Stable Unstable Manifolds and Morse Smale Transversality

Choose a smooth coordinate chart centered at $p$ and represent the vector field by $x^{\mathrm{\prime }}=Ax+R(x)$. Since $d{f}_p=0$, $A=-g_p^{-1}{\mathrm{Hess}}_{p}f$; in the $g_p$ inner product it is self-adjoint and nonsingular. By [F1] and [F3] it has a stable spectral space $E^s$ of dimension $d-\lambda (p)$, an unstable space $E^u$ of dimension $\lambda (p)$, orthogonal projections $P_s,P_u$, and $\lambda >0$ with $\mathrm{\parallel }{e}^{tA}{P}_s\mathrm{\parallel }\le e^{-\lambda t}$ for $t\ge 0$ and $\mathrm{\parallel }{e}^{tA}{P}_u\mathrm{\parallel }\le e^{\lambda t}$ for $t\le 0$. Moreover $R(0)=DR(0)=0$, and $R$ is $C^2$.

Choose a smooth cutoff equal to one on a small ball about zero and zero outside a slightly larger ball, and replace $R$ by the cut-off $R_{\delta }$. Since $DR(0)=0$, the cutoff radius $\delta$ can be chosen so that $R_{\delta }$ is globally Lipschitz with constant $\epsilon$ as small as desired; the derivative of the cutoff contributes at most $O({\sup }_{\mathrm{\mid }x\mathrm{\mid }\le 2\delta }\mathrm{\mid }R(x)\mathrm{\mid }\mathrm{/}\delta )=o(1)$. Fix $0<\beta <\lambda$ and take $\epsilon ((\lambda -\beta {)}^{-1}+(\lambda +\beta {)}^{-1})<1$. For $a\in E^s$ define on the weighted continuous-path Banach space $\mathrm{\parallel }z{\mathrm{\parallel }}_{\beta }={\sup }_{t\ge 0}{e}^{\beta t}\mathrm{\mid }z(t)\mathrm{\mid }$ the map ${\mathcal{T}}_{a}z(t)=e^{tA}a+{\int }_0^t{e}^{(t-s)A}{P}_s{R}_{\delta }(z(s))ds-{\int }_t^{\mathrm{\infty }}{e}^{(t-s)A}{P}_u{R}_{\delta }(z(s))ds\mathrm{.}$ The exponential bounds of step 1.1 make this a contraction with the displayed constant. Iteration from $e^{tA}a$ converges geometrically to its unique fixed point $z_a$, and $\mathrm{\parallel }{z}_a{\mathrm{\parallel }}_{\beta }\le \mathrm{\mid }a\mathrm{\mid }\mathrm{/}(1-\epsilon c_{\beta })$. For small $a$, the entire path stays where the cutoff equals one; differentiating the integral equation shows that it solves the actual $x^{\mathrm{\prime }}=X(x)$ and converges exponentially to $p$.

Its initial value is $z_a(0)=a+h(a)$ with $h(a)\in E^u$. The map $a\mapsto z_a$ is $C^1$ in the weighted-path norm. Indeed, $R_{\delta }$ is $C^2$ with bounded first and second derivatives on the chosen finite-dimensional support, so the integral map in step 2.1 is $C^1$ on the weighted path space; its path derivative has norm below one uniformly. Differentiating the fixed-point equation and summing the resulting Neumann series gives a continuous derivative in $a$. At $a=0$, $z_0=0$ and $D{R}_{\delta }(0)=0$, hence $Dh(0)=0$. Consequently $\{a+h(a):a\text{ small in }{E}^s\}$ is a $C^1$ embedded disk tangent to $E^s$ at $p$. Differentiating $z_a(t)$ with respect to $a$ also shows that every tangent variation to this disk is bounded by $C{e}^{-\beta t}$ along its forward orbit.

Conversely, any actual orbit that stays in a sufficiently small chart neighbourhood for all $t\ge 0$ and converges to $p$ satisfies the integral equation of step 2.1 with $a=P_{s}x(0)$: variation of constants determines the stable part, while boundedness forces the unstable terminal term to cancel. The same integral operator is a contraction in the ordinary bounded-path norm after making $\epsilon <\lambda \mathrm{/}2$, so that orbit equals $z_a$ and lies in the disk. The disk is therefore the local stable set, not just a selected family of decaying solutions. Time reversal gives the unstable disk, its dimension and tangent space, and exponential decay of its tangent variations as $t\to -\mathrm{\infty }$.

