Thom Spaces Normal Data and Collapse Maps

Define $r_{h,k}(0_b)=0_b$ and $r_{h,k}(v)=(\mathrm{\parallel }v{\mathrm{\parallel }}_h\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_k)v$ for $v\mathrm{\ne }0$, as in [F2]. Homogeneity gives $\mathrm{\parallel }{r}_{h,k}v{\mathrm{\parallel }}_k=\mathrm{\parallel }v{\mathrm{\parallel }}_h$, so $r_{h,k}$ maps the $h$-disk to the $k$-disk and the $h$-sphere to the $k$-sphere. The formula is continuous on $E\setminus \{0\}$, where both norms are continuous and nonzero, and it is continuous at $0$ because $\mathrm{\parallel }{r}_{h,k}v{\mathrm{\parallel }}_k=\mathrm{\parallel }v{\mathrm{\parallel }}_h\to 0$; its inverse is $r_{k,h}$.

The pair homeomorphism descends to a based quotient homeomorphism ${\mathrm{Th}}_h(E)\to {\mathrm{Th}}_k(E)$. For $v\mathrm{\ne }0$, substituting the formulas gives $r_{k,l}(r_{h,k}(v))=\frac{\mathrm{\parallel }{r}_{h,k}(v){\mathrm{\parallel }}_k}{\mathrm{\parallel }{r}_{h,k}(v){\mathrm{\parallel }}_l}{r}_{h,k}(v)=\frac{\mathrm{\parallel }v{\mathrm{\parallel }}_k}{\mathrm{\parallel }v{\mathrm{\parallel }}_l}\cdot \frac{\mathrm{\parallel }v{\mathrm{\parallel }}_h}{\mathrm{\parallel }v{\mathrm{\parallel }}_k}v=r_{h,l}(v),$ and all three maps fix $0$, so the composition law holds exactly. The positive-definite family $h_t=(1-t)h+tk$ gives the radial isotopy $r_{h,h_t}$ from the identity to $r_{h,k}$. Empty bases and rank zero have identity maps with the conventions of [F1].

With the product metric and the supplied trivialization, [F1] identifies the disk/sphere pair of $B\times {\mathbb{R}}^r$ with $(B\times D^r,B\times S^{r-1})$ and computes the quotient as $B_{+}\wedge (D^r\mathrm{/}{S}^{r-1})$, naturally in $B$. Under Disk bundle, sphere bundle, and Thom space: the differential topology interface this quotient is exactly $\mathrm{Th}(B\times {\mathbb{R}}^r)$, with the same based convention and the same compactly generated quotient topology.

Since $D^r\mathrm{/}{S}^{r-1}=S^r$ — with the conventions $S^{-1}=\mathrm{\varnothing }$ and $D^0\mathrm{/}{S}^{-1}=\{\mathrm{p}\mathrm{t}{\}}_{+}=S^0$ when $r=0$ — step 1.1 gives $\mathrm{Th}(B\times {\mathbb{R}}^r)\cong B_{+}\wedge S^r={\mathrm{\Sigma }}^r{B}_{+}$. For $r=0$ the empty sphere bundle and the convention $X\mathrm{/}\mathrm{\varnothing }=X_{+}$ leave $B_{+}$; for empty $B$ both sides are the one-point based space; for $r=1$ the boundary is the two endpoints. The identity formula commutes with pullback along every map $B^{\mathrm{\prime }}\to B$, which is the asserted naturality. This is the AT result restated for the framed DT target.

If $N_0+N_1=0$, all tangent and normal fibres are zero and the unique zero-bundle isomorphism proves the assertion; assume the ambient dimension is positive below. In ${\mathbb{R}}^{N_0}\oplus {\mathbb{R}}^{N_1}$ set $j_t(x)=(\cos (\pi t\mathrm{/}2)i_0(x),\sin (\pi t\mathrm{/}2)i_1(x))$. At least one coefficient is nonzero; hence $j_t$ is injective and $d{j}_t$ is injective on each tangent space, because the coefficient that does not vanish already forces equality of $x$, respectively vanishing of the tangent vector. Let $Q_t(x)$ be the orthogonal projection onto $d{j}_t(T_{x}M)$ and $P_t=I-Q_t$. In a smooth local frame $e_1,\dots ,e_n$ of $TM$ from [F3], the matrix $A(t,x)$ with columns $d{j}_t(e_i(x))$ has full column rank, and $Q_t(x)=A(A^{T}A{)}^{-1}{A}^T$ is smooth in $(t,x)$; the same formula applies in boundary charts. For a zero-dimensional $M$, the frame is empty and the formula reads $Q_t=0$, $P_t=I$. Thus $P_t$ is a smooth family of orthogonal projections with $\ker P_t=d{j}_t(TM)=d{j}_t(T_{x}M)$ fibrewise. At $t=0$ one has $\mathrm{im}{P}_0=d{i}_0(T_{x}M{)}^{\perp }\oplus {\mathbb{R}}^{N_1}$ and at $t=1$ one has $\mathrm{im}{P}_1={\mathbb{R}}^{N_0}\oplus d{i}_1(T_{x}M{)}^{\perp }$, and [F1] identifies the orthogonal summands with ${\nu }_{i_0}$ and ${\nu }_{i_1}$ smoothly over $M$, in the half-space form at boundary points.

