The Whitney Trick and Surgery Below the Middle Dimension

The parametrization $\theta \mapsto (\theta ,2\theta <mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>1)$ is injective on $S^1=\mathbb{R}\mathrm{/}\mathbb{Z}$ because its first coordinate is, so $B$ is an embedded circle with tangent spanned by $(1,2)$, while $T_{p}A$ is spanned by $(1,0)$; hence $A\cap B=\{(\theta ,0):2\theta \equiv 0\}=\{(0,0),(1\mathrm{/}2,0)\}$, at both of which $T_{p}A+T_{p}B=T_p{T}^2$, so the two intersections are transverse, and the isomorphism $T_{p}A\oplus T_{p}B\to T_p{T}^2$ with the factor $A$ first has, in the basis $({\mathrm{\partial }}_{\theta },{\mathrm{\partial }}_y)$, the matrix with columns $(1,0)$ and $(1,2)$ and determinant $2>0$, so both local signs equal $+1$.

Summing the two local signs $+1$ over the transverse intersection as in [F1], and counting its two points modulo two as in [F3], gives $I(A,B)=2$ and $I_2(A,B)=0$; the two values are congruent modulo $2$, as [F4] requires, and the example therefore has $I_2(A,B)=0$ while $I(A,B)\mathrm{\ne }0$.

Suppose a smooth homotopy of $i_A$ ended at a smooth map $F_1$ whose image meets $B$ in no point; then $F_1$ is transverse to $B$ with empty preimage, so $I(F_1,B)=0$ by [F1], while the homotopy-invariance consequence [F5], applied to the two transverse maps $i_A$ and $F_1$ in the same homotopy class, gives $I(i_A,B)=I(F_1,B)$, which with step 2.1 is the contradiction $2=0$. Since an isotopy of $A$ is such a homotopy, no isotopy can make $A$ disjoint from $B$; in particular no Whitney cancellation of the pair — an isotopy removing the two points and creating no new ones — is available, and the necessary opposite-sign hypothesis is violated because both local signs equal $+1$ by step 1.1. The homotopy-invariance input [F5] assumes ${\mathrm{A}\mathrm{C}}_{\omega }$, inherited here through The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$), while steps 1.1-2.1 are choice-free.

By [F1] every point of $A$ has a neighbourhood basis of path-connected open sets, so $A$ is locally path-connected; since $A$ is connected, [F2] makes $A$ path-connected, so there is a continuous path $\eta :I\to A$ with $\eta (0)=p$ and $\eta (1)=q$.

The compact image $\eta (I)$ has a finite coordinate-ball cover. A sufficiently fine partition $0=t_0<\cdots <t_m=1$ has $\eta ([t_{i-1},t_i])\subset U_i$ for some coordinate ball $U_i$. Each overlap $U_i\cap U_{i+1}$ contains $\eta (t_i)$ and is nonempty and open; in positive dimension it cannot be contained in the finite set $F$. Choose $z_i\in (U_i\cap U_{i+1})\setminus F$, with $z_0=p$ and $z_m=q$. Applying [F3] successively to the finitely many forbidden points in each ball shows $U_i\setminus F$ is path-connected. Join $z_{i-1}$ to $z_i$ there and concatenate these finitely many paths. This gives a path in $A\setminus F$ from $p$ to $q$, without any assumption that ${\eta }^{-1}(F)$ is finite.

Since $p,q\in A\setminus F$ were arbitrary, step 2.1 shows that $A\setminus F$ is path-connected; it is nonempty and, by [F2], connected.

Thus $A\setminus F$ is a nonempty connected smooth $a$-manifold without boundary with $a\ge 2$; by [F4] there is a proper smooth embedding $e:A\setminus F\to {\mathbb{R}}^{2a+1}$ (this is where ${\mathrm{A}\mathrm{C}}_{\omega }$ is used), whose image is an embedded submanifold by [F4] and is closed: if $e(x_j)\to x$ in ${\mathbb{R}}^{2a+1}$, then $\{x\}\cup \{e(x_j):j\ge 1\}$ is compact, its preimage under the proper map $e$ is compact and contains all $x_j$, and a convergent subsequence $x_{j_k}\to x_{\ast }$ has $e(x_{\ast })=x$ by continuity.