An orbit converging to $p$ eventually enters the local chart and stays there, so after some finite time it belongs to the local stable disk by step 4.1. The global stable set is the union of finite-time backward flow images of that disk. By [F2] each image is an immersed $C^1$ disk with the same dimension; their structures agree on overlaps by flow uniqueness. A tangent vector at any point transports by the derivative of a finite-time flow into a tangent vector of the local disk, whose variational solution decays exponentially by step 3.1. The same argument with reversed time gives the global unstable manifold and its tangent-decay assertion. No completeness outside the orbits under discussion is required.

The Morse--Smale condition says that the two immersion maps from $W^u(p)$ and $W^s(q)$ are transverse. Thus [F2] makes their fibre product a smooth manifold. Its evaluation image is precisely $W^u(p)\cap W^s(q)$, hence, by the point-marked convention, $\tilde{\mathcal{M}}(p,q)$.

First suppose $\lambda (p)<\lambda (q)$. The source dimensions in [F1] sum to $\lambda (p)+(n-\lambda (q))<n$, so [F2] makes the fibre product empty. If the fibre product is nonempty, [F2] instead gives its dimension as $\lambda (p)+(n-\lambda (q))-n=\lambda (p)-\lambda (q)\ge 0$. These cases prove the stated conditional dimension and emptiness assertions without assigning a negative dimension to an empty space.

Equality $s\cdot \gamma =\gamma$ says $\gamma (t+s)=\gamma (t)$ for every $t$, so if $s\mathrm{\ne }0$ then $\gamma$ is periodic with nonzero period $\mathrm{\mid }s\mathrm{\mid }$.

By [F2] the orbit is nonconstant, while [F1] makes $f\circ \gamma$ strictly decreasing; it therefore cannot be periodic. Hence $s=0$, which is exactly freeness.

Continuity and the endpoint limits give a time $t_0$ with $f(\gamma (t_0))=c$; strict decrease in [F1] makes $t_0$ unique. Translating by $t_0$ therefore gives exactly one representative in the stated slice.

Conversely, two slice representatives in one time orbit differ by a translation, and step 1.1 forces that translation to be zero. The slice lies in the embedded hypersurface of [F2], giving the asserted identification.

Put $P=\tilde{\mathcal{M}}(p,q)$. If $P=\mathrm{\varnothing }$, its orbit set $\mathcal{M}(p,q)$ is empty and has the empty smooth structure; no dimension is assigned by a negative formula. Assume henceforth that $P$ is nonempty. No point of a trajectory is critical: flow uniqueness would then make it constant, contradicting its distinct endpoint limits. In the transported immersion charts of [F3], the action $t\cdot \gamma =\gamma (\cdot +t)$ is smooth and evaluates to ${\mathrm{\Phi }}_t(\gamma (0))$. Its tangent vector $V$ on the fibre product evaluates to $X(\gamma (0))\mathrm{\ne }0$, so $V$ is nonzero. Hence the dimension of the nonempty smooth manifold $P$ is at least one. By [F1], this dimension is $\lambda (p)-\lambda (q)$; in particular an index difference at most zero forces $P$, and therefore its orbit set, to be empty.

In the nonempty case strict decrease and the endpoint limits imply $f(p)>f(q)$. Set $c=(f(p)+f(q))\mathrm{/}2$, without imposing any global regular-value condition on the ambient function $f$, and put $h=f\circ {\mathrm{ev}}_0:P\to \mathbb{R}$. This map is smooth by [F3]. For every $\gamma \in P$, the flow tangent vector of step 1.1 satisfies $d{h}_{\gamma }(V)=d{f}_{\gamma (0)}(X)<0$, so $h$ is a submersion. For each trajectory, continuity and its two endpoint limits give a finite time $t_0$ with $f(\gamma (t_0))=c$, and strict decrease makes $t_0$ unique. In particular $h^{-1}(c)$ is nonempty. Apply [F4] to $h$, whose scalar target has dimension one: this slice is an embedded codimension-one submanifold of $P$, of dimension $\lambda (p)-\lambda (q)-1\ge 0$.

