Homotopy excision for a single relative cell layer

Each of these open cells is open in $X$, since all its attaching boundary is in $C$ and no other relative cell attaches to its interior. Give it the coordinate chart $e^k\cong {\mathbb{R}}^k$ from its supplied characteristic map, using $u\mapsto u\mathrm{/}(1-\mathrm{\parallel }u\mathrm{\parallel })$ on the open disk. Every closed coordinate ball is compact and hence closed in the Hausdorff CW space. For any map $f:I^d\to X$ one can make the following local modification in a selected cell, without changing any point mapped outside that cell. Put $A_0=f^{-1}(B_1)$ and $P_0=f^{-1}(B_2)$ for its coordinate balls. If $A_0$ is empty, leave $f$ unchanged and take a small coordinate cube about zero missed by its image. Otherwise compactness separates $A_0$ from the closed complement of $f^{-1}(\mathrm{int}{B}_2)$ by a positive distance, and gives uniform continuity of the coordinate map on $P_0$. These facts follow by the finite-subcover argument in [F2], without selecting an infinite family of neighborhoods.

Choose a finite cubical mesh of $I^d$ with diameter small enough that the union $K_2$ of cubes meeting cubes that meet $A_0$ lies in $P_0$, and the coordinate oscillation of $f$ on each such cube is less than $1\mathrm{/}4$. Let $K_1$ be the union of cubes meeting $A_0$. Triangulate the cubes by successively coning their faces from their centers. Let $g$ interpolate $f$ affinely on these simplices, and let $\theta$ be the affine function with value one on vertices in $K_1$ and zero at the other vertices of $K_2$. Then $\theta =1$ on $K_1$ and zero on the relative boundary of $K_2$. The homotopy $(1-t\theta )f+t\theta g$ on $K_2$, unchanged elsewhere, is continuous by closed pasting, stays within the selected open cell, and is fixed outside its inverse image. Its endpoint $f^{\mathrm{\prime }}$ is affine on each simplex of $K_1$. Outside $K_1$ its image misses $B_{3\mathrm{/}4}$: on a simplex meeting the complement of $K_1$, take a point whose original image has norm greater than one; all its vertex images, and their convex interpolations, are within $1\mathrm{/}4$ of that image. This finite estimate makes no comparison between $d$ and $k$.

For any finite choice of interior points $P,Q$ in the indicated relative cells, radially deform each punctured characteristic disk onto its boundary as follows. If the removed point has disk coordinate $q$, then for $u\mathrm{\ne }q$ the ray $q+s(u-q)$ meets the boundary at the unique positive parameter $\lambda (u)$; solving its quadratic equation gives a continuous function with $\lambda (u)\ge 1$ and $\lambda (u)=1$ on the boundary. The formula $q+((1-t)+t\lambda (u))(u-q)$ stays in the punctured disk and fixes its boundary. Use it simultaneously on the finitely many selected disks and the identity on $C$. The attachment prescriptions agree on every boundary, and [F2]'s quotient-times-interval theorem makes the descended deformation continuous. It gives deformation retractions $X\setminus P\to B$, $X\setminus Q\to A$, and $A\setminus P\to C$, fixing the named target subspaces. The target $C$ need not be finite: it is simply the unchanged summand of the attachment quotient. Finite point sets are closed in the Hausdorff CW space, so restricting the quotient over their open complement is valid. The $Q$ deformation restricts on $X\setminus (P\cup Q)$ to a deformation retraction onto $A\setminus P$; it fixes $P$'s complement in $A$ throughout. Thus $C\hookrightarrow X\setminus (P\cup Q)$ and $A\hookrightarrow X\setminus Q$ are weak equivalences by [F4].