Put $K_t={\dot{P}}_t{P}_t-P_t{\dot{P}}_t$ and solve ${\dot{U}}_t=K_t{U}_t$, $U_0=I$ in the finite-dimensional ambient space, with $x\in M$ as a parameter. For each fixed $x$ the coefficients $t\mapsto K_t(x)$ are smooth, so [F2] gives a unique solution on all of $[0,1]$; For an interior parameter chart, the coefficients are defined on an open time-state-parameter domain, so [F2] gives local smooth dependence. At a boundary parameter point use [F4] to extend the coordinate functions of $i_0,i_1$ and the local frame to an open Euclidean parameter neighbourhood; shrink it so both embedding differentials and the frame remain full rank. The same formula for $A(t,x)$ then stays full rank on an open time interval containing $[0,1]$, since its sine and cosine coefficients never vanish together. Hence $P_t$ and $K_t$ have smooth extensions on an open time-parameter domain, and the vector field $(t,U,x)\mapsto K_t(x)U$ is smooth on an open time-state-parameter domain as required by [F2]. Apply that theorem on the extension and restrict back to the half-space. On the compact time interval, finitely many local continuations and uniqueness glue these solution maps to a smooth $U_t(x)$; the restricted extensions give smoothness up to the boundary. Since $K_t^T=-K_t$, differentiating $U^{T}U$ gives zero, so every $U_t$ is orthogonal. Differentiating $P_t^2=P_t$ gives $P_t{\dot{P}}_t{P}_t=0$, whence $KP-PK=(\dot{P}P-P\dot{P})P-P(\dot{P}P-P\dot{P})=\dot{P}P+P\dot{P}=\dot{P},$ so $\frac{d}{dt}(U_t^{-1}{P}_t{U}_t)=U_t^{-1}(-K_t{P}_t+{\dot{P}}_t+P_t{K}_t)U_t=0$ and $U_t^{-1}{P}_t{U}_t=P_0$. Passing to $t=1$, the smooth family $x\mapsto U_1(x)$ restricts to a smooth bundle isomorphism $\mathrm{im}{P}_0\to \mathrm{im}{P}_1$ over $M$.

By step 1.1 the initial complement bundle is ${\nu }_{i_0}\oplus {\epsilon }^{N_1}$ and the final one is ${\epsilon }^{N_0}\oplus {\nu }_{i_1}$, so with the reordering of the two orthogonal summands the bundle isomorphism of step 2.1 gives ${\nu }_{i_0}\oplus {\epsilon }^{N_1}\cong {\nu }_{i_1}\oplus {\epsilon }^{N_0}$. This proves the intrinsic nature of the stable normal class and applies verbatim to compact manifolds with boundary, where the same fibrewise orthogonal projections are smooth in boundary charts. For empty $M$ the assertion is the unique map of zero bundles. No summand has been cancelled: both sides retain their unstable ranks. The countable choice of [F1] is the only choice used; the ODE transport selects the solution of a linear equation with Lipschitz coefficients and adds none.

By [F1] fix a tubular chart ${\mathrm{\Phi }}_0$, and by [F2] fix a Riemannian metric $g$ on $M$; then [F3] exhibits $\nu (S)$ as the smooth quotient bundle identified with $C$ and makes ${\pi }^{\perp }$ smooth. Let $\mathrm{\Psi }:\mathrm{\Omega }\subseteq F\to M$ be any chart of a smooth bundle $F\to S$ of rank equal to $\mathrm{codim}S$ with $\mathrm{\Psi }(0_s)=i(s)$. Write $z$ for the zero section and identify $T_{(s,0)}F=d{z}_s(T_{s}S)\oplus F_s$, where $F_s=\ker d{\pi }_F$ is the vertical subspace. Since $\mathrm{\Psi }\circ z=i$, one has $d{\mathrm{\Psi }}_{(s,0)}(d{z}_s(u))=d{i}_s(u)$; hence $d{\mathrm{\Psi }}_{(s,0)}$ sends the horizontal summand isomorphically onto $d{i}_s(T_{s}S)$ and $F_s$ isomorphically onto a complement of it. The quotient class ${\beta }_s(v):=[d{\mathrm{\Psi }}_{(s,0)}(v)]\in \nu (S{)}_s$ is therefore a well-defined linear map $F_s\to \nu (S{)}_s$.

