CW quotients and collapse of a contractible subcomplex

Start with the discrete vertex set consisting of $\ast$ and the vertices of $X\setminus A$. For each positive dimension attach the characteristic disks of the corresponding cells of $X\setminus A$, composing their original boundary maps with the collapse already constructed on $A$ and the lower-dimensional cells. This composition is continuous by induction on dimension. The boundaries are cellular and have finite support: a closed cell of $X$ has finite support by its supplied CW structure, and collapsing its portion in $A$ replaces that portion by at most the one vertex $\ast$. Thus [F1] gives a CW complex $Q$ with precisely the asserted cells. Points of the open cells outside $A$ are not identified with one another or with $\ast$, so its underlying set is exactly the set $X\mathrm{/}A$.

This CW topology is the ordinary quotient topology. A function $h:X\mathrm{/}A\to T$ is continuous for the ordinary quotient exactly when $hq:X\to T$ is continuous, by [F3]. By the characteristic-disk test [F1], the latter means continuity on each characteristic disk of $X$. Disks belonging to $A$ map constantly to $h(\ast )$; the other tests are precisely the characteristic disks used to construct $Q$. Hence the map-out tests agree for every target $T$. Taking $T$ to be the two-point space with open sets $\mathrm{\varnothing },\{1\},\{0,1\}$, the characteristic map of a subset is continuous exactly when that subset is open. Therefore the two topologies agree. In particular $X\mathrm{/}A$ is Hausdorff and CW with the displayed quotient characteristic maps; no separation of an arbitrary quotient was assumed in advance.

Extend $H$, viewed in $X$, by [F2] from the initial map ${\mathrm{i}\mathrm{d}}_X$ to a homotopy $F:X\times I\to X$ with $F(x,0)=x$ and $F{\mathrm{\mid }}_{A\times I}=H$. Thus $F(-,1)$ is constant at $a$ on $A$ and factors continuously as $gq$ for $g:X\mathrm{/}A\to X$ by [F3]. For every $t$, the map $qF(-,t)$ is constant at $\ast$ on $A$, since $H$ stays in $A$. Consequently the jointly continuous map $qF$ descends through $q\times {\mathrm{i}\mathrm{d}}_I$ to a continuous $\bar{F}:(X\mathrm{/}A)\times I\to X\mathrm{/}A$ by [F3]. It starts at the identity and ends at $qg$: the endpoint equality follows after composition with the surjective $q$. We have proved ${\mathrm{i}\mathrm{d}}_X\simeq gq$ and ${\mathrm{i}\mathrm{d}}_{X\mathrm{/}A}\simeq qg$, with the exact identity $\bar{F}(qx,t)=qF(x,t)$.

These homotopies make the induced component functions of $g$ and $q$ inverse, because each point is joined to its image under the corresponding composite. For positive degree at any $x\in X$, put $y=q(x)$ and $\alpha (t)=F(x,t)$, a path from $x$ to $g(y)$. The track of $\bar{F}$ at $y$ is $q\alpha$. Define $L={\beta }_{\alpha }{g}_{\ast }:{\pi }_j(X\mathrm{/}A,y)⟶{\pi }_j(X,x),j\ge 1.$ By [F4] applied to $F$, $L{q}_{\ast }=\mathrm{i}\mathrm{d}$. The radial-shell formula in [F4] gives $q_{\ast }{\beta }_{\alpha }={\beta }_{q\alpha }{q}_{\ast }$, where the right-hand $q_{\ast }$ is based at $g(y)$. Thus $q_{\ast }L={\beta }_{q\alpha }{q}_{\ast }{g}_{\ast }=\mathrm{i}\mathrm{d}$ by [F4] applied to $\bar{F}$. Hence $q_{\ast }$ is an isomorphism at every basepoint, proving weak equivalence without a based-contraction assumption.

If $H$ fixes $a$, then $F(a,t)=a$ by its prescribed restriction, $g(\ast )=a$, and $\bar{F}(\ast ,t)=\ast$ already holds by construction. Thus both maps and homotopies are based as asserted. If $A=X$, the quotient CW consists only of $\ast$, and the same contraction gives the claimed equivalence. If $A$ is a singleton, the quotient identifies no distinct points and the identity contraction is available. Empty $A$ is excluded because the displayed collapse has a specified quotient vertex; no empty-set contraction is postulated. Zero-dimensional relative cells are retained as separate vertices, higher-dimensional cells retain their supplied attaching identifications, and no regularity of those maps was used. Only the one given contraction and the specified choice-free HEP construction enter steps 3.1–4.1. This proves all assertions without AC.