Let $f: [0,1]^2 \to \mathbf{R}$ be a continuous function such that, for each $y \in [0,1]$, there exists a unique $x=g(y) \in [0,1]$ for which $f(\cdot,y)$ is strictly increasing in $[0,x]$ and strictly decreasing in $[x,1]$.
In addition, we know that $f$ is supermodular, i.e., for all $a,b \in [0,1]^2$ it holds $$ f(a)+f(b) \le f(a\vee b)+f(a\wedge b), $$ where the order on $\mathbf{R}^2$ is the one induced by the product order on $\mathbf{R}$.
Fact 1: $g:[0,1]\to [0,1]$ is a continuous nondecreasing function.
Fact 2: If $f$ belongs also to $\mathcal{C}^2(\mathbf{R}^2)$ then $g$ is strictly increasing in all $y \in [0,1]$ such that $g(y)$ is an interior point of the image of $g$.
Question. Could we obtain the same conclusion, i.e., $g$ being strictly increasing, without the hypothesis the $f$ is also twice differentiable? [You can also use Fact1 without proving it]
[Edit: a useful lemma could be that every function of bounded variation is differentiable almost everywhere]