With $j:C\to \nu (S)$ the identification of [F3] and ${\tilde{\beta }}_s:={\pi }^{\perp }\circ d{\mathrm{\Psi }}_{(s,0)}{\mathrm{\mid }}_{F_s}$ one has $\beta =j\circ \tilde{\beta }$, because $d\mathrm{\Psi }(v)-{\pi }^{\perp }d\mathrm{\Psi }(v)$ lies in $d{i}_s(T_{s}S)$. In local frames of $F$ and $TM{\mathrm{\mid }}_S$ the components of $d{\mathrm{\Psi }}_{(s,0)}{\mathrm{\mid }}_{F_s}$ are smooth functions of $s$, since $\mathrm{\Psi }$ is smooth, and ${\pi }^{\perp }$ has smooth local matrices by [F3]; so [F4] makes $\tilde{\beta }$ and $\beta$ smooth bundle maps over ${\mathrm{i}\mathrm{d}}_S$. Fibrewise, ${\pi }^{\perp }d\mathrm{\Psi }(v)=0$ forces $d\mathrm{\Psi }(v)\in d{i}_s(T_{s}S)$, hence $v=0$ by the splitting and injectivity of $d\mathrm{\Psi }$; since $\mathrm{rank}F=\mathrm{codim}S=\mathrm{rank}C$, each ${\tilde{\beta }}_s$ and ${\beta }_s$ is bijective. By [F5], $\beta$ is a smooth bundle isomorphism.

Apply step 2.1 to $F=\nu (S)$ and $\mathrm{\Psi }={\mathrm{\Phi }}_0$: the induced map ${\beta }_0$ is a smooth bundle automorphism of $\nu (S)$. Put $\gamma ={\beta }_0^{-1}\circ \alpha :E\to \nu (S)$ and $\mathrm{\Phi }={\mathrm{\Phi }}_0\circ \gamma :{\gamma }^{-1}({\mathrm{\Omega }}_0)\to U_0$. Since $\gamma$ is a smooth bundle isomorphism over ${\mathrm{i}\mathrm{d}}_S$, the set ${\gamma }^{-1}({\mathrm{\Omega }}_0)$ is open and contains $0_S$, and $\mathrm{\Phi }$ is a diffeomorphism with $\mathrm{\Phi }(0_s)={\mathrm{\Phi }}_0(0_s)=i(s)$. For $v\in E_s$ one has $d{\gamma }_{(s,0)}(v)={\gamma }_s(v)$, because $\gamma$ is fibrewise linear over ${\mathrm{i}\mathrm{d}}_S$; hence by the chain rule [F6] the induced map of $\mathrm{\Phi }$ at $s$ is ${\beta }_{0,s}\circ {\gamma }_s={\beta }_{0,s}\circ {\beta }_{0,s}^{-1}\circ {\alpha }_s={\alpha }_s$.

Taking $E=\nu (S)$ and $\alpha =\mathrm{i}\mathrm{d}$ in step 3.1 gives the chart ${\mathrm{\Phi }}_0\circ {\beta }_0^{-1}$, which induces the identity, so the normal bundle admits a chart in the specified compatible class. If $S=\mathrm{\varnothing }$ then $E$, $\nu (S)$, ${\mathrm{\Omega }}_0$ and $U_0$ are empty and the condition is vacuous; if $E$ has rank zero then $\mathrm{codim}S=0$ and ${\beta }_s$ is the unique isomorphism between zero spaces, so step 3.1 still applies. The isomorphism $\gamma$ is determined by the supplied chart and $\alpha$, so no object is selected beyond [A1]; the metric of [F2] only exhibits the smooth structure and does not enter ${\beta }_0$.

