Hurewicz Whitehead Freudenthal and Cw Approximation

The image $L$ is compact in the open-cover sense. Indeed, the inverse images of any open cover of $L$ cover $K$; a finite subcover of $K$ yields a finite subcover of $L$ by the same covering members. This argument does not impose a metric on the target. Since $X$ is Hausdorff, [F2] makes $L$ closed in $X$.

Every nonempty compact subset $T\subseteq {\mathbb{R}}^d$ has a uniquely specified lexicographically least point. For $d\ge 1$, minimize its first coordinate using [F5] and restrict to the minimum level set. That set is nonempty, closed in $T$ and compact by [F3]. Minimize the next coordinate on it and continue through the finite ordered coordinate set. After $d$ steps all coordinates are fixed, and the nonempty final set is a singleton. Each minimum value and each level set is unique; no minimizing point is selected until the final singleton. For $d=0$, the sole possible nonempty subset of ${\mathbb{R}}^0$ is already a singleton. This is a finite prescription defined for every such $T$, not a family of arbitrary existential choices.

Let $e$ be a positive-dimensional open cell meeting $L$, with characteristic map ${\chi }_e:D^d\to X$. For $r\ge 0$ let $B_{d,r}$ be the concentric closed ball of radius $1-1\mathrm{/}(r+2)$ in $D^d$. These balls exhaust its interior. Thus some $B_{d,r}$ meets ${\chi }_e^{-1}(L)$; let $r_e$ be the least such integer. The set $T_e=B_{d,r_e}\cap {\chi }_e^{-1}(L)$ is a nonempty compact subset of ${\mathbb{R}}^d$: it is closed in the compact ball by step 1.1, continuity and [F3], [F4]. Let $v_e$ be its uniquely specified point from step 1.2 and set $x_e={\chi }_e(v_e)\in L\cap e$. For an occupied zero-cell use that point itself. The least integer, the finite sequence of coordinate minima, and the supplied characteristic map specify $x_e$ uniquely for every occupied cell; the resulting function is defined by this formula on the set of occupied cells.

Put $S=\{x_e:e\cap L\mathrm{\ne }\mathrm{\varnothing }\}$. Distinct occupied cells give distinct points, since their interiors are disjoint. For every subset $T\subseteq S$ and every closed cell $\bar{a}$, closure finiteness in [F1] says that $\bar{a}$ meets only finitely many open cells. Hence $T\cap \bar{a}$ is finite, with at most one point from each of those cells. A finite subset of a Hausdorff space is closed: singleton complements are open by the Hausdorff separation axiom, and finite unions of closed sets are closed. The weak topology in [F1] now makes $T$ closed in $X$. In particular $S$ is closed in $X$, hence closed in the compact $L$. By [F3], $S$ is compact.

For $s\in S$, the set $S\setminus \{s\}$ is closed in $X$ by step 3.1. Its complement intersects $S$ in $\{s\}$, so $S$ is discrete. Its singleton cover is an open cover of $S$ and therefore has a finite subcover. Thus $S$ is finite, without first extracting a countably infinite subset from an arbitrary infinite set. The bijection $e\mapsto x_e$ from occupied cells to $S$ shows that only finitely many cells meet $L$.

If there are occupied cells, start with their finite set. Add every cell meeting the boundary of a cell already in the set, and repeat downward in dimension. At each stage only finitely many cells are added by closure finiteness [F1]. A cell boundary lies in the preceding skeleton, so the dimensions strictly decrease along every newly required boundary chain. The finite starting set has a maximum dimension $N$, and after at most $N$ such downward stages no more are required. The union of these cells contains the entire closure of each member, hence is a finite CW subcomplex by [F1]. It contains $L$, because every point of $L$ belongs to an occupied open cell.

If $K$ or $L$ is empty, the empty subcomplex suffices and no minima are taken. For a point image, the closure process starts at its one occupied cell; it need not itself be a zero-cell. Zero-dimensional cells and the zero-dimensional Euclidean coordinate space were handled without a norm or empty-coordinate minimum. The radii in step 2.1 are strictly between zero and one and approach one, so all chosen preimages lie in cell interiors and no boundary point is mistaken for a point of that open cell. Nonregular characteristic maps cause no problem, since only their interior restrictions are used for the selected points. Steps 1.2 and 2.1 specify every selection uniquely, while steps 4.1 and 5.1 use compactness and finite closure operations; no AC is used anywhere.

Each $B_a\subset e$ is compact by [F2] and the coordinate homeomorphism, hence closed in $Z$. Its open-ball interior is open in $Z$: the preimage in the attachment disk is open away from its boundary and the preimage in $W$ is empty. Consequently $P=f^{-1}(B_2)$ and $A=f^{-1}(B_1)$ are compact subsets of $I^n$, and $C=I^n\setminus f^{-1}(\mathrm{int}{B}_2)$ is closed and disjoint from $A$. If $A$ is empty, $f$ already misses the coordinate origin; retain $f$ and proceed to the radial construction below.