Translating each trajectory by its unique time $t_0$ from step 2.1 gives exactly one representative in $h^{-1}(c)$. Conversely, two representatives in the same time orbit can both lie in that slice only if their translation is zero. This is the level-representative bijection, using only the continuity and strict-decrease part of [F2]. Put $S=h^{-1}(c)$. The same equation $h(t\cdot \gamma )=c$, now with $\gamma$ varying in a chart of the entire $P$, has nonzero time derivative at every solution. By [F5] its unique hitting time $\tau (\gamma )$ is smooth locally on $P$; uniqueness patches these local functions globally. It satisfies $\tau (s\cdot \gamma )=\tau (\gamma )-s$, so $r(\gamma )=\tau (\gamma )\cdot \gamma$ is a smooth orbit-invariant map $P\to S$ restricting to the identity on $S$. Let $q:P\to P\mathrm{/}\mathbb{R}$ carry the quotient topology [F6]. The induced inverse $\bar{r}:P\mathrm{/}\mathbb{R}\to S$ is continuous: for every open $U\subseteq S$, $q^{-1}({\bar{r}}^{-1}(U))=r^{-1}(U)$ is open, and [F6] applies. Conversely $q{\mathrm{\mid }}_S$ is continuous, and these two maps are inverses. Thus the slice bijection is a homeomorphism for the quotient topology. Transport its smooth structure to $\mathcal{M}(p,q)$; its underlying topology is precisely that quotient topology, and its dimension is the one in step 2.1. Any other $c^{\mathrm{\prime }}$ strictly between the endpoint values gives the same smooth structure: the equation $h(t\cdot \gamma )=c^{\mathrm{\prime }}$ has nonzero derivative in $t$, so [F5] makes the unique hitting time smooth locally on the first slice; uniqueness patches these local times, and the smooth flow action gives mutually inverse smooth change-of-slice maps. Thus the construction and the empty case prove the full statement without Sard's theorem, an ambient regular level, or an added choice hypothesis.

If $\mathcal{M}(p,q)$ were nonempty, [F1] would give it dimension $\lambda (p)-\lambda (q)-1\le -1$.

A nonempty smooth manifold has a nonnegative integer dimension, contradicting step 1.1. Thus the moduli space is empty.

Use an equivalent norm from the inner product in the statement. By [F1], ${\mathbb{R}}^d=E^s\oplus E^u$ with spectral projections $P_s,P_u$ and some $\lambda >0$ such that $\mathrm{\parallel }{e}^{tL}{P}_s\mathrm{\parallel }\le e^{-\lambda t}$ for $t\ge 0$ and $\mathrm{\parallel }{e}^{tL}{P}_u\mathrm{\parallel }\le e^{\lambda t}$ for $t\le 0$. For the constant path $L$ define $(R_{L}v)(t)={\int }_0^t{e}^{(t-s)L}{P}_{s}v(s)ds-{\int }_t^{\mathrm{\infty }}{e}^{(t-s)L}{P}_{u}v(s)ds\mathrm{.}$ Both integrals converge. Their norms are bounded by $2{\lambda }^{-1}\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}$; splitting the integrals into a compact initial interval and a tail where $v$ is small shows $R_{L}v(t)\to 0$. Differentiation gives $(R_{L}v{)}^{\mathrm{\prime }}=L{R}_{L}v+v$, so the derivative also tends to zero. Thus $R_L:C_0^0\to C_0^1$ is bounded and $F_L{R}_L=I$.

Write $B(t)=A(t)-L$. If $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}$ is sufficiently small, $F_A{R}_L=I-B{R}_L$ on $C_0^0$, and the Neumann series gives the bounded right inverse $R_A:=R_L(I-B{R}_L{)}^{-1}$. It tends to $R_L$ in operator norm as $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$.

