Mnev's Universality Type Theorem In order to state properly Mnev's universality type theorems, one has to understand the definition of stable equivalence. I have some questions to the definition.
Here is the definition as in Oriented Matroids from Björner et.al.

Let $V \subseteq \mathbb{R}^n$ and $W \subseteq \mathbb{R}^{n+d}$ be
semi-algebraic sets with $\pi(W) = V$, where 
$\pi : \mathbb{R}^{n+d} \rightarrow \mathbb{R}^n$ is the canonical
projection that deletes the last $d$ coordinates. 
$V$ is a stable projection of $W$ if $W$
has the form
$$W = \{(v,v') \in \mathbb{R}^{n+d} : v \in V,\ \phi_i(v) \cdot v' > 0; \
\psi_j(v) \cdot v' = 0 \textrm{ for } i \in X; j \in Y \} .$$
Here $X$ and $Y$ denote finite (possibly empty) index sets. 
For $i \in X$ and $j \in Y$
the functions $\phi_i$ and $\psi_j$ have to be polynomial functions
$$ \phi_i= ( \phi_i^1 , . . . , \phi_i^d ) : \mathbb{R}^n \rightarrow (\mathbb{R}^d)^* \text{ with } \phi_i^k \in \mathbb{Z}[x_i, \ldots , x_n] \quad \mbox{and}$$
$$\psi_j = (\psi_j^1,\ldots,\psi_j^d) :\mathbb{R}^n \rightarrow (\mathbb{R}^d)^*
\text{ with } \psi_i^k \in \mathbb{Z}[x_i, \ldots , x_n],$$
that associate to every element of $\mathbb{R}^n$ a linear functional on $\mathbb{R}^d$. Two semialgebraic
sets $V$ and $W$ are rationally equivalent
if there exists a homeomorphism
$f : V \rightarrow W$ 
such that both $f$ and $f^{-1}$ are given by rational functions.
Two semialgebraic sets $V$ and $W$ are stably equivalent if 
they are in the same
equivalence class with respect to the equivalence 
relation generated by stable projections and rational equivalence.

Question 1:
What is exactly meant by homeomorphism in this case?
What is exactly meant by rational function in this case?

Question 2:
OK, to unwrap and really understand the concept, how
can I show that the following two sets are
stably-equivalent?
$S = \{(x,y,z)\in \mathbb{R}^3 : x y = z\}$
$T = \{(x,y,z,a)\in \mathbb{R}^4 : x y = z ; a = (x+y)^2\}$.

Question 3:
Intuitively the following two sets should be stably
equivalent: $S\subset \mathbb{R}^n$ and 
$S'= \{(x,1)\in \mathbb{R}^{n+1} : x\in S\}.$
I don't see how to show this.

Question 4:
Are the following two sets stably-equivalent?
$S = \{x\in \mathbb{R} : x>0\}$
$T = \{(x,y)\in \mathbb{R}^2 : x y^2 - 1 = 0\}.$

thanks Till
 A: Ok, it seems I can answer the first question.
In this context, it seems the following is meant:
A homeomorphism is a function that is continuous and is invertible and
has a continuous inverse.
A rational function is a function that can be written as ratio of
polynomials, i.e. f(x_1,...,x_n) = p(x_1,...,x_n)/q(x_1,....,x_n)
where p and q are polynomials.
A: I found an answer to the third question.
Consider the sets $S\subset \mathbb{R}^n,$
$$S'= \{(x,0)\in \mathbb{R}^{n+1} : x\in S\}\quad \text{and}$$
$$S''= \{(x,1)\in \mathbb{R}^{n+1} : x\in S\}.$$
In order to show that $S''$ and $S$ are stably equivalent, we will show that
(1) $S'$ and $S''$ are rationally equivalent. Furthermore, we will show that 
(2) there is a stable projection from $S'$ to $S$. 
We start with (1). The homeomorphism $\phi : S'\rightarrow S''$ maps 
$(x,\ldots,x_n,t) \mapsto  (x,\ldots,x_n,t+1)$. 
It is easy to see that $\phi$ and its inverse are a rational functions.
This shows (1).
For (2), we define $\psi_1$ as the polynomial that is everywhere constant to $1$.
It holds that 
$$S' = \{(x_1,\ldots,x_n,t) \in \mathbb{R}^n : (x_1,\ldots,x_n) \in S \text{ and } \psi(x_1,\ldots,x_n) t = 0\} =$$
$$ = \{(x_1,\ldots,x_n,t) \in \mathbb{R}^n : (x_1,\ldots,x_n) \in S \text{ and } 
1\cdot  t = 0\} =$$
$$ = \{(x_1,\ldots,x_n,0) \in \mathbb{R}^n : (x_1,\ldots,x_n) \in S \}.$$
This finishes the proof and answers my third question.
A: To show that 
$$S = \{(x,y,z)\in \mathbb{R}^3 : x y = z\}$$
and 
$$T = \{(x,y,z,a)\in \mathbb{R}^4 : x y = z ; a = (x+y)^2\}$$
are stably equivalent, we show that they are in fact rationally equivalent.
We define $f : S\rightarrow T$ as follows
$$f(x,y,z) = (x,y,z,(x+y)^2).$$
Note that its inverse is
$$g(x,y,z,a) = (x,y,z).$$
Both functions are bijective, continuous and rational.
Thanks Michael Dobbins for clarifying this for me.
A: OK, Let me also answer Question 4. 
It may be a little silly, but the two sets are not stably-equivalent as they have a different number of connected components as can be seen in the picture below.

