Benedetti and Risler's "Real algebraic and Semi-algebraic sets" book on Semialgebraic Geometry has the following theorem:
Theorem 2.6.14 Let $f:V \to Y$ be a continuous semialgebraic mapping defined on the compact semialgebraic set $V \subset \mathbb{R}^{n}$ onto $Y \subset \mathbb{R}^{m}$. Consider its graph $X \subset \mathbb{R}^{m+n}$. Set $s = m+n$, let $(x_{1},...,x_{m},y_{1},...,y_{n}) = (x,y)$ be the coordinates of $\mathbb{R}^{s}$ with
$$\mathbb{R}^{s} \supseteq \mathbb{R}^{s-1} = \{y_{n}=0\} \supseteq \mathbb{R}^{s-2} = \{y_{n-1} = y_{n} = 0\} \supseteq \cdots \supseteq \mathbb{R}^{m}$$
and let $\pi_{j}: \mathbb{R}^{j+1} \to \mathbb{R}^{j}$, for $j=s-1,...,m$. Assume that, for every $j$, all the polynomials involved in some representation of $X_{j} = \pi_{j} \circ \pi_{j+1} \circ ... \circ \pi_{s-1}(X) (X_{s} = X, X_{m} = Y)$ have constant leading coefficients with respect to the indeterminate $Y_{j}$. Then there exists a triangulation $G:Y \to |H|$, a triangulation $F:X \to |K|$ and a simplicial map $f':|K| \to |H|$ such that
$F$ and $G$ satisfy the conditions of the Triangulation Theorem
$G \circ f = f' \circ F$.
There are some questions here. The first is, $f$ is defined on $V$, and not on $X$. So either the $f$ on item 2 is the natural projection, or the triangulation is of $V$, not $X$. In the first case, the theorem is immediate, since $X$ is compact (because $V$ is), and thus there is a triangulation of $X$ that is compatible with the projections. In particular, the projections of the vertices of the polyhedron have vertices as images, and thus the natural projection is simplicial. But is doesn't seem to be the case here, because then we wouldn't need the hypothesis about the polynomials involved in some representation of $X$.
I can't understand why this hypothesis is necessary, and how it could be used in a proof. The book does not prove it. Instead, it says right after proving the Triangulation Theorem for compact semialgebraic sets that "by the way, we actually proved this theorem" and states it.
The book even has a example to show why it is necessary, which I'll write here(by the way, all simplices considered in this question are open):
"Set $V = \mathbb{S}^{2}$, $S = V \cap \{x=0\}, f:V \to \mathbb{R}^{4}, (x,y,z) \mapsto x^{2}(x,y,z,1), W=f(V)$. Note that
(a) $f$ is an analytic isomorphism of $V\setminus S$ onto $W\setminus \{0\}$ (b) $f(S) = 0$
That is, $f$ "collapses" $S$ onto the point $0 \in \mathbb{R}^{4}$. We claim that $f$ cannot be made simplicial as in Theorem 2.6.14.
In fact, any such triangulation of $V$ should have $S$ as a subpolyhedron, and there would exist at least one $2$-simplex $T$ of $V$ with exactly one $1$-face $T'$ contained in $S$. The image of $T$ via $f$ cannot be a simplex of $W$"
As for why the image of $T$ cannot be a simplex, let $<a,b>$ be the $T'$, and $v$ the other vertex of $T$. Since $f$ collapses $S$ onto $0$, $f(a) = f(b)$, but $f(T)$ has dimension $2$, thus the resulting set is not a simplex.
I appreciate any help on the understanding of this theorem. I'll put the Triangulation Theorem for compact semialgebraic subsets for reference:
Triangulation Theorem: Let $X$ be a compact semialgebraic subset of $\mathbb{R}^{n}$, and $X_{1},...,X_{h}$ a finite semialgebraic partition of $X$. There exists a triangulation $F:X \to |K|$ such that:
- Each $X_{i}$ equals the union of some $F^{-1}(T_{j})$, where $T_{j}$ are simplices of $|K|$.
- The collection of sets of the form $F^{-1}(T_{j})$ is a stratification of $X$
- The restriction of $F$ to $F^{-1}(T_{j})$ is a real analytic isomorphism of analytic manifolds.