Equip $A\setminus F$ with the pullback $g:=e^{\ast }\delta$ of the Euclidean metric $\delta$ of ${\mathbb{R}}^{2a+1}$; since $e$ is the smooth embedding of step 4.1, [F6] makes $g$ a Riemannian metric and $e$ a Riemannian isometry onto the embedded submanifold $e(A\setminus F)$ with its induced metric, so $e$ preserves distances by [F6]. That submanifold is closed in the complete manifold $({\mathbb{R}}^{2a+1},\delta )$ by step 4.1, hence complete in the induced metric by [F5], and therefore $(A\setminus F,d_g)$ is complete: a $g$-Cauchy sequence maps under the distance-preserving bijection $e$ to a Cauchy sequence in a complete space, which converges, and its preimage converges in $A\setminus F$.

Assume $p\mathrm{\ne }q$. Then $(A\setminus F,g)$ is a nonempty connected boundaryless complete Riemannian manifold, so [F7] supplies $v\in T_p(A\setminus F)$ with ${\exp }_p(v)=q$, $\mathrm{\mid }v{\mathrm{\mid }}_{g_p}=d_g(p,q)$ and $\gamma (t):={\exp }_p(tv)$ of length $d_g(p,q)$ on $[0,1]$; by [F8] the distance between the distinct points $p$ and $q$ is positive, so $\mathrm{\mid }v\mathrm{\mid }=d_g(p,q)>0$, and [F7] makes the speed $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}\mathrm{\mid }$ constant, hence equal to $\mathrm{\mid }v\mathrm{\mid }>0$, so $\gamma$ is an immersion.

The curve $\gamma$ is injective: if $\gamma (s)=\gamma (t)$ with $0\le s<t\le 1$, then the concatenation of $\gamma {\mathrm{\mid }}_{[0,s]}$ with $\gamma {\mathrm{\mid }}_{[t,1]}$ is a piecewise $C^1$ curve from $p$ to $q$ whose length is $s\mathrm{\mid }v\mathrm{\mid }+(1-t)\mathrm{\mid }v\mathrm{\mid }=(1-(t-s))d_g(p,q)<d_g(p,q)$ by the constant speed and the additivity of length [F8], while [F8] also says that every piecewise $C^1$ curve from $p$ to $q$ has length at least $d_g(p,q)$, a contradiction.

Consequently $\gamma :I\to A\setminus F\subseteq A$ is smooth, injective, an immersion and a homeomorphism onto its image ($I$ is compact, $A$ is Hausdorff, and a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism), so by [F9] it is a smooth embedded arc from $p$ to $q$, and $\gamma ((0,1))\cap F=\mathrm{\varnothing }$ because its image lies in $A\setminus F$.

The remaining clauses follow: for $p=q$ the path-connectedness assertion is step 3.1 and the constant degenerate arc satisfies the arc assertion; for arbitrary $p,q\in A$, applying the established case to the finite set $F\setminus \{p,q\}$, which no longer contains $p$ or $q$, produces a smooth embedded arc from $p$ to $q$ whose interior avoids $F\setminus \{p,q\}$, hence whose image meets $F$ only in the endpoints; and if $A$ is a connected embedded submanifold of a smooth manifold, [F10] equips it with the induced smooth structure, so the same argument applies verbatim to that manifold. Countable choice is used exactly through the proper embedding [F4], the completeness statements [F5] and Hopf-Rinow with the geodesic speed [F7]; steps 1.1-3.1 and 7.1 add no choice.

By [F2] we may take $D$ neat, so the closed tubular neighbourhood $N$ of [F3] is diffeomorphic to $D\times D^2$, with $N\cap S^3=\mathrm{\partial }D\times D^2$ and with $\mathrm{\partial }N$ split into the two solid tori $\mathrm{\partial }D\times D^2\subseteq S^3$ and $D\times S^1$, glued along the torus $\mathrm{\partial }D\times S^1$. Then $Y:=\overline{B^4\setminus \mathrm{int}N}$, with its corners rounded, is a compact $4$-manifold with boundary and $B^4=N\cup Y$ with $N\cap Y=D\times S^1$, which is homotopy equivalent to $S^1$ with generator the meridian circle $\{x\}\times S^1$.