For this small-perturbation case choose $0<\beta <\lambda$ and, for $a\in E^s$, consider the integral equation $u(t)=e^{tL}a+{\int }_0^t{e}^{(t-s)L}{P}_{s}B(s)u(s)ds-{\int }_t^{\mathrm{\infty }}{e}^{(t-s)L}{P}_{u}B(s)u(s)ds\mathrm{.}$ On the Banach space of continuous paths with ${\sup }_{t\ge 0}{e}^{\beta t}\mathrm{\mid }u(t)\mathrm{\mid }<\mathrm{\infty }$, the integral operator has norm at most $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}((\lambda -\beta {)}^{-1}+(\lambda +\beta {)}^{-1})$. Shrink the permitted $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}$ so this is below one. Iterating from $e^{tL}a$ converges geometrically to a unique solution $u_a$; differentiating the equation gives $u_a^{\mathrm{\prime }}=A{u}_a$. Its initial value is $a+h_A(a)$, where $h_A(a)\in E^u$ and $\mathrm{\parallel }{h}_A\mathrm{\parallel }\to 0$ as $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Conversely, variation of constants for any homogeneous solution tending to zero gives this integral equation with $a=P_{s}u(0)$: the unwanted unstable term is fixed by its terminal value at infinity. Uniqueness also holds in the unweighted bounded-path norm under the same smallness condition. Hence $S_A$ is exactly the graph of $h_A:E^s\to E^u$, has dimension $\dim E^s$, and all its solutions decay at least as $e^{-\beta t}$.

Choose a smooth scalar $\rho$ supported in $(0,\mathrm{\infty })$ with integral one. For each $a\in E^u$ put $v_a(t)=-\rho (t)e^{tL}a$. Then $v_a(0)=0$ and step 1.1 gives $R_L{v}_a(0)=a$. For a basis of $E^u$, step 2.1 makes the vectors $R_A{v}_a(0)$ close to that basis, while step 3.1 makes $S_A$ a graph close to $E^s$. Thus these vectors together with $S_A$ span ${\mathbb{R}}^d$ for sufficiently small $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}$; this follows by invertibility of the finite matrix close to the identity in the fixed splitting $E^s\oplus E^u$.

For general $A(t)\to L$, choose $T$ so large that the shifted path $A_T(s)=A(T+s)$ meets the smallness bounds of steps 2.1–4.1. Apply those steps on $[T,\mathrm{\infty })$. For $v\in C_0^0([0,\mathrm{\infty }))$, set the solution's value at $T$ to $R_{A_T}(v(T+\cdot ))(0)$ and solve $u^{\mathrm{\prime }}=Au+v$ backwards on $[0,T]$ by [F2]. This constructs a bounded linear right inverse $R_A:C_0^0\to C_0^1$; boundedness on the finite interval follows from the variation-of-constants formula and the finite maximum of $\mathrm{\parallel }A(t)\mathrm{\parallel }$ there. Homogeneous evolution carries $S_A$ isomorphically onto $S_{A_T}$, so $\dim S_A=\dim E^s$ and exponential decay persists. To obtain the spanning assertion at $0$, choose $w$ on the shifted tail with $w(0)=0$ and extend it by zero to $[0,T]$. The right inverse so constructed has $R_{A}v(0)={\mathrm{\Phi }}_A(0,T)R_{A_T}w(0)$, while $S_A={\mathrm{\Phi }}_A(0,T)S_{A_T}$; step 4.1 then spans all initial values. Finally $t\mapsto -t$ changes $u^{\mathrm{\prime }}-A(t)u=v$ on a negative half-line into $z^{\mathrm{\prime }}-(-A(-s))z=-v(-s)$ on a positive half-line; negative eigenvalues of $-A(-\mathrm{\infty })$ are positive eigenvalues of $A(-\mathrm{\infty })$.

The convergence of $A(t)$ at both ends makes it bounded, so $D_A$ is a bounded map from the indicated supremum $C^1$ norm to the supremum $C^0$ norm. Apply [F1] to $A{\mathrm{\mid }}_{[0,\mathrm{\infty })}$ and, after time reversal, to $A{\mathrm{\mid }}_{(-\mathrm{\infty },0]}$. Denote the bounded right inverses by $R_{+},R_{-}$. The two homogeneous initial-value spaces are $E_{+}^s,E_{-}^u$, with $\dim E_{+}^s=d-\dim E^u(A_{+})$ and $\dim E_{-}^u=\dim E^u(A_{-})$.

A homogeneous whole-line solution is determined by its value at zero by [F2] and belongs to $C_0^1$ exactly when that value lies in both $E_{-}^u$ and $E_{+}^s$. These spaces are finite-dimensional, hence closed, and the evaluation map gives $\ker D_A\cong E_{-}^u\cap E_{+}^s$.

