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If $E$ is connective spectrum in the sense of a connective spectrum, ${\pi }_{0}E\cong \mathbb{Z}$, and $X$ is $(n-1)$-connected with $n\ge 2$, then ${\tilde{E}}_n(X)\cong {\pi }_n(X)$. In the homological Atiyah-Hirzebruch spectral sequence, the only nonzero term in total degree $n$ is ${\tilde{H}}_n(X;{\pi }_{0}E)$. Incoming differentials have negative coefficient degree, and outgoing differentials have spatial degree below $n$. The Hurewicz theorem identifies this term with ${\pi }_{n}X$.

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.

For a compact connected Riemannian manifold with nonnegative Ricci curvature, its finitely generated fundamental group acts freely and properly discontinuously on the complete universal cover. The preceding polynomial growth of properly discontinuous isometry groups bound applies. The degree-one Hurewicz theorem identifies first integral homology with the abelianization of the fundamental group. If its free rank is $b_1(M)$, this group maps onto ${\mathbb{Z}}^{b_1(M)}$. Choosing lifts of a basis gives at least a constant times $m^{b_1(M)}$ distinct elements in a word ball of radius a constant times $m$. The upper bound of order $m^{\dim M}$ forces the displayed inequality.

The integral homology of the loop space of a sphere implies that the unit $S^{r-1}\to \Omega S^r$ has vanishing relative homology through degree $2r-3$. Both spaces are simply connected for $r\ge 3$. The absolute Hurewicz theorem first identifies their degree-two homotopy groups with homology; the relative Hurewicz theorem then rules out a first nonzero relative homotopy group in degrees three through $2r-3$. The relative homotopy sequence makes the unit an isomorphism on ${\pi }_i$ through degree $2r-4$. Loop-suspension adjunction identifies this map with suspension, yielding ${\pi }_i(S^{r-1})\cong {\pi }_{i+1}(S^r)$ in that range.

Moving the basepoint around a loop acts on a higher homotopy group. Since ${\pi }_n(X)$ is an abelian group for $n\ge 2$, this action extends linearly to its integral group ring, making ${\pi }_n(X)$ a module. For a connected CW complex, its universal cover gives an equivalent description by deck transformations, together with change of basepoint upstairs. Left and right conventions are converted by inversion of the loop. The Hurewicz theorem respects this action whenever it identifies a higher homotopy group with the corresponding homology of the cover.

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.

A sphere map $f:S^n\to X$ is sent to $f_{\ast }[S^n]\in H_n(X;\mathbb{Z})$. After tensoring with $\mathbb{Q}$ this gives the rational Hurewicz homomorphism. It is natural in $X$, sends Whitehead products to zero, and is an isomorphism in the first nonzero homotopy degree of a simply connected space by the Hurewicz theorem.

For $r\ge 3$, the path-loop Serre fibration has contractible total space. The relative homology of a fibration over a sphere gives $H_j(\Omega S^r)\cong H_{j-r+1}(\Omega S^r)$ for $j>0$, with negative homology zero and $H_0=\mathbb{Z}$. This proves the displayed groups. The unit map $S^{r-1}\to \Omega S^r$, adjoint to the identity of $S^r$, represents the generator of the first nonzero homotopy group. Naturality of the Hurewicz theorem shows that it induces an integral homology isomorphism in all degrees through $2r-3$.

Interpret the connectivity hypothesis in its usual based sense: $X$ is $(n-2)$-connected, with $n\ge 2$ for the nontrivial suspension range. Form the suspension as two cones $C_{-}X$ and $C_{+}X$ meeting in $X$. Both cones are contractible. Their relative homotopy sequences give ${\pi }_j(C_{\pm }X,X)\cong {\pi }_{j-1}(X)$ in the relevant degrees, so both cone pairs are $(n-1)$-connected.

The homotopy excision theorem says that if two CW pairs meeting in a connected subcomplex are $p$- and $q$-connected, $p,q\ge 1$, the map from the first relative pair to the union relative to the second part is an isomorphism in degrees $i<p+q$ and a surjection in degree $p+q$. Here $p=q=n-1$, givingas an isomorphism for $i<2n-2$ and a surjection for $i=2n-2$. Contractibility identifies the domain with ${\pi }_{i-1}(X)$ and the target with ${\pi }_i(\Sigma X)$. The comparison under these identifications is the suspension map: a sphere representative is coned in each half and the two cones give its suspended sphere representative. Hence the Freudenthal suspension from two cones provesFor connected CW complexes, reduced and unreduced suspension have the same based homotopy type, so either standard suspension convention gives this result. In the low-dimensional range, pointed sets are used where necessary; the displayed positive-degree homomorphism assertion concerns the usual groups.