Mayer-Vietoris [F8] for the collar-thickened open cover of $B^4$ by the interiors of enlarged $N$ and $Y$, retracting to $N,Y,N\cap Y$ respectively, with $N$ and $B^4$ contractible and $N\cap Y\simeq S^1$ gives $H_i(Y;\mathbb{Z})=0$ for every $i\ge 2$, because $H_i(N\cap Y)$ vanishes there and $H_i(B^4)$ vanishes for $i\ge 1$; in degree one it makes $H_1(N\cap Y;\mathbb{Z})\to H_1(Y;\mathbb{Z})$ an isomorphism, since the preceding term $H_2(B^4)$ vanishes and $H_1(N)=0$, so $H_1(Y;\mathbb{Z})\cong \mathbb{Z}$ is generated by the meridian; in reduced degree zero all terms of ${\tilde{H}}_0(N\cap Y)\to {\tilde{H}}_0(N)\oplus {\tilde{H}}_0(Y)\to {\tilde{H}}_0(B^4)$ vanish except possibly the middle, so $Y$ is connected.

The connected manifold $Y$ is path-connected, and its ball or half-ball charts give contractible neighbourhoods, so it is locally path-connected and semilocally simply connected. Fix a basepoint in $Y$. By [F5] the composite ${\pi }_1(Y)\to H_1(Y;\mathbb{Z})\cong \mathbb{Z}\to \mathbb{Z}\mathrm{/}2$ (reduction mod $2$) is a surjection, so it determines a connected double cover $p:\tilde{Y}\to Y$. Over ${\mathbb{F}}_2$ the singular chains of the cover form the short exact sequence of [F4], and its long exact homology sequence together with step 2.1 gives: $H_0(Y;{\mathbb{F}}_2)={\mathbb{F}}_2$, $T_{\ast }:H_0(Y;{\mathbb{F}}_2)\to H_0(\tilde{Y};{\mathbb{F}}_2)$ is zero and $P_{\ast }:H_0(\tilde{Y};{\mathbb{F}}_2)\to H_0(Y;{\mathbb{F}}_2)$ is an isomorphism (the cover is connected), so the connecting map $H_1(Y;{\mathbb{F}}_2)\to H_0(Y;{\mathbb{F}}_2)$ is an isomorphism; consequently $P_{\ast }=0$ in degree one and $T_{\ast }:H_1(Y;{\mathbb{F}}_2)\to H_1(\tilde{Y};{\mathbb{F}}_2)$ is an isomorphism; and $H_i(\tilde{Y};{\mathbb{F}}_2)=0$ for $i\ge 2$.

The preimage in $\tilde{Y}$ of the solid torus $N\cap Y\cong D\times S^1$ is connected, because the meridian has odd class in $\mathbb{Z}\mathrm{/}2$, and the covering restricts to the model $(x,z)\mapsto (x,z^2)$ of $D\times S^1$ onto itself; naturality of the transfer in step 3.1 shows that its upstairs meridian generates $H_1(\tilde{Y};{\mathbb{F}}_2)$: the transfer of a downstairs circle is the sum of its two lifted half-circle paths, hence the single upstairs circle. Glue a copy of $D\times D^2$ to $\tilde{Y}$ by a diffeomorphism of its boundary solid torus onto that upstairs overlap, identifying the upstairs circle coordinate $z$ with itself; its projection downstairs is $(x,z)\mapsto (x,z^2)$; after rounding corners the result $W$ is a compact connected smooth $4$-manifold, and the gluing map exhibits $W\to B^4$ as a branched double cover whose restriction off $D$ is the covering $p$ and whose local model at $D$ is $(x,z)\mapsto (x,z^2)$ in complex normal coordinates; pulling back the orientation of $B^4$ along this branched cover orients $W$, and the boundary $\mathrm{\partial }W$ is the double cover of $S^3$ branched along $\mathrm{\partial }D$.

Mayer-Vietoris [F8] over ${\mathbb{F}}_2$ for collar-thickened open pieces retracting to the displayed pieces of $W=\tilde{Y}\cup (D\times D^2)$ with overlap $D\times S^1$, whose $H_1$ maps isomorphically onto $H_1(\tilde{Y};{\mathbb{F}}_2)$ by step 4.1, while $H_i(D\times D^2;{\mathbb{F}}_2)=H_i(D\times S^1;{\mathbb{F}}_2)=0$ for $i\ge 2$ and $H_i(\tilde{Y};{\mathbb{F}}_2)=0$ for $i\ge 2$ by step 3.1, yields $H_i(W;{\mathbb{F}}_2)=0$ for every $i>0$ and, by the reduced degree-zero segment, ${\tilde{H}}_0(W)=0$, so $W$ is connected; the exact piece in degrees two and one is $0\to H_2(W)\to H_1(D\times S^1)\stackrel{\cong }{\to }{H}_1(\tilde{Y})\to H_1(W)\to 0$, so $H_2(W)=H_1(W)=0$.

