= Solution
The intended <monotone rearrangement> is
$$
\boxed{T^\dagger(x)=G^{-1}(F(x))\quad\mu\text{-almost everywhere}.}
$$
Here $G^{-1}$ is the <quantile function> of $\nu$, agreeing with the ordinary inverse when $G$ is continuous and strictly increasing. With an <atomless measure> $\mu$, its <cumulative distribution function> $F$ is continuous, and the <probability integral transform> makes $F(X)$ uniform on $(0,1)$ for $X\sim\mu$. Thus $G^{-1}(F(X))\sim\nu$. The one-dimensional <monotone rearrangement> theorem says this <transport map> minimizes the cost $d(x-y)$ for <convex> continuous $d$, whenever the cost integrals are well defined. Values at exceptional endpoints may be chosen arbitrarily.
\b[The printed assumptions omit an essential source condition.] Invertibility of $G$ alone does not ensure an admissible <transport map>. For example, $\mu=\delta_0$ and $\nu$ a <standard normal distribution> satisfy the stated condition on $G$, but $T_\#\delta_0$ is always a <Dirac measure>. There is no solution to the <Monge optimal transport problem> in this example. The boxed answer therefore requires the additional assumption that $\mu$ is an <atomless measure>, or an equivalent condition making the displayed map admissible. For arbitrary sources the always admissible monotone <transport plan> is $(F^{-1},G^{-1})_\#\mathcal U(0,1)$, which need not be induced by a map.
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