompose $\mathbf{z}$ into a component parallel to $\mathbf{w}$, $\mathbf{z}_{||}$, and a component perpendicular to $\mathbf{w}$, $\mathbf{z}_{\perp}$. Then, clearly, we have
$$\mathbf{y} = f(\mathbf{z}) = \mathbf{z}_{\perp} + \mathbf{z}_{||} + \mathbf{u} h(\mathbf{w}^T \mathbf{z}_{||} + b).$$
(2) Finding $\mathbf{z}_{\perp}$ (50% of the job done): $$\mathbf{z}_{\perp} = \mathbf{y} - \mathbf{z}_{||} - h(\mathbf{w}^T \mathbf{z}_{||} + b).$$
(3) Finding $\mathbf{z}_{||}$ (the remaining 50% of the job):
Here, they used a clever argument. Since $\mathbf{z}_{||}$ is parallel to $\mathbf{w}$, it can be expressed as
$$\mathbf{z}_{||} = \alpha \frac{\mathbf{w}}{||\mathbf{w}||^2}$$
Therefore, if we multiple both sides of
$$f(\mathbf{z}) = \mathbf{z} + \mathbf{u} h(\mathbf{w}^T \mathbf{z} + b)$$
by $\mathbf{w}$, we get
\begin{array} \mathbf{w}^T f(\mathbf{z}) &= \mathbf{w}^T\mathbf{z} + \mathbf{w}^T\mathbf{u} h(\mathbf{w}^T \mathbf{z} + b) \\ &= \mathbf{w}^T\mathbf{z}_{||} + \mathbf{w}^T\mathbf{u} h(\mathbf{w}^T \mathbf{z}_{||} + b) \\ &=\alpha + \mathbf{w}^T\mathbf{u} h(\alpha + b). \end{array}
The paper says that as long as the right-hand side is a non-decreasing function, then the equation above is invertible with respect to $\alpha$, i.e., its derivative $1 + \mathbf{w}^T \mathbf{u} h'(\alpha+b) \geq 0 \rightarrow \mathbf{w}^T \mathbf{u} \geq - \frac{1}{h'(\alpha + b)}$. Finally, since $0 \leq h'(\alpha + b) \leq 1$, we would have $\mathbf{w}^T \mathbf{u} \geq -1$.
Here's my question. Why can't the right-hand side be a non-increasing function?