Suppose $A\mathrm{\ne }\mathrm{\varnothing }$ and $n\ge 1$. There is a positive $d$ such that every point within distance $d$ of $A$ lies outside $C$. To see this without selecting a neighborhood at every point, take all pairs $(a,r)$ with $a\in A$, $r>0$, and the relative ball $B(a,2r)$ disjoint from $C$. Their balls $B(a,r)$ cover $A$. Compactness gives finitely many such pairs, and the minimum of their radii is a suitable $d$: if $x$ is within $d$ of $a^{\mathrm{\prime }}\in A$, choose a member containing $a^{\mathrm{\prime }}$ and use the triangle inequality. If $C$ is empty any $d>0$ works. Likewise the coordinate map $f:P\to {\mathbb{R}}^k$ is uniformly continuous. For all pairs $(a,r)$, $a\in P$, whose relative radius-$2r$ ball has images within $1\mathrm{/}8$ of $f(a)$, the radius-$r$ balls cover $P$. A finite subcover and the minimum radius give $\delta >0$ such that $x,y\in P$ and $\mathrm{\parallel }x-y\mathrm{\parallel }<\delta$ imply $\mathrm{\parallel }f(x)-f(y)\mathrm{\parallel }<1\mathrm{/}4$. Only finite subcovers and finite minima were used.

Subdivide $I^n$ into a finite uniform grid of closed cubes of diameter $h<\min (d\mathrm{/}3,\delta )$. Let $K_1$ be the union of cubes meeting $A$ and $K_2$ the union of cubes meeting $K_1$. Every point of $K_2$ is within $2h<d$ of $A$, so $K_2\subseteq P$. A point on the relative boundary of $K_2$ cannot belong to $K_1$: otherwise a cube outside $K_2$ containing that boundary point would meet $K_1$ and would have been included. Triangulate the grid compatibly by first leaving its vertices, then coning each face from its center over the already triangulated boundary, in increasing dimension. The resulting finite simplices have diameter at most $h$ and $K_1,K_2$ are subcomplexes of this triangulation.

On $K_2$ let $g$ be the affine interpolation of the coordinate values of $f$ at these finitely many vertices. On every simplex barycentric coordinates are unique, are nonnegative and sum to one; the formulas agree on faces, hence define a continuous $g$. Define the piecewise affine function $\varphi$ to be one at vertices in $K_1$ and zero at the other vertices of $K_2$. Then $\varphi =1$ on $K_1$ and $\varphi =0$ on the relative boundary of $K_2$, by step 3.1. The formula $f_t(x)=(1-t\varphi (x))f(x)+t\varphi (x)g(x)(x\in K_2)$ takes its values in the coordinate cell. Outside $K_2$ retain $f$. The two prescriptions agree on the boundary and paste continuously on $K_2\times I$ and $\overline{I^n\setminus K_2}\times I$, a finite closed cover. Since $K_2\subseteq f^{-1}(e)$, the homotopy fixes $f^{-1}(W)$. Write $F=f_1$; on $K_1$ it is the finite piecewise affine map $g$.

The image under $F$ of $I^n\setminus K_1$ misses the coordinate ball of radius $3\mathrm{/}4$. Outside $K_2$ it misses $B_1$ by definition of $K_1$. For a point $x\in K_2\setminus K_1$, take a simplex $\sigma$ containing it. This simplex is not contained in $K_1$; fix a point $z\in \sigma \setminus K_1$. Then $\mathrm{\parallel }f(z)\mathrm{\parallel }>1$, while uniform continuity and the diameter bound in step 3.1 give $\mathrm{\parallel }f(y)-f(z)\mathrm{\parallel }<1\mathrm{/}4$ for all $y\in \sigma$. Convexity puts $g(x)$ and $F(x)$ in the same radius-$1\mathrm{/}4$ ball about $f(z)$, so $\mathrm{\parallel }F(x)\mathrm{\parallel }>3\mathrm{/}4$. This estimate applies only to points mapped into $e$; points mapped to $W$ already miss all its coordinate balls.

A finite union of affine subspaces of dimension at most $n<k$ cannot fill a nonempty open ball in ${\mathbb{R}}^k$. Here is a finite algebraic verification. For each subspace its spanning vectors have rank less than $k$; row elimination gives a nonzero vector $a$ orthogonal to them, so the subspace lies in a hyperplane $a\cdot x=b$. For the finite list of nonzero normals, substitute $v(t)=(1,t,\dots ,t^{k-1})$. Each $a\cdot v(t)$ is a nonzero polynomial and has finitely many roots: division by $t-t_0$ at a root and induction on degree prove that assertion. Choose an integer $t$ outside the finite union of root sets. The line $s\mapsto sv(t)$ meets each affine hyperplane in at most one point. An interval of sufficiently small $s$ lies in the specified ball centered at zero and contains a point outside that finite list. Thus for the finitely many affine images of simplices of $K_1$, some $p\in \mathrm{int}{B}_{1\mathrm{/}2}$ is omitted by $F(K_1)$; step 5.1 shows that $p$ is omitted by all of $F$. All the linear algebra and selections here are finite.

The cases excluded from the mesh construction also give an omitted point. If $A$ is empty use the coordinate origin, as in step 1.1. If $n=0$, the domain is one point; if its image lies in $W$, the constant homotopy already solves the problem. Otherwise choose one of two fixed distinct points of $e$ unequal to that image, leaving $f$ unchanged. Thus in every case there is a map $F$ homotopic to $f$ rel $f^{-1}(W)$ and a point $p\in e\setminus F(I^n)$.

