Fiber cutting

We now want to obtain a description of the projections of bounded semi-$\C$-sets. By this corollary, we only need consider projections of trivial semi-$\C$-sets, that is, of trivial $\C$-sets. We start with a few remarks about $\C$-manifolds: let $M \subseteq \RR^n$ be a nonempty $\C$-manifold of dimension $m$.

First, let $r$ be a polyradius associated to $M$ as in the definition of $\C$-manifold (with $n-m$ in place of $k$ there). Let $g \in \C_r$ be such that $g$ is strictly positive on $M$ and identically $0$ on $\fr(M)$. The following lemma generalizes the identity theorem for real analytic functions:

Identity Lemma
Let $C$ be a connected component of $M$, and assume there exists an open subset $U$ of $C$ such that $g \rest{U}$ is identically $0$. Then $g \rest{C}$ is identically $0$.

Proof. By Assumption (C3), and after translating and shrinking $r$ if necessary, we may assume that $0 \in U$. The definition of $\C$-manifold and the assumption that the system $\C$ is closed under composition imply that we may even assume (after replacing $n$ with $m$ and again shrinking $r$ if necessary) that $C = B(r)$; in particular, $U$ is an open neighbourhood of $0$. Since $g$ is $C^\infty$, it follows that $Tg = 0$, so $g = 0$ by Assumption (C2), as required. $\qed$

A point $a \in M$ is a critical point of $g \rest{M}$ if the gradient $\nabla g(a)$ of $g$ at $a$ is orthogonal to $T_aM$.


  1. Show that the set $\cr(g\rest{M})$ of critical points of $g \rest{M}$ is a $\C$-set.

Recall from this corollary that every $C$-set has dimension.

Assume that $m>0$ and $g$ is strictly positive on $M$ and $0$ on $\fr(M)$. Then $\dim \cr(g \rest{M}) < m$.

Proof. If $\dim \cr(g \rest{M}) = m$, then $\cr(g\rest{M})$ contains an open subset $U$ of $M$. Without loss of generality, we may assume that $U$ is connected, and we let $C$ be the connected component of $M$ containing $U$. The map $h:M \into \RR^n$ defined by $$h(x):= \Pi_{T_xM}(\nabla g(x))$$ belongs to $(\C_r)^n$ and is equal to $0$ on $U$. So, by the Identity Lemma, $h = 0$ on all of $C$, which means that $g \rest{C}$ is constant.

However, $\dim C > 0$ implies that $\fr(C) \ne \emptyset$. Since $g$ is strictly positive on $C$ and $0$ on $\fr(C)$, it follows that $g\rest{C}$ cannot be constant, a contradiction. $\qed$

We need to apply this corollary to the fibers of $M$ over various projections.


  1. Given a strictly increasing $\iota:\{1, \dots, m\} \into \{1, \dots, n\}$, show that the set $$M_\iota:= \set{x \in M:\ \Pi_\iota\rest{T_xM} \text{ has rank } m}$$ is an open $\C$-subset of $M$ (and hence a $\C$-manifold).
  2. Show that $M = \bigcup_{\iota} M_\iota$, where $\iota$ ranges over all strictly increasing maps $\{1, \dots, m\} \into \{1, \dots, n\}$.

For $k \leq n$ and strictly increasing $\iota:\{1, \dots, m\} \into \{1, \dots, n\}$, we let $m(k,\iota) \in \{0, \dots, m\}$ be maximal such that $\iota(m(k,\iota)) \leq k$, and we put $$\iota_{k}:= \iota \rest{\{1, \dots, m(k,\iota)\}} \quad\text{ and }\quad \iota^{k}:= \iota
\rest{\{m(k,\iota)+1, \dots, m\}}.$$

Let $k \leq n$ and $\iota:\{1, \dots, m\} \into \{1, \dots, n\}$ be strictly increasing, and assume that $M = M_{\iota}$ and $\Pi_{k}\rest{M}$ has constant rank $\mu:= m(k,\iota)$. Let $a \in \Pi_k(M)$.

  1. Show that the fiber $M_a := \Pi_{k}^{-1}(a) \cap M$ is a $\C$-manifold of dimension $m – \mu$, and that $\Pi_{\iota^{k}}\rest{M_a}$ is an immersion.
  2. Show that, for every connected component $C$ of $M_a$, the projection $\Pi_{\iota^{k}} (C)$ is open and $\fr C \neq \emptyset$ if $\mu < m$.

Fiber Cutting Lemma
Let $k < n$ and $\iota:\{1, \dots, m\} \into \{1, \dots, n\}$ be strictly increasing, and assume that $M = M_{\iota}$ and $\Pi_{k}\rest{M}$ has constant rank $\mu:= m(k,\iota) \lt m$. Then there is a $\C$-set $A \subseteq M$ such that $\dim A \lt m$, $\Pi_{k}(M) = \Pi_{k}(A)$ and, for $a \in \Pi_k(M)$, the fiber $A_a$ intersects every connected component of $M_a$.