On the closed tube the continuous map $\mathrm{\Phi }(s,v)\mapsto [(s,v\mathrm{/}\rho )]$ sends its boundary to the basepoint. On the closed complement of its interior the map is constant. These two closed sets cover $X$ and the definitions agree on their intersection, so the closed pasting lemma [F2] proves continuity on $X$; at a disjoint added basepoint continuity is immediate. At the compactification point, the complement of the compact closed tube is a neighborhood mapped constantly to the basepoint; this proves continuity there.

On the inverse image of the nonbasepoint stratum the formula is a smooth tubular inverse followed by fiber scaling. In a bundle trivialization about zero the map reads $(s,v)\mapsto (s,v\mathrm{/}\rho )$, whose derivative in the fibre directions is ${\rho }^{-1}$ times the identity, so it is a submersion there; identifying the normal quotient of $X$ along $S$ with $E$ by ${\alpha }^{-1}$, that vertical derivative is ${\rho }^{-1}{\alpha }^{-1}$ and its zero fibre is exactly $S$. No smoothness assertion at the Thom basepoint is needed.

More generally let $a:[0,\rho ]\to [0,1]$ be smooth on $[0,\rho )$, positive off zero, equal to $\lambda t$ near zero for $\lambda >0$, and equal to $1$ on a neighborhood of $\rho$. Map $v\mathrm{\ne }0$ to $a(\mathrm{\parallel }v\mathrm{\parallel })v\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel }$ and zero to zero, and then take the quotient; outside the tube use the basepoint. This is continuous by step 1.1 and smooth on its nonbasepoint stratum, including zero because its formula there is $\lambda v$. Convex interpolation between this radius profile and $t\mathrm{/}\rho$ stays positive for $t>0$, is linear with positive coefficient near zero, and equals $1$ at the boundary. The same pasting argument on $X\times I$ proves the based homotopy. Thus a cutoff supplies the contracted smooth representative near regular values without assigning a smooth structure at the collapsed point.

Shrink around $0_S$ so $\psi ={\mathrm{\Phi }}_1^{-1}{\mathrm{\Phi }}_0$ and its inverse are defined. It fixes $0_S$ and induces the identity on the normal quotient by [F1]. For fiber dilation ${\delta }_t$, define ${\psi }_t={\delta }_{1\mathrm{/}t}\psi {\delta }_t$ for $t>0$, and ${\psi }_0=\mathrm{id}$. In bundle charts write $\psi (x,v)=(b(x,v),w(x,v))$, using a target trivialization near $x$. Then $b(x,0)=x$, $w(x,0)=0$ and ${\mathrm{\partial }}_{v}w(x,0)=I$. Taylor's integral formula writes $w(x,tv)\mathrm{/}t={\int }_0^1{\mathrm{\partial }}_{v}w(x,utv)vdu$, while $b(x,tv)\to x$. These formulas prove smooth extension at $t=0$ with value $(x,v)$. They are coordinate-compatible because the dilation is intrinsic.

Apply the same formulas to ${\psi }^{-1}$. Compactness of $S\times I$ gives a single small disk on which the two families and their composites are defined; shrinking again if necessary, their compositions are the identity, by the identity for $t>0$ and continuity at zero. Thus each ${\psi }_t$ is a diffeomorphism onto its image, and ${\mathrm{\Phi }}_1{\psi }_t$ is a smooth path of compatible tubular charts from ${\mathrm{\Phi }}_1$ to the germ of ${\mathrm{\Phi }}_0$. Its induced normal derivative remains the identity: in the above local expression ${\mathrm{\partial }}_v(w(x,tv)\mathrm{/}t){\mathrm{\mid }}_{v=0}=I$, while the horizontal derivative is irrelevant to the normal quotient. A common small closed tube exists by compactness.

Use that common radius to collapse along ${\mathrm{\Phi }}_1{\psi }_t$. The inverse tube coordinates vary smoothly. The track of the closed tubes is a compact subset of $X\times I$, its boundary maps to the basepoint, and closed pasting as in [F2] proves joint continuity; at a compactification point the compact track gives a uniform constant neighborhood. This supplies the based homotopy of the common-radius endpoint collapses. Interpolate each endpoint radius to its original radius within that endpoint chart: the formula is fiber scaling on the varying tube, agrees with the basepoint at its boundary, and is jointly continuous by the same compact-track argument. Radial cutoffs are covered by [F2].

Finally interpolate metrics by $h_t=(1-t)h_0+t{h}_1$, use their canonical radial maps [F3] to express all targets in the $h_0$ model, and choose a common small tube uniformly in $t$. The formulas and pasting of step 3.1 give the corresponding homotopy. Concatenation proves precisely the stated choice independence. If a chart is instead precomposed with a normal automorphism, step 1.1 limits to that automorphism rather than the identity; for a point and a reflected normal line the resulting sphere maps have opposite degree. Hence preserving the specified normal data is essential to this proof and statement.

