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For a connective spectrum with ${\pi }_{0}E\cong \mathbb{Z}$, an $E$-homology isomorphism of simply connected spaces is a weak homotopy equivalence. Its simply connected homotopy mapping cone is $E$-acyclic. If it had a first nonzero integral homology group, the Hurewicz theorem and the bottom-degree generalized Hurewicz isomorphism would produce a nonzero $E$-homology group. Thus the cone is integrally acyclic and the homological Whitehead theorem applies.

The cellular boundary ${\mathbb{Z}}^2\to \mathbb{Z}$ is $(m,n)$, with kernel $\mathbb{Z}$ and zero cokernel when the degrees are coprime. The van Kampen theorem gives trivial fundamental group. The Hurewicz theorem supplies a map from the sphere representing a homology generator, and the homological Whitehead theorem makes it a homotopy equivalence. For degrees $2,3$ the space cannot be homeomorphic to a sphere: an embedded circle in a sphere bounds its two complementary regions, so both characteristic boundary maps would have degree of absolute value one.

Start with a finite $n$-dimensional $n$-connected approximation. Its degree-$n$ homology kernel is finite free. Realize a basis via the Relative Hurewicz theorem and attach $(n+1)$-cells; their boundary is injective, so they kill precisely that kernel without adding top homology. The homological Whitehead theorem gives a finite weak equivalence of dimension at most $n+1$.

A homotopy sphere is a closed manifold with the homotopy type of a sphere of the same dimension. In dimensions at least two, a closed simply connected integral homology sphere is a homotopy sphere: successive applications of the Hurewicz theorem produce a map from the sphere inducing an isomorphism on integral homology, and the homological Whitehead theorem makes it a homotopy equivalence. The topological generalized Poincare theorem identifies its topological type, while prescribed smooth structures require a separate question.

Choose basepoints and replace the spaces and map by CW complex models. If the hypothesis is stated for unreduced represented homology theory, the basepoint coefficient summands split naturally, so it also gives an isomorphism in reduced represented homology theory. Let $C$ be the homotopy mapping cone of $f$. Its reduced represented homology theory is zero in every degree, by the exact sequence of the corresponding cofiber sequence of spectra.

The mapping cone $C$ is a simply connected space: the mapping cylinder retracts onto $Z$, and coning off the path-connected space $Y$ adds no fundamental-group generators; the van Kampen calculation gives the quotient of ${\pi }_{1}Z$ by the image of ${\pi }_{1}Y$, which is trivial. Suppose some positive-degree integral homology group of $C$ were nonzero, and let $n$ be the least such degree. Then $n\ge 2$. Starting with ${\pi }_{1}C=0$, the Hurewicz theorem shows inductively that ${\pi }_{i}C=0$ for $i<n$: at each first possible degree $i$, its homotopy group is identified with the zero group $H_i(C;\mathbb{Z})$. It then identifiesApplying the result of part (a) to the $(n-1)$-connected based space $C$ gives ${\tilde{E}}_n(C)\cong {\pi }_{n}C\ne 0$, contradicting the vanishing of its represented homology theory.

Thus $C$ is integrally acyclic, and the ordinary homology exact sequence says that $f$ is an integral homology isomorphism. The allowed homological Whitehead theorem for simply connected spaces makes the map of CW complex models a homotopy equivalence. Returning to the original spaces therefore givesThe simply connected hypothesis is used both in the first-nonzero-degree Hurewicz theorem argument and in the final homological Whitehead theorem.

A simply connected space is path-connected and has trivial fundamental group, equivalently every based loop is null-homotopic. The Whitehead theorem says that a map of CW complexes inducing isomorphisms on all homotopy groups is a homotopy equivalence. Its homological Whitehead theorem form says that a map between simply connected CW complexes, in particular finite ones, is a homotopy equivalence if it induces isomorphisms on integral homology. The latter implication follows from the relative Hurewicz theorem and the ordinary Whitehead theorem; homotopy equivalences necessarily induce homology isomorphisms.

For the attached-cell space $X$, the cellular homology chain complex in positive degrees isDualizing gives the cellular cohomology complex $\mathbb{Z}\stackrel{0}{\to }\mathbb{Z}\stackrel{t\mapsto (2t,3t)}{\to }{\mathbb{Z}}^2$. The last map is injective and its image is primitive because $\gcd (2,3)=1$. Its cokernel is $\mathbb{Z}$, explicitly via $(a,b)\mapsto 3a-2b$. HenceThe van Kampen theorem gives ${\pi }_1(X)=⟨a\mid a^2=1,a^3=1⟩=1$. Moreover $H_2(X)=\ker (23)$ is generated by $(3,-2)$. Since $X$ is simply connected, the Hurewicz theorem gives ${\pi }_2(X)\cong H_2(X)$. Choose a map $S^2\to X$ representing that generator. It induces an isomorphism on $H_2$, and on $H_0$, while all other reduced homology groups vanish on both sides. The homological Whitehead theorem proves $\boxed{X\simeq S^2}$. This is a special case of coprime two-cell attachments to a circle.