The double $DW=W{\cup }_{\mathrm{\partial }W}W$ of $W$ along its collared boundary is a compact smooth $4$-manifold without boundary, hence by [F7] has the homotopy type of a CW complex whose compact image lies in a finite subcomplex $K^{\mathrm{\prime }}$; the folding retraction $r:DW\to W$ collapsing the second copy onto the first through the collar satisfies $r{\mathrm{\mid }}_W={\mathrm{i}\mathrm{d}}_W$, so composing an equivalence, its inverse and $r$ exhibits $W$ as a homotopy retract of the finite CW complex $K^{\mathrm{\prime }}$. Therefore every $H_i(W;\mathbb{Z})$ is finitely generated. For $i>0$, the universal coefficient sequence of [F6] injects $H_i(W;\mathbb{Z})\otimes {\mathbb{F}}_2$ into $H_i(W;{\mathbb{F}}_2)=0$ by step 5.1, so $H_i(W;\mathbb{Z})$ has no nontrivial free part; tensoring with $\mathbb{Q}$ gives $H_i(W;\mathbb{Q})=H_i(W;\mathbb{Z})\otimes \mathbb{Q}=0$ for $i>0$. This completes the proof; full AC is used for the bundle homotopy-invariance, CW and universal-coefficient suppliers, and supplies the countable choice needed for collaring and tubes.

The inclusion of the original mapping cylinder of $f$ into $Z$ fixes $M$ and is a homotopy equivalence: both cylinders retract onto the same $X$, and their maps onto $X$ commute. Its inverse up to homotopy relative to $M$ can be constructed by extending the cylinder contraction across the characteristic disk with its specified map $h$; equivalently use the mapping-cylinder homotopy-extension property. Thus ${\pi }_i(Z,M)={\pi }_i(f)$, with the disk of $W$ mapping to the specified class $x$. By definition ${\pi }_i(Z,W)={\pi }_i(F)$.

We verify the only relative-cell input directly. A disk map into $(W,M)$ has all its boundary in $M$. In the open attached cell smooth its Euclidean coordinate map near the inverse image of a small closed ball about the cell centre, using finitely many interior source bumps and Euclidean relative approximation. Choose a regular point in that small ball by Sard. For a source dimension less than $n$ its preimage is empty, since a transverse inverse image would have negative dimension. Radially retract the punctured cell from that point onto its boundary and leave $M$ fixed; this retracts $W$ minus that point onto $M$. Consequently ${\pi }_i(W,M)=0$ for $i<n$. No smoothness of the attaching map or of $X$ is needed; all smoothing takes place inside the open Euclidean cell and off the source boundary.

For a source $n$-disk the regular-point preimage in step 1.2 is finite, since it is discrete and contained in a compact interior set. Small disjoint disks about these points map locally diffeomorphically to the cell, with orientation degree $+1$ or $-1$. On their complement retract the map into $M$ by the same punctured-cell retraction. Join these small disks to the marked outer boundary point by a finite embedded tree of thin tubes in the source, arranged disjointly away from their common endpoint. Cutting along this tree is the standard disk description of relative multiplication: each small disk contributes the characteristic relative disk or its inverse, and the image of its tube, after retraction, supplies a path in $M$ carrying its marked value to the basepoint. Comparing that path with the fixed characteristic-disk path gives the ${\pi }_1(M)$ action. For $n\ge 3$ the relative group is abelian, so this is a sum of signed translates. For $n=2$ it is an ordered product; changing the cutting order or tube produces conjugates, so the image in any target relative group is the normal subgroup generated by those translates. Conversely every translate and its inverse is represented by the corresponding whiskered characteristic disk, and every conjugate maps to that normal subgroup. This proves the exact image description needed below, not freeness of a degree-two relative group.

