Converse in supermodularity and single-crossing Suppose I denote $$f(x,t)=g(x,t)+h(x)$$ where $X\in \mathbb{R}$, $t\in [0,1]$ and $h(x)$ is any function.

I want to prove that if the single-crossing property holds in $f(x,t)$, then $g(x,t)$ must be supermodular over $X\times T$.

I know in general supermodular means single-crossing but the converse is not true, but my guess adding a random function $h(x)$ should guarantee single-crossing property will show supermodularity.
I'm not sure there are enough assumptions here to move forward, in particular I would think you would need some conditions on your function $h$, at the very least. It's pretty easy to show that $g(\cdot,\cdot)$ will have increasing differences, which implies the single-crossing property for $g$ (this can be quickly seen by letting $h$ be identically zero). But I don't see how this is enough, since single-crossing doesn't imply supermodularity. Maybe I'm missing something, though.