Suppose I have iid Uniform $[0, 1]$ random variable $U_i, 1\leq i\le m$. Let $n=m(m-1)/2$. I am interested in a measurable function $F:[0, 1]^m\to [0, 1]^n$ such that $Y:=F((U_i)_{i=1}^m)$ has absolutely continuous law with respect to $Leb_n$. I know that the answer to this question is affirmative, because $([0, 1]^m, Leb_m)$ and $([0, 1]^n, Leb_n)$ are equivalent. That is, there is a measure preserving isomorphism between these two spaces. If $F$ is such a measure preserving isomorphism, then the Law of $Y$ is simply $n$-dimensional Lebesgue measure on $[0, 1]^n$.
However, I am interested in a s special kind of function $F$. That is, I want $$F((U_i)_{1\leq I\leq m})=(f(U_i, U_j)_{1\leq i<j\leq m}),$$ for some function $f:[0, 1]^2\to \mathbb{R}$. The question is does there exist such a function $F$ so that $F((U_i)_{1\leq I\leq m})$ is absolutely continuous with respect to $n$-dimensional Lebesgue measure. (This is where it becomes important that $n=m(m-1)/2$.)