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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.

For $n\ge 1$, let $\#$ denote concatenation in the first cube coordinate and let $\cdot$ denote pointwise multiplication of maps into the topological group. Both operations descend to the homotopy group, and the constant map is their common identity. They satisfy the interchange rulewhich follows by considering the two halves of the first coordinate. Using the identity class $0$ givesTherefore $[\alpha \beta ]=[\alpha ]+[\beta ]$. The equality is on based homotopy classes; the usual reparametrization homotopies justify the unit identities for concatenation. This is the Eckmann-Hilton argument, and proves homotopy-group addition in a topological group even for $n=1$.

For the unit quaternions, the explicit inverse isBoth compositions cancel in the displayed order, without commuting the quaternions. Multiplication, inversion and all integer powers are continuous, so this proves that ${\Phi }_{k,l}$ is a homeomorphism for all integers $k,l$.

Put $m=k+l$. Let $a,b$ be the standard generators of $H_3(S^3\times S^3;\mathbb{Z})$, represented by the first and second factors. The previous pointwise-multiplication result and the Hurewicz theorem imply that the mapping degree of $q\mapsto q^m$ on $S^3$ is $m$, including negative integers. Restricting $\Phi$ to the two factors therefore givesThe map on $H_0$ is the identity. For the top homology group, take the dual degree-three cohomology classes $x,y$. We have ${\Phi }^{\ast }x=mx+y$, ${\Phi }^{\ast }y=x$. Since these classes have odd degree, graded commutativity of the cup product givesThus ${\Phi }_{\ast }$ is multiplication by $-1$ on $H_6$, and all other homology groups of the product are zero.

For the gluing, regard $\Phi$ as the attaching identification from $\partial (D^4\times S^3)$ to $\partial (S^3\times D^4)$. In these coordinates its first input is the boundary coordinate of $D^4$, its second input the fibre coordinate; its second output is the boundary coordinate of the other $D^4$. This makes the two pieces the usual two trivializations of a three-sphere bundle over the four-sphere.

Let $A=S^3\times D^4$ and $B=D^4\times S^3$, and parametrize their common boundary $T=S^3\times S^3$ using the $B$ coordinates. Collar neighbourhoods give an open cover with the same homotopy types, so the Mayer–Vietoris sequence applies. Both pieces retract onto $S^3$. In degree three, the map into the homology of the pieces isIndeed, inclusion into $A$ retains the first output coordinate, while inclusion into $B$ retains the second input coordinate. Exactness now givesThe second relation eliminates the second generator of the cokernel, leaving a single generator with relation $m$ times that generator equal to zero. Thus $\mathrm{coker}L=\mathbb{Z}/m\mathbb{Z}$. The kernel is zero if $m\ne 0$ and is generated by $(1,0)$ if $m=0$.

The degree-six boundary class gives $H_7(X_{k,l})\cong H_6(T)=\mathbb{Z}$. All remaining positive-degree groups outside degrees $3,4,7$ vanish by the same Mayer–Vietoris sequence; connectedness gives $H_0=\mathbb{Z}$. Consequently the complete integral answer isHere $\mathbb{Z}/0\mathbb{Z}$ means $\mathbb{Z}$, and a negative $m$ gives the same cyclic group as $|m|$. In particular, $m=\pm 1$ gives the integral homology of a seven-sphere, while $m=0$ gives the integral homology of $S^3\times S^4$.

A Serre fibration has the homotopy lifting property for disks: for every $k\ge 0$, a map $D^k\to E$ and a homotopy $D^k\times I\to B$ starting at its composite with $p:E\to B$ admit a compatible lift $D^k\times I\to E$. Equivalently, it has that lifting property for CW complexes. Relative lifting for CW pairs follows by attaching cells.

Let $i:F\hookrightarrow E$ be the inclusion of the chosen fiber. The long exact sequence of homotopy groups of a fibration isIts low-dimensional end isThe maps $i_{\ast }$ and $p_{\ast }$ are induced by inclusion and projection on based maps; at degree zero they send a path component to its containing or image component. The last part is an exact sequence of pointed sets, not in general a sequence of groups.