For $S^r$ with $r\ge 2$, cellular approximation makes every sphere map of degree below $r$ null-homotopic: it can be homotoped into the point skeleton of the CW structure having cells only in degrees zero and $r$. The Hurewicz theorem therefore identifies its first possible nonzero homotopy group with homology. Its cellular chain complex gives $H_r(S^r;\mathbb{Z})=\mathbb{Z}$. For $r=1$, lifting a based loop to the universal cover $\mathbb{R}\to S^1$ classifies it by an integer endpoint displacement, so ${\pi }_1(S^1)=\mathbb{Z}$; the degree-one Hurewicz map is abelianization and is an isomorphism here. ThusAlternatively, after the $r=2$ Hurewicz base case, the suspension isomorphisms above propagate the generator to every higher sphere.

Let $\Gamma ={\pi }_1(M)$. A compact connected smooth manifold has finite CW complex homotopy type, hence a finitely generated fundamental group. Its universal cover with the lifted Riemannian metric is complete by the lifted-geodesic argument in part 2(b) and has nonnegative Ricci curvature. The deck transformations act freely and properly discontinuously by isometries. Part (a) givesThe degree-one Hurewicz theorem identifies $H_1(M;\mathbb{Z})$ with the abelianization of $\Gamma$. This finitely generated abelian group has free rank $b=b_1(M)$, so there is a surjection $\Gamma \to {\mathbb{Z}}^b$. Choose lifts ${\gamma }_1,\dots ,{\gamma }_b$ of a basis and let $A\ge 1$ bound their word lengths. For every $(z_1,\dots ,z_b)$ with $\sum |z_i|\le m$, the product ${\gamma }_1^{z_1}\cdots {\gamma }_b^{z_b}$ has word length at most $Am$ and maps to that vector. Distinct vectors therefore give distinct group elements. The integer cube $|z_i|\le \lfloor m/b\rfloor$ shows, for $b>0$, thatfor some $c>0$ and all sufficiently large $m$. Comparing with the degree-$n$ upper bound forces $b\le n$; the case $b=0$ is immediate. ThusThis is the first Betti-number bound from polynomial fundamental-group growth.

The standard connectedness convention for a Riemannian manifold is needed here. If disconnected manifolds are allowed, the bound applies to each connected component; a disjoint union of two flat $n$-tori has first Betti number $2n$ and contradicts a global bound by $n$.

A simply connected manifold is connected and orientable: its orientation character is a group homomorphism ${\pi }_1(X)\to \{\pm 1\}$, and this group is trivial. Choose an orientation. The Hurewicz theorem in degree one gives $H_1(X;\mathbb{Z})=0$. The universal coefficient theorem for cohomology then gives $H^1(X;\mathbb{Z})=0$ andbecause the possible $\mathrm{Ext}(H_1(X),\mathbb{Z})$ term is zero.

The homology groups are finitely generated for a closed manifold. The displayed dual group is therefore free abelian. Poincare duality identifies $H_2(X;\mathbb{Z})$ with $H^2(X;\mathbb{Z})$, so $H_2$ itself has no torsion. Since the second Betti number is one, both groups are isomorphic to $\mathbb{Z}$. This torsion argument is essential: rank one by itself would not rule out finite summands.

Again by Poincare duality, $H^3(X;\mathbb{Z})\cong H_1(X;\mathbb{Z})=0$, and $H^4(X;\mathbb{Z})\cong H_0(X;\mathbb{Z})=\mathbb{Z}$. Thus the only nonzero cohomology groups are $\mathbb{Z}$ in degrees $0,2,4$.

Let $u$ generate $H^2$. The integral intersection formis unimodular. Indeed, cap product with $[X]$ is the Poincare duality isomorphism to $H_2$, and the preceding universal-coefficient identification makes evaluation the full integral dual. Composing these isomorphisms identifies the pairing with an isomorphism $H^2\to \mathrm{Hom}(H^2,\mathbb{Z})$. In rank one its matrix must therefore be $[1]$ or $[-1]$. Consequently $⟨u^2,[X]⟩=\pm 1$, so $u^2$ generates $H^4$.

All products in degrees above four vanish. Hence the assignment of the degree-two generator giveswhich is the cohomology ring of complex projective space for ${\mathbb{CP}}^2$. If the intersection form is negative, this ring isomorphism reverses the chosen top orientation generator; preserving that additional orientation is not part of the requested ring isomorphism.