Among the finitely many affine maps on the simplices of $K_1$, discard their images of rank less than $k$ by choosing a small closed coordinate cube $\mathrm{\Delta }$ with nonempty interior inside $B_{1\mathrm{/}2}$ disjoint from all those images. Such a cube exists: each deficient image lies in a proper affine hyperplane. For the finitely many nonzero normals $a_j$, choose $v=(1,t,\dots ,t^{k-1})$ with every $a_j\cdot v\mathrm{\ne }0$, avoiding the finitely many roots of the resulting nonzero polynomials. A short line segment parallel to $v$ inside the ball meets each affine hyperplane at most once, so it contains a point outside their finite union. The positive distance from that point to the closed finite union permits the required small cube. On each remaining simplex $\sigma$, the restriction to its affine hull has rank $k$. Therefore for every $z\in \mathrm{\Delta }$, $\sigma \cap (f^{\mathrm{\prime }}{)}^{-1}(z)$ is compact and convex, given by finitely many linear equations and inequalities, and lies in an affine space of dimension $\dim \sigma -k\le d-k$. The preimage of $\mathrm{\Delta }$ itself is also a finite union of compact convex polyhedra on which $f^{\mathrm{\prime }}$ is affine. If $d<k$ all simplex images were deficient, so this preimage is empty. Apply steps 1.1–3.1 successively in the finitely many relative cells. Each modification stays inside its selected cell and fixes its complement, so the preceding other-cell data are retained. Denote the final map again by $f$. These preliminary homotopies preserve every prescribed subcomplex-valued face and every face constant at $c$.

It follows from [F3] and step 2.2 that the inclusions of pairs $(A,C)\to (X\setminus Q,X\setminus (P\cup Q))$ and $(X,B)\to (X,X\setminus P)$ induce bijections on every positive relative homotopy set, and isomorphisms in group degrees. The resulting square with horizontal inclusions commutes, since all four maps are inclusions. All deformations fix $c$. These are the two vertical comparisons for the graph deformation.

Suppose $1\le d\le a+l-2$, and let $q$ be any point of the interior of the cube ${\mathrm{\Delta }}_B$ selected in the $l$-cell. Its inverse image is a finite union of compact convex sets contained in affine subspaces of dimension at most $d-l$, by step 3.1; it is empty if $d<l$. Let $\pi :I^d\to I^{d-1}$ forget the last coordinate and set $T={\pi }^{-1}(\pi (f^{-1}(q)))$. Each of the finitely many pieces of $T$ is contained in an affine subspace of dimension at most $d-l+1$; projection cannot increase the dimension of an affine span, and restoring one coordinate increases it by at most one. On each affine simplex describing $f^{-1}({\mathrm{\Delta }}_{\alpha })$, the image of its intersection with $T$ lies in an affine subspace of dimension at most $d-l+1<a\le k_{\alpha }$. There are finitely many such spans. The same finite-hyperplane argument as step 3.1 gives $p_{\alpha }\in \mathrm{int}{\mathrm{\Delta }}_{\alpha }$ outside all these images. Consequently $\pi (f^{-1}(p_{\alpha }))$ is disjoint from $\pi (f^{-1}(q))$ for every $\alpha$. This uses only the dimension of affine spans; no transversality theorem is assumed. Put $P=\{p_{\alpha }\}$ and $Q=\{q\}$.

For a relative representative $f:(I^d,\mathrm{\partial }{I}^d,J)\to (X,B,c)$, the compact set $f^{-1}(q)$ misses $J$. Thus its projection $E\subset I^{d-1}$ misses $\mathrm{\partial }{I}^{d-1}$, and its last coordinates have a maximum less than one. Its projection is disjoint from the compact set $D_P={\bigcup }_{\alpha }\pi (f^{-1}(p_{\alpha }))$ by step 4.1. If $E$ is nonempty, choose $h<1$ larger than all those last coordinates and a continuous function $\eta :I^{d-1}\to [0,1]$ equal to one on $E$ and zero on a neighborhood of $D_P\cup \mathrm{\partial }{I}^{d-1}$. Explicitly the two compact sets have positive distance when the second is nonempty; choose a positive $ϵ$ smaller than that distance and put $\eta (x)=\max (0,1-\mathrm{dist}(x,E)\mathrm{/}ϵ)$. If the second set is empty any positive $ϵ$ works. Set $\phi =h\eta$. If $E$ is empty set $\phi =0$. Every $q$-preimage lies strictly below the graph of $\phi$; every $p_{\alpha }$-preimage lies above it, because its projection has $\phi =0$ and its last coordinate is positive, the bottom face mapping to $B$. Also $\phi =0$ on $\mathrm{\partial }{I}^{d-1}$ and $\phi <1$ everywhere.

Define $f_t(x,s)=f(x,t\phi (x)+(1-t\phi (x))s)\mathrm{.}$ The formula preserves the top and side faces at $c$. On the bottom face it avoids $P$ for every $t$, by the graph inequalities in step 5.1. At $t=1$ the whole image avoids $Q$. Therefore this is a homotopy of relative representatives in $(X,X\setminus P,c)$ from the original representative to one lying in $(X\setminus Q,X\setminus (P\cup Q),c)$. It need not be a homotopy in $(X,B)$; the comparison in step 3.2 accounts for this change of subspace.

