I have a semi-algebraic set defined by the equality constraints $\{f_i (x_0, \dots, x_n) = 0\}_i$ and inequality constraints $\{g_j(x_0, \dots, x_n) \leq 0\}_j$ for real polynomials $f_i, g_j \in \Bbb R[x_0,\dots,x_n]$. I transform all of the points in the locus of the constraints linearly with a matrix $T$. By Tarski-Seidenberg, the resulting set should also be semi-algebraic. My goal is to derive the semi-algebraic mapping that maps the initial set of constraints to the final set of constraints.
This can be done rather easily for the case where $T$ is invertible, you just apply the algebra homomorphism $x_i \rightarrow (T^{-1}x)_i$. However, I have a special case where the matrix is not full rank, specifically for $n=4$ :
$$ T = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$
Here's what I got so far:
Any linear map can be expressed as a composition of invertible linear maps and projections, and any projection can be expressed as a chain of rank $n-1$ projections. So tackling rank n-1 projections should be sufficient. I was given some intuition that what I should be looking towards is the discriminants of the polynomials. For example the matrix above maps $(a_0,a_1,a_2,a_3) \rightarrow (a_0,a_1,a_2,0)$, if $(a_0,a_1,a_2,0)$ is in the image, $f_i(a_0,a_1,a_2,x)=0$ should have a solution in x. However, I'm not very experienced in this field so I don't know how to proceed with this line and/or tackle the inequalities.
I could not make the line of approach with discriminants work but what seems to be working right now is that, under the projection map given above, eliminating/solving for $x_3$ and then adding $x_3=0$ to the constraints gives the correct set of constraints for the new semi-algebraic set. Any other non full rank linear map can be decomposed into a series of such projections and invertible matrices using the Singular Value or Smith decompositions.