Use the based path space and the path-loop fibration $\Omega S^r\to P{S}^r\to S^r$, where paths begin at a fixed basepoint and the projection evaluates their final endpoint. The total space is contractible: the homotopy $\gamma (t)\mapsto \gamma (st)$, $s\downarrow 0$, contracts every path to the constant path. The fiber is path-connected because $S^r$ is simply connected. Hence $H_0(\Omega S^r)=\mathbb{Z}$ and all positive homology of $P{S}^r$ vanishes.

The exact sequence from part (i), for $i\ge 2$, gives $H_{i-r}(\Omega S^r)\cong H_{i-1}(\Omega S^r)$. Put $d=r-1$. Taking negative-degree homology to be zero, this says $H_j\cong H_{j-d}$ for every $j\ge 1$. Starting with $H_0=\mathbb{Z}$ determines every group:There are no torsion groups hidden in the recurrence, since each step is an actual integral isomorphism. This is the integral homology of the loop space of a sphere.

Choose the unit map $\eta :S^{r-1}\to \Omega S^r$ adjoint to the identity under $\Sigma S^{r-1}\cong S^r$. Loop adjunction gives ${\pi }_j(\Omega S^r)\cong {\pi }_{j+1}(S^r)$, so the loop space is $(r-2)$-connected and ${\pi }_{r-1}(\Omega S^r)\cong \mathbb{Z}$. The adjoint of $\eta$ is the degree-one identity, so $\eta$ sends the generator of ${\pi }_{r-1}(S^{r-1})$ to that generator. The Hurewicz theorem for both $(r-2)$-connected spaces, and its naturality, then show that ${\eta }_{\ast }$ is an isomorphism on $H_{r-1}$. It also is on $H_0$, since both spaces are connected. All other positive degrees below $2r-2$ vanish in both spaces. ThusThis proves the required existence with a specific map, without using the suspension theorem in advance.

Use the unit $\eta$ just constructed and replace it by the inclusion $A=S^{r-1}\hookrightarrow Y=\mathrm{Cyl}(\eta )$ into its mapping cylinder; $Y\simeq \Omega S^r$. Put $m=2r-3$. The homology isomorphisms from question 4(ii) and the relative exact sequence give $H_j(Y,A;\mathbb{Z})=0$ for $j\le m$.

Both $A$ and $Y$ are simply connected for $r\ge 3$. For the degree-two start, the absolute Hurewicz theorem identifies their ${\pi }_2$ with $H_2$; naturality and the homology isomorphism show ${\eta }_{\ast }$ is an isomorphism on ${\pi }_2$, so ${\pi }_2(Y,A)=0$. The relative fundamental group also vanishes. If there were a first nonzero relative homotopy group in some degree $3\le k\le m$, the Relative Hurewicz theorem for a $(k-1)$-connected simply connected pair would identify it with $H_k(Y,A)$, which is zero. Induction therefore gives ${\pi }_j(Y,A)=0$ for all $j\le m$.

The relative homotopy sequence now makes ${\eta }_{\ast }:{\pi }_i(S^{r-1})\to {\pi }_i(\Omega S^r)$ an isomorphism when $i+1\le m$, and a surjection at $i=m$. Under loop-suspension adjunction, the adjoint of $\eta \circ a$ is $\Sigma a$, so this particular isomorphism is the suspension homomorphism. Consequently the requested Freudenthal suspension theorem isFor $i=0$ this is read as the identification of connected-component sets; all degree-one groups here also vanish. The proof is the homology comparison proof of Freudenthal suspension.

To compute the specified unstable groups, $S^1$ has contractible universal cover, so ${\pi }_2(S^1)=0$. The Hopf fibration $S^1\to S^3\to S^2$ has exact segment $0={\pi }_3{S}^1\to {\pi }_3{S}^3\to {\pi }_3{S}^2\to {\pi }_2{S}^1=0$, giving ${\pi }_3{S}^2=\mathbb{Z}$. The supplied starting value is ${\pi }_4{S}^3=\mathbb{Z}/2$. To pass from the group over $S^{n-1}$ to that over $S^n$, use $r=n,i=n$; the condition $n\le 2n-4$ holds exactly when $n\ge 4$. Thus every later step is an isomorphism, andThis is the first homotopy group above the dimension of a sphere. The stable range deliberately excludes the step from $n=2$ to $n=3$, so the different groups there do not conflict with the theorem.