Around $x\in P$ choose a bundle chart of $E$ over $U\subseteq B$ and use [F1] to view it as a smooth chart of the target near $f(x)$; on $W$ write $f=(g,h)$ with $h$ valued in ${\mathbb{R}}^r$ and $g$ valued in $U$, so that $P\cap W=h^{-1}(0)$ and the normal space of the zero section at $f(x)$ is identified with the fibre $E_{g(x)}={\mathbb{R}}^r$. By [F2] applied to the smooth map $f{\mathrm{\mid }}_W$ and the embedded zero section, transversality at $x$ is exactly surjectivity of $d{h}_x$. If another trivialization replaces $h$ by $T(g(x))h(x)$ with $T$ a smooth invertible matrix function, then at $h=0$ its derivative is $T(g(x))d{h}_x$, so surjectivity is chart-independent and the transition acts on the normal quotient by the same matrix.

Restricted to $W\cap \mathrm{Int}X$, the map $f$ takes values in the smooth stratum and, by step 1.1, is transverse to the embedded zero section there. The published transverse preimage theorem [F3] therefore makes $P$ an embedded submanifold of codimension $r$ of that open set, with $T_{x}P=\{v\in T_{x}X:d{f}_x(v)\in T_{f(x)}{0}_B\}=\ker d{h}_x$. Since the interior points of $P$ are covered by these open sets, the interior part of $P$ is an embedded submanifold with the asserted tangent space.

The differential $d{h}_x$ factors through the quotient to a linear isomorphism $T_{x}X\mathrm{/}{T}_{x}P\to E_{g(x)}$. Step 1.1 shows that these local isomorphisms transform by exactly the transition matrices of $E$, so they glue to a smooth bundle isomorphism $\nu (P\subset X)\to (g{\mathrm{\mid }}_P{)}^{\ast }E$ over $P\cap \mathrm{Int}X$; smoothness of the normal bundle is [F5].

Let $x\in P\cap \mathrm{\partial }X$ and use boundary coordinates $(u,t)$ with $t\ge 0$. By [F7], the fibre coordinate $h(u,t)$ extends smoothly across $t=0$ near $x$. Boundary transversality says $d_{u}h(u,0)$ is surjective at $x$; hence, after reordering the $n-1$ tangential coordinates, an $r\times r$ minor in the first $r$ coordinates of $u$ is invertible. The map $(u,t)\mapsto (h(u,t),u_{r+1},\dots ,u_{n-1},t)$ has invertible derivative, so [F6] makes it a local diffeomorphism. Its last coordinate is exactly the original $t$, so it maps the source half-space to $\{t\ge 0\}$, without assuming an arbitrary nonlinear image of a half-space is linear. In these coordinates $P$ is precisely $\{h=0,t\ge 0\}$, with boundary $\{h=0,t=0\}$, and its tangent space is $\ker dh$. These charts prove neateness. The same local normal quotient map as step 3.1 is a smooth bundle isomorphism at boundary points, and $T_x\mathrm{\partial }X\mathrm{/}{T}_x\mathrm{\partial }P\to T_{x}X\mathrm{/}{T}_{x}P$ is an isomorphism since $dh{\mathrm{\mid }}_{T_x\mathrm{\partial }X}$ is surjective. Thus its restriction is exactly the boundary normal identification. When $r=0$, the coordinate map is the identity and the same conclusion holds.

For $r>0$ and nonempty $B$, the zero section is closed in $D(E)$ and disjoint from $S(E)$; its saturation under the sphere collapse is itself, so the quotient topology makes its image closed. Passing to the compactly generated topology preserves this closed set. If $r=0$, [F1] uses the based empty-subspace quotient $B\mathrm{/}\mathrm{\varnothing }=B_{+}$, so $B$ and its added isolated basepoint are separate clopen pieces. Thus $0_B$ is closed for every rank, and $P=f^{-1}(0_B)$ is closed and therefore compact for compact $X$; in rank zero it is also open. For empty $B$, $0_B=\mathrm{\varnothing }$ and every conclusion is vacuous. No choice beyond [A1] is used.