The natural exact sequence of the triple $M\subset W\subset Z$ reads ${\pi }_n(W,M)\to {\pi }_n(Z,M)\to {\pi }_n(Z,W)\to {\pi }_{n-1}(W,M)$. The last term vanishes by step 1.2, hence the middle map is surjective and its kernel is the image computed in step 2.1, namely the stated orbit-generated subgroup, with normal closure in degree two. Both neighbouring groups ${\pi }_i(W,M),{\pi }_{i-1}(W,M)$ vanish in lower group degrees, so the triple sequence gives the asserted lower isomorphisms. In degree one, the image of ${\pi }_1(M)$ in ${\pi }_1(X)$ is unchanged: attaching a cell of dimension at least two only adds relations represented by loops whose image under $f$ is already null by the disk extension. The relative degree-one coset description therefore gives the same pointed set. If $f_{\ast }$ is an isomorphism, identify the action group with ${\pi }_1(X)$; for $n\ge 3$ the group is abelian and its action is precisely a group-ring module action.

For a 1-cell, van Kampen says the new core adjoins a loop to the fundamental group of the connected $M$. Under $F$ its image is $\mathrm{\ell }$, so the image subgroup in ${\pi }_1(X)$ becomes $⟨H,\mathrm{\ell }⟩$. Paths from the basepoint classify the relative fundamental pointed set by cosets of that image, yielding the stated low-degree formulation. This uses no nonexistent relative degree-zero group or abelian module structure. The case $x=0$ in higher degrees gives a zero orbit-generated kernel, and all degree ranges and choices were specified.

By [F1] the integral homology groups of $W$ and $M$ are finitely generated. By [F5], the rational homology hypothesis makes $H_i(W;\mathbb{Z})\otimes \mathbb{Q}=0$ for $i>0$; finite generation and the decomposition of [F3] therefore make those integral groups finite. Apply [F3] also with coefficient module $\mathbb{Q}$: the Hom terms vanish in positive degrees and the Ext terms for finite cyclic groups vanish because multiplication by their orders is onto on $\mathbb{Q}$. Thus $H^i(W;\mathbb{Q})=0$ for $i>0$ and $H^0(W;\mathbb{Q})=\mathbb{Q}$. Duality [F2] gives $H_i(W,M;\mathbb{Q})=0$ for $i\mathrm{\ne }4$, and $H_4(W,M;\mathbb{Q})=\mathbb{Q}$. The pair sequence now gives $H_3(M;\mathbb{Q})=\mathbb{Q}$, $H_1(M;\mathbb{Q})=H_2(M;\mathbb{Q})=0$, and $H_0(M;\mathbb{Q})=\mathbb{Q}$. Hence $M$ is a rational homology sphere and $H_1(M)$ is finite. Integral duality on $M$, followed by [F3], gives $H_2(M)\cong H^1(M)=0$: the Ext term for $H_0(M)=\mathbb{Z}$ and the Hom term for finite $H_1(M)$ both vanish. Likewise $H_3(W,M)\cong H^1(W)=0$. These are exactly the vanishings needed in the order calculation.

Put $A:=H_2(W)$ and $B:=H_1(W)$; both are finite, because $W$ is rationally a point and the groups are finitely generated by [F1]. Duality [F2] and the cohomology universal coefficient sequence [F3] give $H_2(W,M)\cong H^2(W)\cong \mathrm{Ext}(B,\mathbb{Z})$ and $H_1(W,M)\cong H^3(W)\cong \mathrm{Ext}(A,\mathbb{Z})$, the Hom terms vanishing because $A,B$ are finite. For a finite abelian group $F$ decomposed into cyclic summands, the sequence $0\to \mathbb{Z}\stackrel{n}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}n\to 0$ computes $\mathrm{Ext}(\mathbb{Z}\mathrm{/}n,\mathbb{Z})=\mathbb{Z}\mathrm{/}n$, so $\mathrm{\mid }\mathrm{Ext}(F,\mathbb{Z})\mathrm{\mid }=\mathrm{\mid }F\mathrm{\mid }$.

