Is the inverse of a causal function causal? I am wondering if the inverse of a causal function is causal. I'll illustrate what I mean with an example: 
Assume $f$ is a bijection of $\mathbb R^2$ in $\mathbb R^2$. I assume $f$ is causal in the following sense $f(x) = (f_1(x_1), f_2(x_1,x_2))$, or in other words, $x_1 = y_1 \Rightarrow f_1(x) = f_1(y)$. Can I say something about the causality of $f^{-1}$ ?
I could not find a very definitive answer to this question, even though some literature on operators on Hilbert space might suggest it is not the case. here 
I would like to either prove or find a counterexample of causality of the inverse, i.e. in this case, denoting $g = f^{-1}$, is it true that $x_1 = y_1 \Rightarrow g_1(x) = g_1(y)$.
I know it holds for linear functions in finite dimensional spaces and I would also be interested in a reference to a simple proof of that. 
Thanks in advance
 A: Without further assumptions on $f$ this is false. Here is a counterexample: Let $\eta_1 : (\infty, 0] \to \mathbb{R}$ and $\eta_2 : (0, \infty) \to \mathbb{R}$ be bijections, and define
Define
$$
f(x, y) = \left\{
     \begin{array}{ll}
       \big(\eta_1(x), \eta_1^{-1}(y)\big) & : x \in (-\infty, 0] \\
       \big(\eta_2(x), \eta_2^{-1}(y)\big) & : x \in (0, \infty)
     \end{array}
   \right.
$$
Then $f$ is a causal bijection, but $f^{-1}$ is not causal (since $f_1$ is not injective).
Edit: We give a proof that if $f$ is continuous, then $f^{-1}$ is causal. 
It suffices to show that $f_1$ is injective; it follows easily that $f_1$ is a bijection and $g_1 = f_1^{-1}$.
For each $x \in \mathbb{R}$, the restriction $f_2(x, \cdot)$ of $f_2$ to the fiber $\lbrace x \rbrace \times \mathbb{R}$ gives a continuous injection $\mathbb{R} \to \mathbb{R}.$ This forces the image $V_x := f_2(x, \mathbb{R})$ to be open in $\mathbb{R}$ (an open interval, in fact). 
Since $f$ is injective, $V_x$ and $V_{x'}$ are disjoint when $f_1(x) = f_1(x')$ and $x \neq x'$, and since $f$ is surjective, $\mathbb{R}$ is covered by the disjoint open intervals $V_{x'}$ with $f_1(x') = f_1(x)$. But $\mathbb{R}$ is connected, so it is not covered by a nontrivial family of disjoint open sets, and this implies that $f_1$ is injective.