For $v\in C_0^0(\mathbb{R},{\mathbb{R}}^d)$, write $v_{\pm }$ for its half-line restrictions and define the bounded linear map $T(v)=[R_{+}{v}_{+}(0)-R_{-}{v}_{-}(0)]\in {\mathbb{R}}^d\mathrm{/}(E_{-}^u+E_{+}^s)\mathrm{.}$ Every decaying solution on the positive half-line has the form $R_{+}{v}_{+}+u_{+}$, with $u_{+}(0)\in E_{+}^s$; the analogous form on the negative half-line has $u_{-}(0)\in E_{-}^u$. The two half-line solutions can be matched at zero exactly when $T(v)=0$. When matched, their first derivatives also agree there because each satisfies $u^{\mathrm{\prime }}=Au+v$ and $v$ is continuous. Thus $\mathrm{ran}{D}_A=\ker T$, which is closed.

The map $T$ is onto. By [F1], $E_{+}^s$ together with $\{R_{+}w(0):w(0)=0\}$ spans ${\mathbb{R}}^d$. Extend any such $w$ by zero to the negative half-line; the extension is continuous at zero, belongs to $C_0^0(\mathbb{R})$, and has $R_{-}{v}_{-}(0)=0$. Its $T$-images therefore span the quotient by $E_{-}^u+E_{+}^s$. Consequently $C_0^0\mathrm{/}\mathrm{ran}{D}_A\cong {\mathbb{R}}^d\mathrm{/}(E_{-}^u+E_{+}^s)$, a finite-dimensional space.

Set $a=\dim E_{-}^u$, $b=\dim E_{+}^s$, and $c=\dim (E_{-}^u\cap E_{+}^s)$. Steps 2.1 and 3.1 give $\dim \ker D_A=c$ and $\mathrm{codim}\mathrm{ran}{D}_A=d-(a+b-c)$. Hence $D_A$ is Fredholm and $\mathrm{ind}{D}_A=c-[d-a-b+c]=a+b-d=\dim E^u(A_{-})-\dim E^u(A_{+})$ by step 1.1. The quotient in step 3.1 vanishes exactly when the two initial-value spaces span ${\mathbb{R}}^d$.

Differentiating $\dot{\gamma }+{\mathrm{grad}}_{g}f(\gamma )=0$ in a decaying variation gives the displayed operator by [F1]. Since $\gamma$ converges to the critical points and $\mathrm{\nabla }{\mathrm{grad}}_{g}f$ is continuous, its coefficient along $\gamma$ is bounded; hence $D_{\gamma }:E_{\gamma }\to F_{\gamma }$ is bounded.

By [L2], $\gamma (t)$ approaches $p$ and $q$ exponentially at the two ends. Parallel transport a $g$-orthonormal frame along $\gamma$. In fixed endpoint coordinates its transport equation has coefficients bounded by a constant times $\mathrm{\mid }\dot{\gamma }(t)\mathrm{\mid }$, which are integrable on both tails; hence the frame converges at each end to an orthonormal frame. In this frame, ${\mathrm{\nabla }}_t$ is ordinary differentiation, so $D_{\gamma }$ becomes $z^{\mathrm{\prime }}-A(t)z$, with $A(t)$ converging to $-g_p^{-1}{\mathrm{Hess}}_{p}f$ at $-\mathrm{\infty }$ and $-g_q^{-1}{\mathrm{Hess}}_{q}f$ at $+\mathrm{\infty }$. These limits are self-adjoint for the endpoint inner products and invertible because $f$ is Morse. Parallel transport preserves norms and takes ${\mathrm{\nabla }}_t\xi$ to $z^{\mathrm{\prime }}$, so it identifies exactly the displayed $C_0^1$ and $C_0^0$ spaces, not a weighted Sobolev replacement.