Let $q\in \mathrm{int}{D}^k$ be the unique characteristic preimage of $p$. For $u\in D^k\setminus \{q\}$ set $v=u-q$ and $\lambda (u)=\frac{-q\cdot v+\sqrt{(q\cdot v{)}^2+(1-\mathrm{\parallel }q{\mathrm{\parallel }}^2)\mathrm{\parallel }v{\mathrm{\parallel }}^2}}{\mathrm{\parallel }v{\mathrm{\parallel }}^2}\mathrm{.}$ This is the positive solution of $\mathrm{\parallel }q+\lambda v\mathrm{\parallel }=1$, by expanding the square. Since $q$ is interior and $u$ is in the disk, $\lambda \ge 1$, with equality for $u$ on its boundary. The homotopy $R_t(u)=q+((1-t)+t\lambda (u))(u-q)$ lies on the ray segment between $u$ and its boundary endpoint, stays in the convex disk, never equals $q$, and fixes the boundary. All formulas are continuous since $\mathrm{\parallel }v{\mathrm{\parallel }}^2>0$.

The attachment quotient restricted over $Z\setminus \{p\}$ is still quotient: this subset is open, its inverse image is saturated and open, and any set open in that inverse image is open upstairs, so the quotient test descends it. On its domain, the homotopy given by step 8.1 on the punctured disk and the identity on $W$ agrees on the attaching identifications. It descends continuously even with the ordinary product topology on time. Indeed, for any quotient $Q:E\to V$, a map $H:V\times I\to T$ continuous after $Q\times \mathrm{i}\mathrm{d}$ has a well-defined transpose; [F3] makes its composite with $Q$ continuous, the quotient test makes the transpose continuous, and [F3] makes $H$ continuous. Applied here, this proves a deformation retraction of $Z\setminus \{p\}$ onto $W$. Compose it with $F$ and concatenate with the first homotopy. The result ends in $W$ and fixes $f^{-1}(W)$.

No infinite family of witnesses has been selected. The grid and its triangulation are finite; neighborhood families were taken in their entirety before finite subcovers; omitted-point linear algebra involves only finitely many hyperplanes. The cases $n=0$ and $f^{-1}(B_1)=\mathrm{\varnothing }$ were treated separately. The inequality $n<k$ is used exactly to find proper affine hyperplanes; no equal-dimension claim is made. At times zero and one the stated endpoint maps follow from the explicit formulas. Boundary fibers of a nonregular attaching map remain fixed, so the quotient argument does not require their injectivity. This proves the claimed choice-free relative homotopy.

For any map $u:(D^n,S^{n-1})\to (Y,Y^{n-1})$, [F3] gives a finite target subcomplex $T$ containing its image. A cube and a Euclidean disk are homeomorphic as pairs: after centering the cube, the radial map sends a nonzero vector $v$ to $v\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_2$, with inverse $w\mathrm{\parallel }w{\mathrm{\parallel }}_2\mathrm{/}\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}$ and zero mapped to zero. Thus [F4] applies to disk domains as well. If $T$ has cells of dimension greater than $n$, take a cell of maximum dimension. Removing its interior leaves a subcomplex $W$, since no boundary of any remaining cell can meet that maximum-dimensional interior. Apply [F4] to move the disk off this cell. Its boundary remains fixed because its image is in $Y^{n-1}$ and misses that cell. Repeat in the smaller finite subcomplex until no cell of dimension greater than $n$ remains. The resulting homotopy is rel boundary and ends in $Y^n$. This is a finite argument with finite selections, including when the finite subcomplex has cells not met by $u$. For $n=0$ the boundary is empty and [F4] moves the one-point map to a vertex.

We will use the following continuity criterion. A function $H:Z\times I\to Y$ on a CW complex is continuous if its composite with every characteristic disk cylinder is continuous. First, its pointwise transpose to $C(I,Y)$ is well defined, since every point lies in a characteristic disk. By [F5] the transpose is continuous after every characteristic map. The latter maps are quotient onto their closed-cell images: each is a continuous surjection from a compact disk to a Hausdorff space and is closed, as compact images of closed disk subsets are closed. Thus the transpose is continuous on every closed cell. The weak topology [F2], applied to inverse images of closed subsets of $C(I,Y)$, makes it continuous on $Z$, and [F5] uncurries it. The compact-image and closedness argument, with no metric assumed on a closed cell, is also given explicitly in the proof of [F1]. The same criterion applies to subcomplexes.

Suppose a current map $f_{n-1}:X\to Y$ is cellular on $D_{n-1}$ and agrees with $f$ on $A$. On each $n$-cell outside $A$, apply step 1.1 to $f_{n-1}$ composed with its characteristic map. Its boundary is mapped into $Y^{n-1}$ by the induction hypothesis. The resulting disk homotopies fix all boundary fibers, so they descend and agree with the stationary homotopy on $D_{n-1}$. They give a homotopy on $D_n$, ending cellularly on $D_n$ and fixed on $A\cup X^{n-1}$. On a closed cell in $A$ it is constant; on each new $n$-disk it is the specified deformation; on lower cells it is constant. Step 1.2 proves continuity even when $A$ has arbitrarily high-dimensional cells. Extend this homotopy to $X$ using [F1] for the subcomplex $D_n$, and call its endpoint $f_n$. The extension is still fixed on $A\cup X^{n-1}$.