In case (a), F1 makes $W$ closed and compact for every rank. The collars are constant in time. At their zero points the fibre differential has zero time component, so surjectivity of the full fibre differential is exactly surjectivity on the $X$ directions; hence the boundary restrictions are transverse as well. Apply F1 to the smooth neighbourhood of $W$ in $X\times I$: $W$ is neat, $\mathrm{\partial }W=W_0\bigsqcup W_1$, and its specified normal isomorphism is the pullback of $E$ and restricts to the endpoint normal isomorphisms. This is the asserted compact normal cobordism.

For case (b), if $W=\mathrm{\varnothing }$, retain $H$; smoothness and transversality near the empty zero preimage are vacuous and $F$ is untouched. If $r=0$, the zero stratum $B$ is clopen in $B_{+}$ by F1. For each $x$, the inverse image of $B$ under the continuous time path is clopen in the connected interval, so it is either all of $I$ or empty. Thus $W=W_0\times I$, with $W_0$ clopen in compact $X$, and $F$ lies in its complement. Extend $H{\mathrm{\mid }}_{W_0\times I}$ constantly past both endpoints to $W_0\times (-1,2)\to B$. This map is smooth on endpoint time collars because $H_0,H_1$ are smooth near their whole zero preimages $W_0$. Apply [F2] relative to the closed union of smaller extended endpoint collars to obtain a smooth map into $B$ and a homotopy fixed there. Restrict to $W_0\times I$ and paste with the unchanged basepoint map on the clopen complement, using [F7]. This gives (b), fixes $F$ pointwise and all original basepoint values, and stays constant on smaller endpoint collars. Transversality to the rank-zero zero section, the entire smooth stratum, is automatic. This includes $W_0=X$ and $W_0=\mathrm{\varnothing }$; empty $B$ was already covered by $W=\mathrm{\varnothing }$.

Now assume $r>0$ and $W\mathrm{\ne }\mathrm{\varnothing }$. Choose $0<\delta <1\mathrm{/}4$ so $H$ is constant in time on $[0,\delta ]$ and $[1-\delta ,1]$. Choose open neighbourhoods $O_i$ of $W_i$ in $X$ where $H_i$ is smooth with values in $E$. On the boundaryless source $X\times (0,1)$ choose an open set $U$ containing its part of $W$, disjoint from $F$, with $H(U)\subseteq E$, and such that $U\cap \{t\le \delta \mathrm{/}2\}\subseteq O_0\times (0,\delta )$ and $U\cap \{t\ge 1-\delta \mathrm{/}2\}\subseteq O_1\times (1-\delta ,1)$. Such $U$ is obtained by intersecting $H^{-1}(E)\setminus F$ with the open collar/central unions; it contains the interior zeros since collar zeros lie in $O_i$. Set $J=[\delta \mathrm{/}4,1-\delta \mathrm{/}4]$. The set $K=W\cap (X\times J)$ is compact, unlike the entire interior part of $W$. Choose a compact neighbourhood $C$ of $K$ and open $V$ with $K\subseteq \mathrm{int}C\subseteq C\subseteq V$ and $\bar{V}\subseteq U$ compact. The bump in [F4] gives a smooth $\lambda$ equal to one on a neighbourhood of $C$, supported in $V$. Its zero extension is smooth and vanishes near the endpoints and on $F$.

Apply [F8] to the smooth total space $E$ to obtain a proper embedding $j:E\hookrightarrow {\mathbb{R}}^d$. Its image $j(E)$ is closed, so [F8] supplies a smooth retraction $\rho :T\to j(E)$ on an open neighbourhood $T$; set $R=j^{-1}\rho :T\to E$, with $Rj=\mathrm{id}$. The relative approximation [F3] is applied on $U$ to $jH$ with closed protected set $A=U\cap (\{t\le \delta \mathrm{/}3\}\cup \{t\ge 1-\delta \mathrm{/}3\})$. It is smooth near $A$ by step 1.3. Define the compact relevant buffer $Q=(\bar{V}\setminus \mathrm{int}C)\cap (X\times J)$. It misses $W$ because $K\subseteq \mathrm{int}C$; thus $jH(Q)$ lies in the open set $T_0=R^{-1}(E\setminus 0_B)$. Choose a positive continuous error function on $U$ smaller than half the distance of $jH(p)$ to ${\mathbb{R}}^d\setminus T$, and uniformly small enough that every error ball over $Q$ lies in $T_0$; compactness of $jH(Q)$ supplies this uniform bound. Empty complements or empty $Q$ require only any fixed positive bound. Then [F3] gives smooth $\mathrm{\Psi }$ with this error, equal to $jH$ on a neighbourhood of $A$.