For any $n\ge1$, let us equip the power set of the additive group $(\Bbb{Q}^{n},+)$ with the product, or pointwise, topology induced by the natural bijection between $\mathcal{Pow}(\Bbb{Q}^{n})$ and $2^{\Bbb{Q}^{n}}$.
It's well-know that $GL_{n}(\Bbb{Q})$ acts in a Borel (i.e. measurable) way on the space $S(\Bbb{Q}^{n})$ of all subgroups of $\Bbb{Q}^{n}$.
By restriction, $SL_{n}(\Bbb{Z})$ acts as a Borel (i.e.measurable) map on a Borel subset $X_{n}$ of $S(\Bbb{Q}^{n})$ whose elements only have the trivial automorphisms $a\mapsto a, a\mapsto -a$.
Suppose that for $m\le n$ we have defined a Borel map $f\colon X_{n+1}\to X_m$ s.t. whenever $x,y\in X_{n+1}$ belong to the same $SL_{n+1}(\Bbb{Z})$-orbit, their images belong to the same $GL_m(\Bbb{Q})$-orbit.
If $I_m$ denotes the identity matrix of order $m$ , let us consider the well-defined map $\alpha$ from $SL_{n+1}(\Bbb{Z})\times X_{n+1}$ to the quotient group $GL_m(\Bbb{Q})/\{\pm I_m\}$ assigning to each pair $(g,x)$ the unique (in the quotient group) $h$ s.t. $f(g\cdot x)=h\cdot f(x)$.
Finally my question: how do I show the map $\alpha$ is Borel?