Apply [L1] to step 2.1. Its negative-end decaying initial-value space has dimension $\lambda (p)$, and its positive-end space has dimension $\dim M-\lambda (q)$; therefore $D_{\gamma }$ is Fredholm of index $\lambda (p)-\lambda (q)$. By [L2], differentiating the flow along $W^u(p)$ gives decaying negative-end variational solutions, and differentiating along $W^s(q)$ gives decaying positive-end ones. Both tangent spaces have exactly the corresponding dimensions, so inclusion implies equality with the two ODE initial-value spaces at $\gamma (0)$. The cokernel is consequently isomorphic, through the chosen half-line right inverses, to $T_{\gamma (0)}M\mathrm{/}(T_{\gamma (0)}{W}^u(p)+T_{\gamma (0)}{W}^s(q))$.

The quotient in step 3.1 vanishes exactly when the two tangent spaces span $T_{\gamma (0)}M$, which is transversality. By [L1], that quotient vanishes exactly when $D_{\gamma }$ is onto. This proves the biconditional and the index assertion.

If $M$ is empty, the empty family works. Otherwise fix $x\in M$ and use [F1] to obtain $W$, a source coordinate diffeomorphism $T:W\to U\times A$, and a target chart in which $P(T^{-1}(u,v))=(u,g(u,v))$. Write $T(x)=(u_x,v_x)$. Choose radii $a,b>0$ small enough that the closed balls ${\bar{B}}_a(u_x)\subset U$ and ${\bar{B}}_b(v_x)\subset A$. Set $C_x:=T^{-1}({\bar{B}}_a(u_x)\times {\bar{B}}_b(v_x))\mathrm{.}$ Then $C_x$ is closed relative to $W$, and its interior in $M$ contains $x$.

The restriction $P{\mathrm{\mid }}_{C_x}:C_x\to V$ is proper. To see this, let $K\subset V$ be compact and take a sequence $(z_i)$ in $C_x\cap P^{-1}(K)$. After a subsequence $P(z_i)$ converges in $K$; in target coordinates its first components $u_i$ therefore converge to some $u\in {\bar{B}}_a(u_x)$. The $v_i$ lie in the compact finite-dimensional ball ${\bar{B}}_b(v_x)$, so a further subsequence converges to $v\in {\bar{B}}_b(v_x)$. Continuity of $T^{-1}$ and $P$ gives $z_i\to T^{-1}(u,v)\in C_x\cap P^{-1}(K)$. This subset is metrizable through $T$, so sequential compactness implies compactness. The argument uses no compactness of a ball in the Banach range coordinate.

The interiors of the $C_x$ form an open cover of $M$. By [F2], under AC every open cover has a countable subcover: for each member of a countable base that is contained in some cover member, choose one such member. Keep the corresponding countably many $C_x$, and index them by a subset of $\mathbb{N}$. Their interiors still cover $M$, and each retains the properties proved in steps 1.1–2.1.

In normal-form coordinates, $DP$ is onto exactly when $D_{v}g$ is onto. Surjectivity of a linear map between the finite-dimensional spaces $K,Q$ is an open condition on its matrix entries. Thus $A:=C\cap \mathrm{Crit}(P)$ is closed in $C$.

The image $B:=P(A)$ is closed in $V$. Indeed, if $y_i\in B$ converges to $y\in V$, choose $x_i\in A$ with $P(x_i)=y_i$. The set consisting of $y$ and the $y_i$ is compact in the metrizable chart $V$. Properness makes its inverse image in $C$ compact. A subsequence of $x_i$ therefore converges in $C$ to $x\in A$, and continuity gives $P(x)=y$. Sequential closedness equals closedness in the metrizable chart.

If $Q=\{0\}$, every $D_{v}g$ is onto and $B=\mathrm{\varnothing }$. Otherwise suppose a nonempty target-coordinate open rectangle $U_0\times Q_0$ lay inside $B$. Fix $u\in U_0$. For the $C^h$ map $g_u:v\mapsto g(u,v)$ on the open finite-dimensional kernel-coordinate domain, every $q\in Q_0$ would be a critical value: each $(u,q)\in B$ comes from some $(u,v)\in C$ with $D_{v}g(u,v)$ not onto. If $h=\mathrm{\infty }$, choose any finite integer $r>\max \{\dim K-\dim Q,0\}$; otherwise take $r=h$. Then [F1] applies to the $C^r$ slice and says its critical values are null, so they cannot contain the nonempty open set $Q_0$. Thus $B$ has empty interior in $V$, and step 2.1 makes it nowhere dense there.