Start with the prime decomposition of a closed orientable three-manifold. Its fundamental group is a free product of the prime-summand groups. A nontrivial abelian group cannot be a free product of two nontrivial groups. The same is true of a nontrivial finitely generated nilpotent group: it has nontrivial center, whereas such a free product has trivial center. Thus all nontrivial fundamental group comes from a single prime summand; simply connected summands do not affect the argument.

A prime but nonirreducible orientable summand is $S^2\times S^1$, giving $\Gamma \cong \mathbb{Z}$. Otherwise the infinite-group prime summand is an irreducible three-manifold. The sphere theorem for three-manifolds gives ${\pi }_2=0$. Its simply connected universal cover is noncompact, so its top ordinary homology vanishes; its first and second homology also vanish by simple connectivity and the Hurewicz theorem. It is therefore acyclic and contractible by the Whitehead theorem. The summand is aspherical, and its closed orientable three-dimensional classifying space gives integral cohomological dimension of a group equal to three: the cellular resolution gives the upper bound and top Poincare duality gives the lower bound.

If $\Gamma$ is torsion-free abelian, finite generation makes it ${\mathbb{Z}}^r$. Its integral cohomological dimension of a group is $r$, as one sees from the standard $r$-torus. The aspherical case therefore has $r=3$. We obtainBoth occur, on $S^2\times S^1$ and the three-torus.

For the infinite torsion-free nilpotent case, the same reduction gives either $\mathbb{Z}$ or an aspherical group of cohomological dimension three. By the Malcev completion of a torsion-free nilpotent group, the latter is a lattice in a simply connected three-dimensional nilpotent Lie group. A nonabelian three-dimensional nilpotent Lie algebra has abelianization of dimension at least two: a nilpotent algebra generated modulo its derived algebra by one element would be generated by that element and thus be abelian. Its derived algebra is therefore one-dimensional, and nilpotence makes that line central. Hence it has a basis with a single nonzero bracket $[Y,T]=X$; the other possibility is an abelian algebra. In the nonabelian case the lattice's center is an isolated infinite cyclic subgroup, generated by $x$, and the quotient by it is ${\mathbb{Z}}^2$. Choose lifts $y,t$ of a basis of that quotient. Their commutator is $x^b$ for a nonzero integer $b$, and normal forms giveThe subgroup $A=⟨x,y⟩\cong {\mathbb{Z}}^2$ is normal and ${\Gamma }_b/A\cong \mathbb{Z}$. In its ordered basis $x,y$, conjugation by $t$ acts byThis is the unipotent torus-bundle group. Indeed, its normal forms multiply according toProjection to the $n,k$ coordinates, followed by inspection of the central $m$ coordinate, proves torsion-freeness. All commutators lie in $⟨x^b⟩$, which is central, proving nilpotence. The mapping torus of the torus automorphism with this matrix realizes every integer $b$ as a closed orientable three-manifold; $b=0$ gives ${\mathbb{Z}}^3$, while $b\ne 0$ gives the Nil cases. Thus the complete list of groups under the nilpotent hypotheses is $\mathbb{Z}$ and the groups ${\Gamma }_b$.

Here is the center argument, with its literal qualification. If a nilpotent group with abelian quotient $\Gamma /A$ were nonabelian, the last nonzero term ${\gamma }_c\Gamma$ of its lower central series would satisfyConsequently $Z(\Gamma )\cap A=1$ forces the group to be abelian. For the specific nontrivial rank-two $A$ in the printed family the hypothesis cannot actually occur: when $b\ne 0$, $x\in Z({\Gamma }_b)\cap A$; when $b=0$, the intersection is all of $A$. The implication is valid, but should not be used to invent an example satisfying this impossible extra condition.

A torsion-free finite extension of a lattice with quotient $C_4$ is realized by the quarter-turn flat three-manifold. On ${\mathbb{R}}^3$ letLet $c$ be unit translation in the third coordinate. Then $sa{s}^{-1}=b$, $sb{s}^{-1}=a^{-1}$, $sc{s}^{-1}=c$, and $s^4=c$. The lattice $\Lambda =⟨a,b,c⟩\cong {\mathbb{Z}}^3$ is normal and the quotient is $C_4$. A representative $a^m{b}^n{c}^k{s}^j$, $0\le j<4$, with $j\ne 0$ has third-coordinate displacement $k+j/4\ne 0$, so has no fixed point; for $j=0$ a nonidentity element is a nonzero translation. This proves torsion-freeness. The action is properly discontinuous, cocompact and orientation preserving, so it gives a closed orientable flat three-manifold with precisely the required extension.

