Spheres of adjacent critical levels have product neighbourhoods

If $n=0$, the compact zero-manifold is finite and every point is critical; regularity therefore gives $f^{-1}(v)=\mathrm{\varnothing }$. The field is zero, every trajectory is constant, and all crossing sets are empty, so all three assertions hold with the stated empty-map convention. For the rest of the proof assume $n\ge 1$. For each $q\in Q$, choose a sufficiently small Morse chart and ${\delta }_q>0$ so that its local unstable sphere at level $c^{\mathrm{\prime }}-{\delta }_q$ is $\{v_q=0,\mathrm{\mid }{u}_q{\mathrm{\mid }}^2={\delta }_q\}$ and $v<c^{\mathrm{\prime }}-{\delta }_q$. Likewise the local stable sphere at $p\in P$ is $\{u_p=0,\mathrm{\mid }{v}_p{\mathrm{\mid }}^2={\delta }_p\}$ at $c+{\delta }_p<v$. Their dimensions are $\mathrm{ind}(q)-1$ and $n-\mathrm{ind}(p)-1$. The central point is retained in the local disk; its constant trajectory does not cross the intermediate level.

The compact bands from $v$ to $c^{\mathrm{\prime }}-{\delta }_q$ and from $c+{\delta }_p$ to $v$ have no critical points. Put $Z=X\mathrm{/}(-df(X))$, so $df(Z)=-1$; it has the same descending trajectories as $X$. This is the downward version of the regular-product construction in [F2], whose displayed flow increases $f$. On each compact regular band $-df(X)$ has a positive minimum. Consequently $Z$ is smooth on a neighbourhood of the band, and compactness and finite-time continuation give its flow $\mathrm{\Psi }$ for every time needed to reach the other endpoint; along it $f({\mathrm{\Psi }}_t(x))=f(x)-t$. Smooth dependence and reverse flow give mutually inverse smooth level maps, including the endpoints, by the same inverse argument as [F2]. Transport the unstable sphere and its thickening forward by time $c^{\mathrm{\prime }}-{\delta }_q-v$, and the stable sphere and its thickening backward by time $v-c-{\delta }_p$. Their images $A_q,B_p$ are embedded compact spheres; a local disk together with its transported spherical collar is still a disk. All local unstable points other than the centre eventually cross the local sphere in forward time, so their crossing set is exactly $A_q$, and the reversed assertion gives $B_p$.

The local unstable sphere has an explicit product tube in its regular level: for small $z\in {\mathbb{R}}^{n-\mathrm{ind}(q)}$, use $(\omega ,z)\mapsto (\sqrt{{\delta }_q+\mathrm{\mid }z{\mathrm{\mid }}^2}\omega ,z)$ in its Morse chart. The $z$ coordinates trivialize its normal bundle. The symmetric formula trivializes the local stable sphere's normal bundle. The regular flow transports these product tubes, proving the product neighbourhood assertion. This uses the displayed trivializations, rather than inferring a trivial normal bundle from the tubular neighbourhood theorem.

A nonconstant trajectory with past limit $q$ eventually lies in its Morse chart, where $u(t)=e^{2t}u(0)$ and $v(t)=e^{-2t}v(0)$ force $v=0$ for convergence as $t\to -\mathrm{\infty }$. It therefore crosses $A_q$. Convergence to $p$ in forward time similarly forces $u=0$ and crossing of $B_p$. Strict descent makes the intermediate crossing unique. Conversely an intersection belongs to the same unique trajectory through both local disks, so its past and future limits are $q,p$. At index zero the unstable disk is a point and $A_q$ is empty; at index $n$ the stable disk is a point and $B_p$ is empty.