A homology equivalence need not be a homotopy equivalence without simple connectivity

The wedge $W$ has one vertex and two open edges, and $r$ is a continuous loop by endpoint agreement and finite pasting. It is a cellular attaching map into that $1$-skeleton with finite support, so [F1] gives the stated finite CW space $X$, with $W$ a closed subcomplex and $\chi$ a homeomorphism on the open disk onto its cell. Every point of $W$ is connected to $v$ along its edge, and every interior-disk point is connected radially to a boundary point and then to $v$. Thus both $W$ and $X$ are path connected. The quotient $W\mathrm{/}{S}_a^1$ is $S_b^1$, with its usual quotient interval topology. The quotient $X\mathrm{/}W$ is $D^2\mathrm{/}{S}^1\cong S^2$: identify the open disk with ${\mathbb{R}}^2$ by $z\mapsto z\mathrm{/}(1-\mathrm{\mid }z\mathrm{\mid })$, follow by inverse stereographic projection, and send the collapsed boundary to infinity. Continuity at that point follows because escaping every bounded set corresponds to approaching the disk boundary; the same description supplies inverse continuity. These are the actual quotient identifications, by [F10], not merely homology equivalences inferred from a cellular complex.

Let $k:S_a^1\to W$ and $s:S_b^1\to W$ be inclusions, and $p_a,p_b$ the maps collapsing the other loop. They are continuous by the quotient interval test and satisfy $p_{a}k=1$, $p_{b}s=1$, with constant cross composites. The pair exact sequence [F3], the vanishing of positive point homology from [F4], and the $H_0$ isomorphism for the path-connected $S_b^1$ from [F9] show directly that $H_j(S_b^1)\to H_j(S_b^1,\ast )$ is an isomorphism for $j>0$. Thus [F2] and [F4] make $H_j(W,S_a^1)=H_j(S_b^1,\ast )$ zero for $j\ge 2$ and $\mathbb{Z}$ for $j=1$. The pair sequence [F3] gives $H_j(W)=0$ for $j\ge 2$. In degree one it gives $0\to \mathbb{Z}\stackrel{k_{\ast }}{\to }{H}_1(W)\stackrel{(p_b{)}_{\ast }}{\to }\mathbb{Z}\to 0$: the connecting arrow to $H_0(S_a^1)$ vanishes because its next map to $H_0(W)$ is an isomorphism by [F9]. Its right inverse is $s_{\ast }$. For any $z\in H_1(W)$, $z-s_{\ast }(p_b{)}_{\ast }z$ lies in the image of $k_{\ast }$, and applying $(p_a{)}_{\ast }$ gives its unique coefficient. Therefore $H_1(W)=\mathbb{Z}\alpha \oplus \mathbb{Z}\beta$, where $\alpha =a_{\ast }[S^1]$ and $\beta =b_{\ast }[S^1]$, with coordinate inverse $((p_a{)}_{\ast },(p_b{)}_{\ast })$. Also $H_0(W)=\mathbb{Z}$. This proves the wedge calculation directly without invoking the higher-sphere wedge lemma outside its $n\ge 2$ hypothesis.

Compute the fundamental group using an actual open cover. In the disk coordinates of $\chi$ put $U=X\setminus \chi (\{\mathrm{\mid }z\mathrm{\mid }\le 1\mathrm{/}3\})$ and $V=\chi (\{\mathrm{\mid }z\mathrm{\mid }<1\})$. The first is open by the quotient criterion: its inverse image is all of $W$ and the relatively open disk annulus $\mathrm{\mid }z\mathrm{\mid }>1\mathrm{/}3$. The second is open because its inverse image is the open disk and the empty subset of $W$. They cover $X$ and intersect in $1\mathrm{/}3<\mathrm{\mid }z\mathrm{\mid }<1$, an annulus. The radial homotopy $ru\mapsto ((1-t)r+t)u$ on the annular part of $U$, fixing $W$, descends by [F10] and retracts $U$ onto $W$. The disk $V$ is contractible and the annulus retracts onto a circle. All three spaces are path connected. Choose a basepoint in the annulus and the radial path from it to $v=r(1)$ on $W$; the retraction's basepoint track is this path, so [F8] supplies the based comparison. Its positive annulus generator retracts to the attaching word $a^2{b}^{-3}$, with this path fixing the whisker. Thus [F6, F7] give ${\pi }_1(X,v)\cong ⟨a,b\mid a^2{b}^{-3}=1⟩\mathrm{.}$ Indeed the pushout is the free group on $a,b$ modulo the normal closure of that word: a map from it to any group is exactly a choice of images of $a,b$ satisfying that relation, which is the van Kampen universal property for this cover.