We verify the cylinder CW structure used in the remaining assertion. Give $I$ its two endpoint vertices and one open edge. The cells of $X\times I$ are $e^r\times \{0\}$, $e^r\times \{1\}$ and $e^r\times (0,1)$, with characteristic domains $D^r$ and $D^r\times I$. The last is a closed $(r+1)$-disk as a pair: center its interval coordinate and use the radial homeomorphism between the unit balls of the Euclidean norm and the norm $\max (\mathrm{\parallel }u{\mathrm{\parallel }}_2,\mathrm{\mid }s\mathrm{\mid })$, extending by zero at the origin. Their boundaries land in the union of lower-dimensional cells, and closure finiteness follows from that of $X$. These cells have exactly the ordinary product topology. Indeed the map from the disjoint union of characteristic disks onto $X$ is quotient by [F2] and the compact-Hausdorff quotient test in step 1.2. Its product with $I$ is quotient: transpose a proposed map out of the product by [F5], descend its transpose through the quotient, then untranspose. Applying this test to characteristic functions into the two-point space with opens $\mathrm{\varnothing },\{1\},\{0,1\}$ proves the assertion for open subsets, hence for the quotient topology itself. Thus the characteristic prisms test closed sets. To check the attachment topology on the $d$-skeleton $P_d$, suppose $C\subseteq P_d$ has closed preimage under each characteristic map of dimension at most $d$. Its intersection with each such closed cell is closed, by the compact-Hausdorff quotient test, hence closed in the whole product. In any other closed cell $Q$, closure finiteness gives finitely many cells of dimension at most $d$ meeting $Q$. The set $C\cap Q$ equals the intersection of $Q$ with the union of $C$ intersected with the closures of those finitely many cells. It is therefore closed in $Q$. The full weak topology now makes $C$ closed in the product. This proves both that $P_d$ is closed and that its topology is tested on its characteristic disks. Testing a map from $P_{d-1}$ and the $d$-disks is consequently exactly the cell-attachment quotient criterion. This verifies the CW topology, not just its set of cells.

If there are finitely many cells outside $A$, use [F6] to make the finitely many disk-deformation and HEP-extension selections required in step 2.1 at each stage, and stop at their maximum dimension $N$. Only finitely many stages and finite selections are required; the existence of each extension is [F1], regardless of the size of $A$. Concatenating the finitely many homotopies gives a homotopy rel $A$ ending in a map cellular on $D_N=X$. If there are no relative cells, use the constant homotopy of $f$, already cellular on $A=X$. Throughout the homotopy, points of $A$ retain their original images in $B$, so every time slice is a map of pairs.

For arbitrary relative cells assume [A1]. There is a set of all problems $(n,u)$ in step 1.1: continuous maps are subsets of the fixed sets $D^n\times Y$, and take the union over $n\in \mathbb{N}$. Each has a nonempty set of boundary-fixed homotopies with endpoint in $Y^n$, by step 1.1. There is likewise a set of all HEP extension problems that can occur in step 2.1: their subcomplex maps, prescribed homotopies and candidate extensions are subsets of fixed products formed from $X$, $I$ and $Y$, and [F1] says that each resulting set of candidate extensions is nonempty. AC supplies choice functions for both families. Using these two fixed functions at every characteristic disk and every HEP step makes the successor construction in step 2.1 specified. Apply [F6] to the state consisting of a stage number and a finite history of maps and homotopies; the collection of these histories is a set. Recursion over $n=0,1,\dots$ gives the maps $f_n$ and homotopies $H_n$ without another selection of a sequence of existential witnesses. The cell family may be arbitrary and dimensions unbounded. The HEP assertion [F1] is choice-free for each individual problem; the global selection of one extension for every problem used by the recursion is part of the stated use of AC.

Run $H_n$ on $[1-2^{-n},1-2^{-(n+1)}]$, rescaled linearly, beginning with $n=0$. For $x\in X^r$, every stage after $r$ fixes $x$, so define $g(x)=f_r(x)$ and put $H(x,1)=g(x)$. These prescriptions agree where skeleta overlap and at adjacent time endpoints. On the image of any characteristic $r$-disk the cylinder map consists of the finitely many stages through $r$ followed by the constant endpoint map. It is continuous, including at time one, by finite pasting. The criterion of step 1.2 therefore makes $H:X\times I\to Y$ continuous. Its endpoint sends $X^r$ into $Y^r$, and it fixes $A$ at every time. This proves the arbitrary-cell conclusion with its stated assumption.

Let $f_0,f_1$ be cellular and let $K:X\times I\to Y$ be a homotopy rel $A$ between them. In the CW structure of step 2.2 the subspace $E=(X\times \{0,1\})\cup (A\times I)$ is a subcomplex. The restriction $K{\mathrm{\mid }}_E$ is cellular: on endpoint cells this is the cellularity of $f_0,f_1$; on $e^r\times (0,1)$ for a cell of $A$, it is the fixed value $f_0(e^r)\subseteq Y^r\subseteq Y^{r+1}$. Apply the first assertion, proved above, to the pair $(X\times I,E)$ with target pair $(Y,Y)$. It gives a cellular map $K^{\mathrm{\prime }}$ agreeing with $K$ on $E$. Hence $K^{\mathrm{\prime }}$ is the required homotopy with exactly the prescribed endpoints and constant track on $A$. There is precisely one relative cell $e^r\times (0,1)$ for each cell outside $A$; therefore the finite and arbitrary choice clauses apply exactly as stated.