It is not homeomorphic to the sphere. Indeed, if it were, its embedded one-skeleton would be a Jordan circle in $S^2$, with the two open cell interiors as its complementary regions. The Jordan curve theorem and its Schoenflies form identify the closures of those regions with discs. A characteristic map, homeomorphic on the interior of such a disc, has local degree $+1$ or $-1$ at an interior point; the boundary homomorphism in relative homology then makes its attaching map on the circle have degree $+1$ or $-1$. This contradicts degrees $2$ and $3$. This argument does not require choosing special covering representatives of the attaching maps.

Finally, two compact examples with the same additive cohomology areBoth have $H^0=\mathbb{Z}$, $H^1={\mathbb{Z}}^2$, $H^2=\mathbb{Z}$ and zero higher cohomology, by their cellular chain complexes. But their fundamental groups are respectively ${\mathbb{Z}}^2$ and the nonabelian free group on two generators, so they cannot be homotopy equivalent. This illustrates that cohomology groups do not determine homotopy type.

Choose a generator $[S^n]$ of ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree of a continuous map is the integer determined byFor $n\ge 1$ this is ordinary top-dimensional homology; using reduced homology also gives a convention for $n=0$. Domain and target use the same orientation, so reversing the chosen generator in both leaves the integer unchanged. Homotopy invariance of homology makes the mapping degree invariant under homotopy.

We interpret the manifold criterion for a suspension in the category of nonempty topological manifolds without boundary, and “sphere” as homeomorphic to the standard sphere. Let $X$ have dimension $d$. If $X\cong S^d$, an explicit homeomorphism from its suspension to $S^{d+1}$ isIt is a continuous map that is a bijection: the two collapsed ends become the poles, and every other point determines $t$ and $x$ uniquely. The compact-to-Hausdorff continuous bijection theorem makes it a homeomorphism.

Conversely suppose $SX$ is a topological manifold. Its open part $X\times (0,1)$ shows its dimension is $d+1$. At a suspension vertex $v$ there is an open cone neighbourhood $U$ on $X$. It is contractible, while $U\setminus \{v\}$ admits a deformation retraction onto $X$. The long exact sequence in relative homology and excision give the local homology from a link calculationA point of a $(d+1)$-dimensional topological manifold without boundary has local homology $\mathbb{Z}$ in degree $d+1$ and zero in all other degrees. ConsequentlyThus $X$ is an integral homology sphere. This conclusion alone does not identify its homeomorphism type.

If $d\ge 2$, it is also simply connected. Choose a cone neighbourhood $U$ of the vertex and an open coordinate ball $B$ around the vertex inside $U$. By compactness of $X$, sufficiently short entire cone neighbourhoods fit inside any neighbourhood of $v$: the inverse image of that neighbourhood contains $X\times \{0\}$, and a finite subcover supplies one uniform collar length. Choose such a shorter cone $V\subset B$. The inclusions of punctured neighbourhoods factor asThe composite induces an isomorphism of fundamental groups, since both cone punctures are products of $X$ with intervals and their inclusion is a homotopy equivalence. The middle punctured ball has the homotopy type of $S^d$ and is simply connected for $d\ge 2$. Hence the composite is also zero on the fundamental group, forcing ${\pi }_1(X)=0$. This uses the nested neighbourhoods, not an unsupported claim that an arbitrary punctured neighbourhood is a sphere.

The Hurewicz theorem, applied successively to the vanishing lower homology groups, now gives ${\pi }_j(X)=0$ for $j<d$ and ${\pi }_d(X)\cong H_d(X)\cong \mathbb{Z}$. A representative of its generator is a continuous map $S^d\to X$ inducing an isomorphism in every integral homology degree. Topological manifolds have the homotopy type of CW complexes, so the homological Whitehead theorem makes it a homotopy equivalence. We have proved that $X$ is a homotopy sphere.

At this point the topological classification theorem is essential: the topological generalized Poincare theorem says that a closed homotopy sphere is homeomorphic to the standard sphere. Applying it completes the converse for $d\ge 2$. For $d=1$, the local calculation makes $X$ a connected compact one-dimensional manifold without boundary, hence a circle. For $d=0$, a compact zero-dimensional manifold is a finite set, and ${\tilde{H}}_0(X)\cong \mathbb{Z}$ forces exactly two points. Thus it is $S^0$. Altogether,The last classification step is not a consequence of homology alone, and we make no assertion of diffeomorphism with a prescribed smooth structure. If the word manifold allows boundary on both sides, the printed assertion needs this correction: the suspension of the interval is a closed disk. For instance $[x,t]\mapsto ((2x-1)(1-|2t-1|),2t-1)$ identifies it with the filled diamond $|u|+|v|\le 1$, although the interval is not a sphere.

To construct sphere maps of arbitrary integer degree, first use $h_m:S^1\to S^1$, $h_m(z)=z^m$ for any integer $m$, including the constant map when $m=0$. Lifting its argument to the real line gives $\theta \mapsto m\theta$, so the fundamental one-cycle is sent to $m$ times itself. Therefore $\mathrm{deg}{h}_m=m$.

