Let $\mathbb{R}^d$ denote the $d$-dimensional Euclidean space, $\mathcal{W}\_2(\mathbb{R}^d)$ denote the $2$-Wasserstein space with respect to the $d$-dimensional Euclidean space $\mathbb{R}^d$. Let $L^2(\mathbb{R}^d)$ denote the Bochner space of all Borel functions $f:\mathbb{R}^d\rightarrow \mathbb{R}^d$ satisfying $\int \, \lVert f(x)\rVert^2dx<\infty$.


Let $\mathcal{X}\subseteq \mathcal{W}\_2(\mathbb{R}^d)$ consist of all measures ${\nu}$ for which there is some $f\in L^2(\lambda,\mathbb{R}^d)$ satisfying:
$$
{\nu}=f\_{\#}\lambda
$$
where $\lambda$ is the uniform measure on $[0,1]^d$.


How is $\mathcal{X}$ related to $\mathcal{W}\_2(\mathbb{R}^d)$? Is $\mathcal{X}$ a dense subset of $\mathcal{W}\_2(\mathbb{R}^d)$?



Building the Wasserstein space by pushforwards
Here is an alternative answer, which is weaker than @Benoît Kloeckner's strong form and only asserts density.


Yes, your set is dense. Indeed, $\mathcal X$ contains finitely atomic measures, which are obviously weakly-$\ast$ dense in the Wasserstein space and therefore also dense in distance (it is well known that the Wasserstein distance $W\_2$ metrize the weak-$\ast$ convergence, modulo minor decay conditions at infinity given by convergence of the second moments).


To see that $\mathcal X$ contains finitely atomic measures, fix an arbitrary integer $N$ and take any $\nu=\sum\_{i=1}^N \nu\_i\delta\_{x\_i}\in \mathcal W\_2(\mathbb R^d)$. Choose now any partition of your unit cube $[0,1]^d=\bigcup\_{i=1}^1 C\_i$ into $N$ disjoint sets with corresponding measures $|C\_i|=\nu\_i$. Then the piecewise constant map $f$ defined as $f(x)=x\_i$ for $x\in C\_i$ pushes forward $\lambda$ to $\nu$.