Proof. Let $r$ be a polyradius associated to $M$ as in the definition of $\C$-manifold (with $n-m$ in place of $k$ there), and let $f, g_1, \dots, g_k \in \C_{r}$ be such that $$M = \set{x \in B(r):\ f(x) = 0, g_1(x) \gt 0, \dots, g_k(x) \gt 0}.$$ Let $g$ be the product of all $g_j$, $j=1, \dots, k$, and all $(x_i+r_i)\rest{B(r)}$ and $(r_i-x_i)\rest{B(r)}$, $i=1, \dots, n$. Then $g \in \C_{r}$, is strictly positive on all of $M$ and—by continuity of $f$ and $g$ in $\bar B(r)$—identically zero on $\fr M$. We now set $$A:= \set{x \in M:\ x \text{ is a critical point of } g\rest{M_a} \text{ and } a = \Pi_k(x)};$$ similarly to Exercise 1 above, we see that $A$ is a $\C$-set. Since, for $a \in \Pi_k(M)$, we have that $A_a = \cr(g\rest{M_a})$, it follows from Exercise 4 and the corollary above that $\dim A \lt m$. Finally, since $\mu \lt m$, Exercise 5 above implies that, for $a \in \Pi_k(M)$ and each connected component $C$ of $M_a$, the map $g\rest{C}$ has critical points, so the lemma is proved. $\qed$

In general, for $k \leq n$, the restriction of $\Pi_k$ to $M$ does not have constant rank. Therefore, for any manifold $N \subseteq \RR^n$, we let $r(N,k)$ be the maximal rank of $\Pi_k \rest{T_xN}$, as $x$ ranges over $N$. Note that $r(N,k) \le m$ and, if $\iota:\{1, \dots, m\} \into \{1, \dots, n\}$ is strictly increasing and $N = N_{\iota}$, then $r(N,k) \ge m(k,\iota)$.

More generally, if $A \subseteq \RR^n$ is a $\C$-set and $k \le n$, we set $$r(A,k):= \max\{r(N,k):\ N \subseteq \RR^q \text{ is a submanifold of } A\}.$$ Since $A$ has dimension, it follows that $r(A,k) \le \dim A$.


  1. Show that, if $A$ is a manifold, then either definition of $r(A,k)$ yields the same value.
  2. Let $A \subseteq \RR^n$ be a $\C$-set, $k \le n$ and $N \subseteq \RR^q$, with $q \ge n$, be a manifold such that $\Pi_n(N)$ is a manifold and $\Pi_n\rest{N}: N \into \Pi_n(N)$ is a diffeomorphism. Show that $r(N,k) \le r(A,k)$.

The fiber cutting lemma can be used to obtain any projection of any bounded semi-$\C$-set as a finite union of immersions of trivial semi-$\C$-sets:

Step 2
Let $A \subseteq \RR^{n}$ be a bounded semi-$\C$-set and $k\leq n$. Then there are $l \in \NN$, $n_i \ge n$ and trivial semi-$\C$-sets $N_i \subseteq \RR^{n_i}$, for $i=1, \dots, l$, such that $$\Pi_{k}(A) = \Pi_{k}(N_1) \cup \dots \cup \Pi_{k}(N_l)$$ and, for each $i$, we have $d_i:= \dim(N_i) \leq k$ and there is a strictly increasing $\iota_i:\{1, \dots, d_i\} \into \{1, \dots, k\}$ such that $\Pi_{\iota_i}\rest{N_i}: N_i \longrightarrow \RR^{d_i}$ is an immersion.

Proof. We proceed by induction on the pair $(m,r) \in \NN^2$, considered with the lexicographic ordering, where $m:= \dim A$ and $r:= r(A,k)$. If $m = 0$, then $A$ is finite by this corollary. So we assume $m > 0$ and the proposition holds for all bounded semi-$\C$-sets $B$ such that $(\dim B,r(B,k)) \lt (m,r)$.

By this corollary again, we may assume that $A$ is a trivial $\C$-set, hence a $\C$-manifold. Replacing $A$ with $A_\iota$, where $\iota:\{1, \dots, m\} \into \{1, \dots, n\}$ is strictly increasing, we may assume that $A = A_\iota$. Now, we set $\mu:= m(k,\iota) = m$ and note that $\mu \le r \le m$. If $\mu = r$, then we are done, so we assume $\mu \lt r$. We set $$A_1:= \set{x \in A:\ \Pi_k\rest{T_xA} \text{ has rank } r}$$ and $A_2:= A \setminus A_1$; note that both $A_1$ and $A_2$ are $\C$-sets.

The set $A_1$ is an open subset of $A$, hence a manifold, and $\Pi_k\rest{A_1}$ has constant rank $r$; the only remaining problem is that $A_1$ may not be trivial. However, by this corollary again, and by Exercise 7 above and the inductive hypothesis, the proposition follows with $A_1$ in place of $A$.

The set $A_2$, on the other hand, satisfies $r(A_2,k) \lt r$, so the proposition with $A_2$ in place of $A$ follows from the inductive hypothesis. $\qed$

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