Two cone neighbourhoods cover a suspension; their intersection admits a deformation retraction onto the original space. The reduced Mayer–Vietoris theorem gives a natural suspension isomorphismNaturality means that the suspension of a continuous map induces the same integer multiplier. This proves degree under suspension. Using the standard identification $S(S^k)\cong S^{k+1}$, setThus every integer occurs in every positive sphere dimension, with negative degrees handled just as well as positive ones.

If “every $n$” includes zero, that final printed assertion is false. The generator of ${\tilde{H}}_0(S^0)$ is the difference of its two points. A map of a two-point set is the identity, the transposition, or one of two constant maps, giving preciselyThese degrees of maps of the zero-sphere are the explicit exception to the arbitrary-degree construction.

Let ${\mathcal{C}}_{\mathrm{fg}}$ be the Serre class of finitely generated abelian groups. A homomorphism is an isomorphism modulo this class when its kernel and cokernel are finitely generated. The Hurewicz theorem modulo a Serre class says that, for a simply connected space and $r\ge 2$, if ${\pi }_i(X)\in {\mathcal{C}}_{\mathrm{fg}}$ for $2\le i<r$, thenThe corresponding lower-dimensional homology condition is equivalent to the lower homotopy condition. In particular, degreewise finite generation of integral homology and homotopy are equivalent for simply connected spaces. The Serre class is closed under subgroups, quotients and extensions; these closure properties are what make the modulo-class formulation useful.

Here is the consequence needed for construction. Inductively, if the lower homotopy groups are finitely generated, the theorem gives a finitely generated kernel for $h_r$. Its image is a subgroup of the finitely generated $H_r(X)$, so the image is also finitely generated. The resulting extension proves ${\pi }_r(X)$ finitely generated. Starting with the ordinary isomorphism ${\pi }_2(X)\cong H_2(X)$ yields this for every positive degree.

We now construct a finite type CW approximation. Choose finitely many maps $S^2\to X$ generating ${\pi }_2(X)$, and let $C_2$ be their wedge. The resulting map $C_2\to X$ is $2$-connected: it is an isomorphism below degree two and a surjection in degree two. In general, call a map $r$-connected when its mapping-cylinder pair has relative homotopy groups zero through degree $r$.

Suppose $f_r:C_r\to X$ is $r$-connected and $C_r$ is a finite simply connected CW complex of dimension at most $r$. The homotopy groups of $C_r$ are finitely generated by the same modulo-class theorem, since a finite CW complex has finitely generated homology. The long exact sequence of relative homotopy groups shows thatis finitely generated: it lies between a quotient of ${\pi }_{r+1}(X)$ and a subgroup of ${\pi }_r(C_r)$. Relative groups here refer to the mapping cylinder of $f_r$.

Represent a finite set of generators by relative disks. Attach their boundary spheres to $C_r$, and extend the map over the disks by their chosen maps into $X$. This adds finitely many $(r+1)$-cells and kills the relative group in that dimension without changing lower relative groups. The new map $C_{r+1}\to X$ is $(r+1)$-connected. Iterating givesThere are finitely many cells of each dimension. Every fixed homotopy degree stabilizes to an isomorphism once sufficiently high-dimensional cells have been added, so this is the required weak equivalence:

For the bounded-homology assertion, assume first $n\ge 2$ and take the finite $n$-dimensional $C_n$ just constructed. The Relative Hurewicz theorem for its $n$-connected map givesThe relative homology sequence and $H_{n+1}(X)=0$ identify the latter withBecause $C_n$ is $n$-dimensional, $H_n(C_n)$ is a subgroup of its free cellular $n$-chain group. It is finite free, and so is $K$.

Choose a basis of $K$, lift it using the Relative Hurewicz theorem, and attach exactly those finitely many $(n+1)$-cells to $C_n$. Denote the resulting complex by $C^{\prime }$. Its new cellular boundary has image $K$ and is injective: the selected cycles are linearly independent in $H_n(C_n)$, and there are no old $(n+1)$-boundaries. ThusLower homology remains unchanged, and all higher homology is zero on both sides. The map $C^{\prime }\to X$ is an integral homology isomorphism between simply connected spaces. The homological Whitehead theorem consequently makes it a weak homotopy equivalence. Hence the finite CW approximation from bounded homology hasIf $n=0$ or $1$, simple connectivity and the homology hypothesis make all reduced homology zero. The Hurewicz theorem, applied at the first possible nonzero homotopy degree, shows that $X$ is weakly contractible, so a point suffices. If the wording requires dimension exactly $n+1$ rather than at most $n+1$, add a contractible cancelling pair of $n$- and $(n+1)$-cells, mapping constantly to the basepoint. This does not change the weak homotopy type.

For maps between simply connected spaces, a rational homology equivalence is a rational homotopy equivalence. It follows from the relative Hurewicz theorem modulo torsion, or from rationalization and the homological Whitehead theorem.