Now let $1\le i\le a+l-2$ and start with a representative of ${\pi }_i(X,B,c)$. Perform steps 1.1–6.1 with $d=i$. The preliminary homotopies do stay in $(X,B,c)$, and the final representative lies in the lower-left pair of step 3.2. Its class therefore comes from a unique class of $(A,C,c)$ under the left vertical bijection. Commutativity and the right vertical injectivity in step 3.2 show that this class maps to the original class in $(X,B,c)$. This proves surjectivity throughout this range.

For injectivity let $w_0,w_1:(I^i,\mathrm{\partial }{I}^i,J)\to (A,C,c)$ have a relative homotopy in $(X,B,c)$, with parameter $v\in I$, and suppose $1\le i<a+l-2$. Regard this homotopy as a map of a $(i+1)$-cube, ordered as $(x,v,s)$ with $x\in I^{i-1}$ and $s$ the last relative coordinate. Apply steps 1.1–4.1 with $d=i+1\le a+l-2$. The preliminary homotopies can change $w_0,w_1$, but only through relative maps in $(A,C,c)$: modifications in the $B$-cell fix their whole images, and modifications in the $A$-cells fix their boundary images in $C$. The $q$-preimage projects in the $(x,v)$ coordinates away from $\mathrm{\partial }{I}^{i-1}\times I$ and away from $v=0,1$, since the endpoints lie in $A$. Its $s$-coordinate is bounded below one. Use the distance formula of step 5.1 with this additional closed endpoint set in the zero set to obtain $\phi (x,v)=0$ at $v=0,1$ as well as on the side boundary. The same graph reparametrization in $s$ then leaves the two modified endpoint cubes unchanged, removes the $q$-preimage, and keeps every bottom face outside $P$. It produces a homotopy of the two modified endpoint cubes in $(X\setminus Q,X\setminus (P\cup Q),c)$. The left vertical bijection in step 3.2 implies their equality in ${\pi }_i(A,C,c)$, hence equality of the original classes as well. This argument proves injectivity even for pointed relative degree one.

This proves the one-$B$-cell result. For finitely many $B$ cells, remove a cell $e^l$ of maximum relative dimension and put $B^{\mathrm{\prime }}=B\setminus e^l$, $X^{\mathrm{\prime }}=A\cup B^{\mathrm{\prime }}$. The complement is a subcomplex: boundaries of other relative cells have smaller dimension, and boundaries of cells in $C$ stay in $C$. Regard the new common subcomplex as $B^{\mathrm{\prime }}$, the new first side as $X^{\mathrm{\prime }}=B^{\mathrm{\prime }}\cup \bigcup e_{\alpha }^{k_{\alpha }}$, and the second side as $B=B^{\mathrm{\prime }}\cup e^l$. The first side's relative cell boundaries still lie in $C\subseteq B^{\mathrm{\prime }}$. The one-cell result says that ${\pi }_i(X^{\mathrm{\prime }},B^{\mathrm{\prime }},c)\to {\pi }_i(X,B,c)$ is bijective for $1\le i<a+l-2$ and surjective for positive $i=a+l-2$. Since $l\ge b$, this is a bijection throughout $1\le i<a+b-2$ and a surjection at positive $i=a+b-2$. Repeat finitely, stopping at $(A,C)$. The composite is the required inclusion map, so composition gives the claimed ranges.

If there are no $B$ cells the map is the identity. If there are no $A$ cells, $A=C$ and $X=B$, so both relative sets are singletons: a relative cube entirely in its subspace contracts to $c$ by increasing its last coordinate to one, preserving $J$. In the argument, an empty $q$-fibre permits $\phi =0$, and empty $p$-fibres cause no restriction. For $i=1$ the projected cube is a point, with empty boundary; the formulas still apply. If $a+b-2=0$ there is no positive endpoint degree and both asserted ranges are empty; no relative degree-zero object is used. Nonregular attaching maps are allowed because radial deformations fix their disk boundaries. Every chosen cube, point, mesh, cutoff and cell-removal order belongs to a finite collection; no selection is made on all of $C$. The inequalities in steps 4.1 and 7.2 explain the surjective endpoint and the one-degree-smaller injective range. This proves all assertions choice-free.