For faithful linear holonomy on the maximal translation lattice, the holonomy groups of closed orientable flat three-manifolds areThe cyclic examples replace the quarter-turn by a planar lattice rotation of order $m=2,3,4,6$ and a translation of $1/m$ along its axis. The square and triangular plane lattices realize these orders; the trace condition $2\cos (2\pi /m)\in \mathbb{Z}$ explains the crystallographic restriction. The remaining case is obtained fromTheir squares translate by one in the first and second directions, and $(\alpha \beta {)}^2$ translates by minus one in the third. Their linear parts generate $C_2\times C_2$. The three nontranslation cosets have half-integral displacement along their respective fixed axes, so every element acts without a fixed point. These constructions realize every member of the orientable three-dimensional Euclidean holonomy classification.

The maximality/faithfulness restriction is essential. Taken literally as a question about arbitrary finite group quotients, the wording has more answers: $\mathbb{Z}$ has every finite cyclic quotient, and ${\mathbb{Z}}^3$ has, for example, quotient $C_5$. An exact criterion for the literal extension question is as follows. Choose an action $Q\to \mathrm{Aut}(A)$ and an extension class $e\in H^2(Q,A)$, where $A=\mathbb{Z}$ or ${\mathbb{Z}}^3$. For every subgroup $⟨q⟩$ of prime order $p$, requireTo prove this, lift $q$ to $g$. Its $p$th power is in $A^q$, and replacing $g$ by $ag$ changes that power by the displayed norm of $a$. Thus a lift of order $p$ exists exactly when this restriction is zero. Any torsion element has a prime-order power, so the condition is both necessary and sufficient. For an oriented flat realization in rank three the linear action must also preserve orientation. This distinguishes a complete extension criterion from the finite faithful-holonomy list.

Take a based CW complex model for $X$. Since it is $(n-1)$-connected and $n\ge 2$, its reduced ordinary homology vanishes below degree $n$, with arbitrary constant coefficients. This also follows from a CW complex model with no non-basepoint cells below degree $n$. The connective spectrum $E$ has ${\pi }_{q}E=0$ for $q<0$.

Filter $X$ by its skeleta. The quotients of consecutive stages are wedges of spheres, so the exact sequences in represented homology theory give the homological Atiyah-Hirzebruch spectral sequenceIt is first quadrant because $E$ is a connective spectrum. In a fixed total degree it converges with a finite filtration: cells of sufficiently high dimension cannot contribute in that degree or affect it through a boundary, by the same connectivity bound.

In total degree $n$, every term with $p<n$ is zero, and terms with $q<0$ are zero. The only possible term isNo differential can leave it: the target has spatial degree $n-r<n$. No differential can enter it: its source would be $(n+r,1-r)$, whose coefficient degree is negative for every $r\ge 2$. Thus it survives unchanged and is the only filtration quotient of ${\tilde{E}}_n(X)$; there is no extension problem. The ordinary Hurewicz theorem now yieldsThe isomorphism uses the stated identification ${\pi }_{0}E\cong \mathbb{Z}$ and is natural in $X$. This is the bottom-degree generalized Hurewicz isomorphism.

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.

Let $p:\tilde{M}\to M$ be the universal cover. Its number of sheets is $|{\pi }_1(M)|$, which is finite, so compactness of $M$ makes $\tilde{M}$ compact. As a covering space of a manifold without boundary, it is also a three-manifold without boundary. It is connected and simply connected by the definition of universal cover.

A simply connected manifold is orientable: transport of a local orientation around loops gives the orientation character ${\pi }_1(\tilde{M})\to \{\pm 1\}$, and the trivial fundamental group leaves no orientation obstruction. The degree-one Hurewicz theorem, equivalently the identification of $H_1$ with the abelianization of the fundamental group, gives $H_1(\tilde{M};\mathbb{Z})=0$. The universal coefficient theorem for cohomology then givesbecause the additional term $\mathrm{Ext}(H_0(\tilde{M};\mathbb{Z}),\mathbb{Z})$ vanishes for $H_0\cong \mathbb{Z}$. Integral Poincare duality gives $H_2(\tilde{M};\mathbb{Z})\cong H^1(\tilde{M};\mathbb{Z})=0$. Connectedness gives $H_0\cong \mathbb{Z}$, and the orientation fundamental class gives $H_3\cong \mathbb{Z}$; higher groups vanish by dimension. ConsequentlyThus the universal cover of a closed three-manifold with finite fundamental group is an integral homology sphere. No homeomorphism classification of three-manifolds is needed for this homological conclusion.