Empty $X$ or zero relative cells give the constant construction, and zero-cells were handled in step 1.1 without a boundary condition. An infinite-dimensional $A$ is harmless in the finite clause because its entire homotopy is fixed. The arbitrary concatenation is checked at its accumulating endpoint on every characteristic disk, not only pointwise. Each intermediate map sends $A$ into $B$, and $K^{\mathrm{\prime }}$ also does so because it is fixed on $A\times I$. No claim is made that all slices of $K^{\mathrm{\prime }}$ preserve every skeleton; its product-cell statement is the one established in step 5.1. This proves every assertion with the indicated choice boundary.

The Euclidean sphere and disk are closed and bounded, as are their products with $I=[0,1]$, regarded as subsets of a finite-dimensional Euclidean space. Thus [F3] makes them compact. $D^0$ is a singleton and $S^0$ consists of two points, which are compact by taking one covering member for each of finitely many points. Their products with $I$ are a single interval or two intervals, also closed bounded Euclidean subsets after the usual embeddings.

The sphere has a finite $n$-dimensional CW structure with its designated basepoint as a vertex. One concrete construction for $n\ge 1$ attaches one $n$-disk to a point by collapsing its entire boundary. To identify the quotient with $S^n$, send $u\in D^n$ of norm $r>0$ to $(\sin (\pi r)u\mathrm{/}r,\cos (\pi r))$, and send zero to the north pole. This is continuous at zero since $\mathrm{\mid }\sin (\pi r)\mathrm{\mid }\le \pi r$, is constant at the south pole on the boundary, and is a bijection from the interior to the complement of that pole. The induced continuous bijection from the compact quotient to the Hausdorff sphere is a homeomorphism: the quotient is compact by pulling open covers back to the disk. A closed subset is compact by [F4], its image is compact by the same cover argument, and that image in the Hausdorff sphere is closed by [F4]. Identify the pole with the designated sphere basepoint. This realizes the usual two-cell based sphere.

Apply [F1] directly to $f$, and separately to the specified $H$. It gives finite subcomplexes containing their images. The finite subcomplex for $H$ automatically contains both endpoint images, since the endpoints are restrictions of $H$. This does not require selecting representatives of a family of homotopy classes, nor choosing simultaneous finite subcomplexes for such a family.

For a given based representative $f:(S^n,\ast )\to (X,x_0)$, its restriction to the source vertex is cellular because $x_0\in X^0$. Apply the finite-relative-source clause of [F2] to $(S^n,\ast )$ and $(X,\{x_0\})$. It produces a based homotopy to $g$ that is cellular. The source has dimension $n$, so $g(S^n)\subseteq X^n$. Since the homotopy fixes the basepoint, $g$ represents precisely the original based class. This applies to each class by beginning with any one representative; it asserts existence for each class and does not select representatives simultaneously.

Each homotopy just obtained, being a specified map on $S^n\times I$, also satisfies step 2.1. A basepoint outside $X^n$ cannot belong to the image of a based map landing in $X^n$, so such a skeletal conclusion would be impossible and has not been asserted. The assertions about compact images have no basepoint restriction. Zero-dimensional compact domains were treated in step 1.1; the skeletal group assertion begins at $n=1$, and no ${\pi }_0$ group law is implied. Only choice-free [F1], [F3] and the expressly choice-free clause of [F2] have been used.

Attachment identifies only boundary points with earlier points, so it does not identify distinct earlier points and is injective on every new disk interior. Consequently the old and new open cells partition the underlying set, and the boundary of a cell of dimension $r$ lands in the union of cells of dimension less than $r$. The topologies specified by successive attachment quotients and the final weak attachment test are exactly the topology final with respect to $A$ and all the new disk maps: a function out of the union is continuous precisely when its composites with those maps are continuous, by the inverse-image test for open sets. Since $A$ itself has its characteristic-disk weak topology, all old and new characteristic disks together test continuity and closed sets on $Z$. For an old closed cell, its characteristic map is quotient because it is a continuous compact-to-Hausdorff surjection, so replacing the old closed-cell tests by disk tests is legitimate.

The inclusion of every earlier stage is a closed embedding. For one attachment step, if $C$ is closed in the earlier space, its inverse image in a new disk is contained in the boundary sphere and is closed there by continuity of the attaching map, hence closed in the disk. Thus $C$ remains closed after the step; the earlier topology is exactly its subspace topology since its inclusion is continuous by the quotient construction and every earlier closed set remains closed. The same proof for each subsequent step and then the final disk test shows that every closed subset of $A$ or $Z_k$ remains closed in $Z$. Taking $C=A$ or $C=Z_k$ also proves their closedness. The identical observation applies to an initial segment with finitely many stages.

We construct a continuous real function separating any two distinct points $x,y\in Z$. First do this on the given CW complex $A$, imposing the specified values only at those of $x,y$ that belong to $A$. On its zero-cells set its value to $-1$ at $x$ if $x$ is such a cell, to $1$ at $y$ if it is such a cell, and to zero otherwise. Suppose values on all lower-dimensional cells have been specified, compatibly and continuously on each characteristic disk. On an old characteristic $r$-disk of $A$ with $r\ge 1$, its boundary has a continuous prescribed function $b:S^{r-1}\to [-1,1]$. Indeed its attaching image meets finitely many lower-dimensional cells by closure finiteness in $A$; close this finite set downward in dimension. On each of these finitely many closed cells the already specified function is continuous, since the characteristic disk is a compact-to-Hausdorff quotient in the given space $A$. Finite closed pasting makes the function continuous on their union, and composition with the attaching map gives $b$. Define $h_0(u)=\mathrm{\parallel }u\mathrm{\parallel }b(u\mathrm{/}\mathrm{\parallel }u\mathrm{\parallel })$ for $u\mathrm{\ne }0$ and $h_0(0)=0$. This is continuous at zero since $\mathrm{\mid }{h}_0(u)\mathrm{\mid }\le \mathrm{\parallel }u\mathrm{\parallel }$, and it extends $b$.