By [F2], the pair $(X,W)$ has $H_2(X,W)=\mathbb{Z}$ and all other positive relative groups zero, using $X\mathrm{/}W=S^2$ and [F4]. The characteristic-disk map $\chi :(D^2,S^1)\to (X,W)$ induces an isomorphism in relative homology: the quotient square of [F2] identifies it with the homeomorphism $D^2\mathrm{/}{S}^1\to X\mathrm{/}W$ from step 1.1. The disk has no positive homology by [F11], so the pair sequence [F3] makes $\delta :H_2(D^2,S^1)\to H_1(S^1)$ an isomorphism. Let $[D^2,S^1]$ be the unique class mapping to the positive circle generator of [F4], and put $c={\chi }_{\ast }[D^2,S^1]$. Naturality of the boundary map [F3] gives $\delta c=r_{\ast }[S^1]$. By the degree-one additivity and inversion formula in [F5] and the actual word defining $r$, this is $2\alpha -3\beta$. Thus the pair sequence reduces in its only nonzero positive segment to $0⟶H_2(X)⟶\mathbb{Z}\stackrel{t\mapsto (2t,-3t)}{\to }{\mathbb{Z}}^2⟶H_1(X)⟶0.$ In degrees at least three its surrounding groups vanish, so $H_j(X)=0$ there. The displayed integer map is injective, since $2t=0$ implies $t=0$ in $\mathbb{Z}$, so $H_2(X)=0$ too.

The homomorphism $\lambda :{\mathbb{Z}}^2\to \mathbb{Z}$, $\lambda (u,v)=3u+2v$, kills $(2,-3)$ and sends $(1,-1)$ to $1$. It induces an isomorphism ${\mathbb{Z}}^2\mathrm{/}\mathbb{Z}(2,-3)\to \mathbb{Z}$: for every $(u,v)$, $(u,v)-(3u+2v)(1,-1)=-(u+v)(2,-3),$ so its inverse sends $k$ to $k[(1,-1)]$. The two formulas compose to the identity, proving both injectivity and surjectivity. Define $f:S^1\to X$ by first following $a$ and then $b^{-1}$, using the two half-intervals. By [F5], $f_{\ast }[S^1]$ has coordinates $(1,-1)$ modulo the attaching vector, so the just-proved isomorphism sends it to $1$. Hence $f_{\ast }$ is an isomorphism on $H_1$. In degree zero it takes a point to the sole component and is an isomorphism by [F9]. In every degree at least two both source and target groups vanish by [F4] and step 3.1. This verifies the promised integral homology equivalence in every degree, including its generator normalization.

Send $a$ to the transposition $A=(12)$ and $b$ to the cycle $B=(123)$, acting as permutations of $\{1,2,3\}$ with rightmost composition first. They obey $A^2=B^3=1$, so step 2.2's universal property gives a homomorphism from ${\pi }_1(X)$ to that permutation group. Its image is nonabelian: $(AB)(1)=1$, whereas $(BA)(1)=3$. In fact it is all $S_3$: $ABA=B^{-1}$, so the six distinct permutations $1,B,B^2,A,AB,A{B}^2$ comprise the subgroup generated by $A,B$ and exhaust the permutations of three letters. Therefore ${\pi }_1(X)$ is nonabelian, while ${\pi }_1(S^1)$ is abelian by [F7]. A homotopy equivalence would induce an isomorphism of these groups by [F8], impossible. Thus the particular $f$ in step 4.1 is not a homotopy equivalence despite all its integral homology isomorphisms.

This is a finite nonempty connected CW witness with exactly one vertex, two edges and one $2$-cell; no wild space or missing CW hypothesis is involved. The zero higher homology groups and the one-dimensional generator map were calculated, and the attaching vector has no kernel despite its mixed signs. The annular cover excludes the disk center but includes all boundary identifications in $U$; its homotopy fixes them, as required for quotient descent. The finite permutation calculation witnesses the failed fundamental-group conclusion. Only finitely many cells, maps and representatives are instantiated, and all supplier clauses used above are choice-free. No arbitrary-cell choice or general Hurewicz isomorphism is used.