 Solution

The transport cost separates into a function of $y$ and a function of $x$, so for every transport planThe value is fixed by the marginals. Hence every transport map from $\mu$ to $\nu$ is optimal, and indeed every coupling is optimal.

For an explicit description of the complete set, let $F(x)={\int }_0^{x}f(s)ds$ and $G(y)={\int }_1^{y}g(s)ds$. Their strictly positive continuous densities make $F:[0,1]\to [0,1]$ and $G:[1,2]\to [0,1]$ increasing homeomorphisms. The full family of deterministic plans iswhere $S:[0,1]\to [0,1]$ is any measurable Lebesgue-measure-preserving map. Indeed $F_{\ast }\mu$ is uniform measure and $G_{\ast }\nu$ is uniform measure, so any such $S$ produces the required pushforward. Conversely, for any transport map $T$, the map $S=G\circ T\circ F^{-1}$ preserves uniform measure. This is the measure-preserving parametrization of one-dimensional transport maps; no monotonicity is required for the linear cost.