If the open cell contains neither $x$ nor $y$, use $h_0$ on this disk. Otherwise their relevant interior preimages form a specified set $P$ of one or two distinct points. For each $a\in P$ put $c_a=-1$ for the preimage of $x$ and $c_a=1$ for that of $y$, and choose the explicitly defined radius ${ϵ}_a=\frac{1}{4}\min (\{1-\mathrm{\parallel }a\mathrm{\parallel }\}\cup \{\mathrm{\parallel }a-a^{\mathrm{\prime }}\mathrm{\parallel }:a^{\mathrm{\prime }}\in P,a^{\mathrm{\prime }}\mathrm{\ne }a\})>0.$ The closed balls of these radii are interior and pairwise disjoint. Set ${\beta }_a(u)=\max (0,1-\mathrm{\parallel }u-a\mathrm{\parallel }\mathrm{/}{ϵ}_a)$ and $h(u)=(1-{\sum }_{a\in P}{\beta }_a(u))h_0(u)+{\sum }_{a\in P}{\beta }_a(u)c_a\mathrm{.}$ At most one bump is nonzero, so this remains in $[-1,1]$, is continuous, agrees with $b$ on the boundary, and takes the required values at the marked points. All prescriptions are determined by the supplied disk coordinates and the two given points; no family of extensions has been selected.

Apply steps 2.2 and 3.1 to all old cells in each dimension and use [F3] to recurse on dimension. This yields a continuous $h_A:A\to [-1,1]$ by the known weak topology of $A$. No Hausdorffness of the newly constructed space has been used: all compact quotient tests here took place inside the original CW complex $A$. On the added zero-cells prescribe the marked value if relevant and zero otherwise. At each attachment stage, the boundary function on every new disk is now continuous by composing its supplied attaching map with the continuous function on the previous stage. Extend it by the same radial formula and the same explicit interior bumps of steps 2.2 and 3.1. The quotient test makes the extension continuous on that stage. Apply [F3] to this specified stage rule; the final test in step 1.1 makes the resulting $h:Z\to [-1,1]$ continuous. It has $h(x)=-1$ and $h(y)=1$. Inverse images of disjoint real neighborhoods separate $x,y$, proving Hausdorffness of $Z$ and of every truncated construction.

Each characteristic disk now maps compactly into a Hausdorff space, so its image is closed by [F4]. That image equals the closure of its open cell: it contains the cell and is closed, while continuity and density of the disk interior put the whole image in the cell closure. It is therefore a compact closed cell and its characteristic map is a closed quotient map. An old closed cell retains its old closure by step 2.1. A new one meets only its own open cell and the finitely many cells met by its attaching map. Hence closure finiteness holds. The diskwise closed-set test from step 1.1 is equivalent, via these quotient maps, to the closed-cell test (W).

For completeness, the skeleta carry their required attachment topology. Suppose $C$ is a subset of the $d$-skeleton whose preimage in every characteristic disk of dimension at most $d$ is closed. By step 5.1 its intersection with each corresponding closed cell is closed there, hence closed in $Z$. In any other closed cell $Q$, only finitely many cells of dimension at most $d$ meet $Q$. The set $C\cap Q$ equals $Q$ intersected with the union of $C$ intersected with the closures of those finitely many cells. It is closed in $Q$. The full weak topology makes $C$ closed in $Z$. Taking $C$ equal to the skeleton shows that it is closed, and the same argument shows its subspace topology is final for its characteristic disks. Testing on the previous skeleton and the $d$-disks is consequently precisely the quotient test for attaching its $d$-cells. The zero-skeleton is discrete, since every subset satisfies the same test. Thus all the filtration and topology conditions in [F2] hold.

Every old cell and every cell in an earlier stage has its whole closure in that stage, so step 2.1 identifies $A,Z_k$ with closed subcomplexes. The map-out assertion was proved directly in step 1.1 and places no separation condition on its target. Empty initial space, empty cell families, a single zero-cell, and zero stages all use the same quotient tests; zero-dimensional disks require no radial extension. The two-point separation construction only runs for distinct points, so singleton spaces are already Hausdorff. Every positive radius in step 3.1 is a minimum of a nonempty finite set of positive numbers, and bounded radial extension handles the origin. The only infinite procedure is the specified dimension recursion, not a selection of extensions. This proves the statements without choice.

Form $Z_0=P\bigsqcup \{v_x:x\in X\}$, with the new vertices discrete, and put $Q_0{\mathrm{\mid }}_P=q$, $Q_0(v_x)=x$. This is continuous because the disjoint pieces are open. All vertices of every later stage will be exactly those of $P$ and these new vertices. No path component or point in a component is selected. Give each sphere used below its finite CW structure with its designated basepoint a vertex. The based quotient model in [F4] gives this structure by one zero-cell and one top cell; $S^0$ is two vertices. A disk boundary and disk can use the corresponding finite CW pair structure.