The long exact sequence of the pair $(W,M)$ in low degrees, using $H_2(M)=0$ and the isomorphism $H_0(M)\to H_0(W)$ of connected spaces, reads $0\to A\to \mathrm{Ext}(B,\mathbb{Z})\to H_1(M)\to B\to \mathrm{Ext}(A,\mathbb{Z})\to 0$. Split it into the short exact sequences $0\to A\to \mathrm{Ext}(B,\mathbb{Z})\to K\to 0$ and $0\to K\to H_1(M)\to B^{\mathrm{\prime }}\to 0$, where $K$ is the image of $\mathrm{Ext}(B,\mathbb{Z})$ and $B^{\mathrm{\prime }}$ the image of $H_1(M)$; then $\mathrm{\mid }K\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }A\mathrm{\mid }$ and $\mathrm{\mid }{H}_1(M)\mathrm{\mid }=\mathrm{\mid }K\mathrm{\mid }\cdot \mathrm{\mid }{B}^{\mathrm{\prime }}\mathrm{\mid }$. The third short exact sequence $0\to B^{\mathrm{\prime }}\to B\to \mathrm{Ext}(A,\mathbb{Z})\to 0$, the last map being surjective by exactness at the final term, gives $\mathrm{\mid }{B}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }\mathrm{Ext}(A,\mathbb{Z})\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }A\mathrm{\mid }$ by step 2.1. Hence $\mathrm{\mid }{H}_1(M)\mathrm{\mid }=(\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }A\mathrm{\mid }{)}^2$. The injection $A\hookrightarrow \mathrm{Ext}(B,\mathbb{Z})$ makes $\mathrm{\mid }A\mathrm{\mid }$ divide $\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }\mathrm{Ext}(B,\mathbb{Z})\mathrm{\mid }$, so $\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }A\mathrm{\mid }$ is a positive integer and $\mathrm{\mid }{H}_1(M)\mathrm{\mid }$ is a square. Choice enters only through [F4] and the cited inputs.

The constraint map $\mathrm{\Phi }(F)=F^{T}F-I_k$ takes values in symmetric $k$-by-$k$ matrices. Its derivative at an orthonormal frame is $A\mapsto F^{T}A+A^{T}F$, which is onto: for symmetric $B$, choose $A=FB\mathrm{/}2$. Thus [F2] makes $V_k({\mathbb{R}}^N)$ an embedded smooth manifold. Every based continuous loop can be represented by a smooth based loop: first reparametrize it to be constant on an arc about the basepoint, using a degree-one circle reparametrization based-homotopic to the identity, then use [F3] relative to a closed smaller arc. A continuous disk filling a smooth boundary loop can likewise be made smooth while keeping its boundary values: first compress the original filling radially into a smaller disk and use the boundary loop, constant in the radial variable, on the remaining annulus. Extend this map outside the unit circle by the same radial-constant formula and apply [F3] on ${\mathbb{R}}^2$ relative to a closed exterior annulus contained in its smooth region. Restrict to the disk. The formula is smooth near the unit circle, so no manifold-with-boundary version of approximation is needed. These operations use [A1] only through [F3].

We record the needed explicit complement construction. Let $P(z)$ be any smooth family of orthogonal projections of fixed rank on a disk, and set $P_z(t)=P(tz)$ for $0\le t\le 1$. Define $A_z(t)={\dot{P}}_z(t)P_z(t)-P_z(t){\dot{P}}_z(t)$, and solve $U_z^{\mathrm{\prime }}=A_z{U}_z$, $U_z(0)=I_N$ using [F4]. The coefficient is skew symmetric, so differentiating $U_z^T{U}_z$ makes it constant, equal to $I_N$. Differentiating $P_z^2=P_z$ gives $P_z{\dot{P}}_z{P}_z=0$ and ${\dot{P}}_z{P}_z+P_z{\dot{P}}_z={\dot{P}}_z$, hence $[A_z,P_z]={\dot{P}}_z$. Therefore $Q_z=U_{z}P(0)U_z^T$ and $P_z$ solve the same linear matrix equation $Q_z^{\mathrm{\prime }}=[A_z,Q_z]$ with the same initial value. Uniqueness, applying [F4] in the coordinate array of matrices, gives $Q_z=P_z$. Consequently transporting an orthonormal basis of $\mathrm{im}P(0)$ by $U_z(1)$ gives an orthonormal basis of $\mathrm{im}P(z)$. These transported vectors depend smoothly on $z$ by [F4], including at $z=0$ since $P(tz)$ is jointly smooth there. Smoothness up to the disk boundary follows by extending the smooth projection a little beyond that boundary. This construction uses unique solutions, not a choice of a solution for each parameter.