Suppose $Z_{k-1}$ is CW and $Q_{k-1}:Z_{k-1}\to X$ is specified, for $k\ge 1$. Let ${\mathcal{E}}_k$ be the set of all pairs $(a,b)$ where $a:S^{k-1}\to Z_{k-1}$ is cellular and $b:D^k\to X$ is continuous, with $b{\mathrm{\mid }}_{S^{k-1}}=Q_{k-1}\circ a\mathrm{.}$ For $k=1$, cellular means that the two boundary points go to vertices. Attach one labeled $k$-disk for every member of ${\mathcal{E}}_k$, using $a$ as its attaching map, and define $Q_k$ on this disk to equal its stored map $b$. It agrees with $Q_{k-1}$ on the boundary, so the attachment quotient makes $Q_k$ continuous. These are sets: each map is a subset of the relevant domain-codomain Cartesian product; continuity, cellularity and the boundary equation cut out subsets of their power sets. Labels distinguish different extension data even when their boundary maps coincide.

Every attaching image in step 2.1 meets finitely many cells by [F3], and it lies in $Z_{k-1}^{k-1}$ because $a$ is cellular. Thus [F2] proves that $Z_k$ is CW with the preceding stage a closed subcomplex. This proves the induction assertion needed to make the next stage legitimate. The construction of its quotient topology, labeled cells and stored map is specified by the preceding data, rather than chosen from possible extensions. Apply [F6] to the finite histories of these constructions (and use a fixed default value on invalid histories) to produce all stages. Taking their union with the weak attachment topology gives a CW complex $Z$ by [F2], and their compatible maps give a continuous $Q:Z\to X$ extending $q$. Every stage is a closed subcomplex of $Z$. The number of cells can grow with $k$; Replacement in [F6] is precisely what collects this set-sized sequence.

Each point of $Z$ can be joined to a vertex. In a positive-dimensional open cell, use its interior characteristic preimage and a line segment to a boundary point of the disk; the image is a path ending in a lower-dimensional cell. Repeat in that cell until dimension zero is reached. This terminates after finitely many decreases, so requires only finitely many existential choices for one specified point. Zero-cells are already vertices. For any vertices $u,v$ whose images can be joined by a path $b:I\to X$, their endpoint map is a cellular $a:S^0\to Z_0$ and $(a,b)\in {\mathcal{E}}_1$, so its attached edge joins $u,v$ in $Z$. Every component of $X$ is met by some $v_x$ (indeed every point is met). If two points of $Z$ have images in the same component, join each to a vertex as above, compose their image paths with a connecting path in $X$, and use the corresponding edge to join the vertices. The original two points are then in the same component of $Z$. Conversely, $Q$ sends any connecting path to a connecting path. This proves the bijection on path components without selecting a vertex for every component simultaneously.

Fix any vertex $v\in Z$ and $n\ge 1$. A based class in ${\pi }_n(X,Q(v))$ has, by [F4] and the cube-disk radial homeomorphism, a representative $b:D^n\to X$ constant at $Q(v)$ on its boundary. The constant map $a:S^{n-1}\to \{v\}\subset Z_{n-1}$ is cellular, so this pair occurs in ${\mathcal{E}}_n$. The characteristic disk of its attached cell has boundary constantly $v$; it therefore descends to a based sphere map into $Z_n$, whose composite with $Q$ is the given representative. Descent is continuous by the quotient definition, and the chosen identification $D^n\mathrm{/}\mathrm{\partial }{D}^n\cong S^n$ is the same for the original and lifted representatives. Thus $Q_{\ast }$ is onto at every vertex in every positive degree.

To prove injectivity, let $u:(S^n,\ast )\to (Z,v)$ have nullhomotopic composite with $Q$. Apply the finite-source clause of [F1] to obtain a based homotopy from $u$ to a cellular $a:S^n\to Z$. Its image lies in $Z^n$, which is contained in $Z_n$: all cells added after stage $n$ have higher dimension, while every old cell of $P$ was present initially. Since $Z_n$ embeds with its subspace topology, $a$ is a continuous cellular map into $Z_n$. The homotopy composed with $Q$ followed by the specified nullhomotopy gives a based nullhomotopy of $Qa$. It defines a disk map $b:D^{n+1}\to X$ extending $Qa$: collapse the terminal sphere of $S^n\times I$ to obtain its cone, identified with the disk by $(z,t)\mapsto (1-t)z$. The quotient is compact by pulling covers back to the compact sphere cylinder. Its closed subsets are compact and their images in the Hausdorff disk are closed by [F3], so this continuous bijection is a homeomorphism; the nullhomotopy therefore descends continuously to the disk. Consequently $(a,b)\in {\mathcal{E}}_{n+1}$. Its attached disk extends $a$ in $Z$. If $s_0$ is the marked boundary point of that disk, composing its characteristic map with $(z,t)\mapsto (1-t)z+t{s}_0$ for $z\in S^n$ contracts $a$ to $v$ while fixing the marked point. Hence $a$, and therefore $u$, is based-nullhomotopic. Postcomposition with $Q$ commutes with cubical concatenation, so [F4] makes $Q_{\ast }$ a homomorphism. Its kernel is trivial, which proves injectivity.

Now fix any $z\in Z$ and one path $c$ from a vertex $v$ to $z$, whose existence was proved in step 4.1. By [F5], transport gives isomorphisms from the groups based at $z$ to those based at $v$, and from the groups based at $Q(z)$ to those based at $Q(v)$. The square with the maps induced by $Q$ commutes: the radial-shell transport formula is a representative map on a cube using the original representative on its core and the path on its shell, so composing with $Q$ replaces the path by $Qc$ and the core by its composite. The bottom vertex-based map is an isomorphism by steps 4.2 and 4.3; conjugating it by these two transport isomorphisms proves the same for $Q_{\ast }:{\pi }_n(Z,z)\to {\pi }_n(X,Q(z))$. This chooses one path only after one basepoint has been fixed; it is not a simultaneous choice of paths for all points.

Taking $P=\mathrm{\varnothing }$ gives the asserted $\mathrm{\Gamma }X$ and $\gamma$. For a pair $(X,A)$ with prescribed approximation $q_A:P\to A$, apply exactly the same construction to its composite with the inclusion $A\hookrightarrow X$. The resulting $Q$ agrees literally with that composite on the unchanged subcomplex $P$, so it is a map of pairs with the required restriction, and steps 4.1–5.1 give its whole-space weak equivalence. This argument only uses that the prescribed source $P$ is CW; it does not require $A$ or $X$ to be Hausdorff. No mapping-cylinder theorem with a narrower category of spaces is used.

If $X=\mathrm{\varnothing }$, the existence of $q$ forces $P=\mathrm{\varnothing }$, and there are no vertices or extension data, so $Z=\mathrm{\varnothing }$; the component assertion and all basepoint assertions have their stated vacuous meanings. Empty extension sets at any stage simply attach no cells. For $n=1$, step 4.2 attaches loops from all actual path loops and step 4.3 attaches disks for their actual nullhomotopies; trivial kernel implies injectivity for this possibly nonabelian group as well. Degree zero was proved by actual connecting paths rather than by a group argument. Finite-source cellular approximation and canonical indexing of all data preserve the choice-free claim. The zero and endpoint conditions on every attached disk are its stored boundary equation, not additional extension assumptions.

These are finite CW pairs with the basepoint a vertex. For the absolute model use the sphere structure with one vertex and one $n$-cell, as realized by its collapsed-boundary disk. For the relative model use the sphere boundary structure on $R\cong S^{n-1}$ and then attach its one disk interior by the identity boundary map. The boundary has a vertex at the marked point; $n\ge 2$ ensures it is a positive-dimensional sphere. The wedge identifies only these vertices and retains the finite cell structures. Thus $B$ and $T=B\cup i_1(Q)$ are subcomplexes of $Y$, and $q_j,i_j$ are continuous maps of pairs by the wedge quotient test.

The nested subcomplexes $B\subseteq T\subseteq Y$ give a degreewise short exact sequence $0⟶C_{\ast }(T)\mathrm{/}{C}_{\ast }(B)⟶C_{\ast }(Y)\mathrm{/}{C}_{\ast }(B)⟶C_{\ast }(Y)\mathrm{/}{C}_{\ast }(T)⟶0.$ Indeed the singular simplices in a subspace are subsets of the basis of singular simplices in the larger space, so the first quotient includes injectively and its image is exactly the kernel of the last quotient. The boundaries preserve these subgroups. By [F2], the sequence $H_n(T,B)⟶H_n(Y,B)\stackrel{r}{\to }{H}_n(Y,T)$ is exact at its middle term. CW excision [F3] identifies $H_n(T,B)$ with the first copy $H_n(Q,R)$, using $T=i_1(Q)\cup B$ and intersection $i_1(R)$. It identifies $H_n(Y,T)$ with the second copy, using $Y=i_2(Q)\cup T$ and intersection $i_2(R)$. These identifications are induced by the actual inclusions.

On the original cube define the pinch by sending $(t_1,t^{\mathrm{\prime }})$ to the first copy represented by $(2t_1,t^{\mathrm{\prime }})$ when $t_1\le 1\mathrm{/}2$, and to the second represented by $(2t_1-1,t^{\mathrm{\prime }})$ when $t_1\ge 1\mathrm{/}2$. At the common face these are the collapsed basepoint in each model: coordinate-one end faces are contained in the collapsed set. In the relative model this uses $n\ge 2$, since the distinguished face is in the last coordinate, not coordinate one. Finite closed pasting and quotient descent give $c$. The boundary subset goes into $B$, so it is a map of pairs. The exact formulas in [F4], together with [F1]'s spherical transport, identify these positive rescalings with absolute and relative concatenation.

The map $(q_2{)}_{\ast }:H_n(Y,B)\to H_n(Q,R)$ factors through $r$, because $q_2(T)$ is contained in $R$. On $H_n(Y,T)$ its factor is inverse to the second excision inclusion in step 2.1: the composite $q_2{i}_2$ is the identity of $(Q,R)$. Thus $\ker (q_2{)}_{\ast }=\mathrm{im}(i_1{)}_{\ast }$. Both identities $q_j{i}_j=\mathrm{i}\mathrm{d}$ hold, and $q_j{i}_{3-j}$ is constant in $R$, hence induces zero on the relative chain quotient. For $z\in H_n(Y,B)$, subtract $(i_2{)}_{\ast }(q_2{)}_{\ast }z$ to get an element of this kernel, say $(i_1{)}_{\ast }a$. Applying $(q_1{)}_{\ast }$ shows $a=(q_1{)}_{\ast }z$. Therefore $z=(i_1{)}_{\ast }(q_1{)}_{\ast }z+(i_2{)}_{\ast }(q_2{)}_{\ast }z,$ and uniqueness follows by applying the two projections. This proves the relative splitting and its exact inverse, without asserting that singular chains on a wedge themselves split.