For $\partial :{\pi }_q(B)\to {\pi }_{q-1}(F)$, represent a class by a based cube $a:I^{q-1}\times I\to B$, constant at $b_0$ on its boundary. Lift it from the constant map $e_0$ on the bottom and side faces, using relative disk lifting. Its top face lands in $F$ and is constant on that face's boundary; its class is $\partial [a]$. The choice of lifts does not change the resulting class. For $q=1$, lift a based loop starting at $e_0$ and take the path component of its endpoint in $F$. This fixes the boundary-map convention and defines every map in the displayed sequence.

For the splitting assertion, first take $n\ge 1$. We use the cohomological Serre spectral sequence with coefficient group $G$:Since $B$ is simply connected, these coefficient systems are constant. The Eilenberg–MacLane space $F=K(G,n)$ has $H^t(F;G)=0$ for $0<t<n$, andFor $n=1$ this uses $H_1(F)=G$, with $G$ abelian; for $n>1$ it follows from the Hurewicz theorem and the universal coefficient theorem for cohomology. Let $u$ correspond to ${\mathrm{id}}_G$, the universal cohomology class of an Eilenberg–MacLane space.

In bidegree $(0,n)$ there are no incoming differentials. For $2\le r\le n$, an outgoing differential lands in a vanishing fiber-cohomology row. The only remaining possible differential isThus $u$ survives. The edge map, which is restriction to $F$, supplies a classBy representability of cohomology by Eilenberg–MacLane spaces, choose a based map $q:E\to K(G,n)$ representing $c$. On $F$, $q$ induces the identity on ${\pi }_n=G$ and is an isomorphism on every homotopy group, since the other positive groups vanish.

Consider $h=(p,q):E\to B\times K(G,n)$. This is a map of fibrations over $B$, whose map on the fiber is the weak equivalence just obtained. The map on the base is the identity. Comparing the two instances of the long exact sequence of homotopy groups of a fibration and applying exactness, or the Five lemma in the group-valued range, shows that $h$ induces all homotopy isomorphisms. The spaces are path-connected because the base and fiber are. Hence the splitting of a simply connected Eilenberg–MacLane fibration givesThe crucial use of $H^{n+1}(B;G)=0$ is to lift the fiber's identity class; existence of a section alone was not assumed to prove a product splitting.

If $K(G,0)$ is allowed, its components have trivial positive homotopy groups. The simply connected base has no monodromy on this discrete set. Obstruction theory on the CW base gives a section in each fiber component, since all positive-dimensional fiber obstructions vanish and the component system is constant. Together these sections give $B\times G\to E$, a weak equivalence on each component. Thus the same conclusion also covers that interpretation.

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.

Write $u=u_n\in H^n(K(\mathbb{Z}/2,n);{\mathbb{F}}_2)$ for the fundamental class. We use the path-loop Serre spectral sequence, its multiplicative structure, and the Kudo transgression theorem: transgression of the fundamental classes and their compatible Steenrod squares is natural, equivalently cohomology suspension commutes with these stable operations. The initial ring is $H^{\ast }(K(\mathbb{Z}/2,1);{\mathbb{F}}_2)={\mathbb{F}}_2[t_1]$.

Here is the low-degree calculation. In the first path fibration, $t$, $t^2$ and $t^4$ transgress respectively to $u_2$, ${\mathrm{Sq}}^1{u}_2$ and ${\mathrm{Sq}}^2{\mathrm{Sq}}^1{u}_2$, in degrees $2,3,5$. These supply all indecomposable base classes through degree five; the other classes there are products. Repeating the path-loop calculation raises the degree of the transgressive classes by one. For $n=3$, the class ${\mathrm{Sq}}^2{u}_3$ transgresses from $u_2^2$, and ${\mathrm{Sq}}^2{\mathrm{Sq}}^1{u}_3$ from the degree-five generator. For $n=4$, ${\mathrm{Sq}}^3{u}_4$ transgresses from $u_3^2$. For larger $n$ all four displayed low-degree operations are transgressive indecomposables. Acyclicity of the path-space total cohomology forces these transgressions and excludes additional classes in this range.

More systematically, this is the range up to $n+3$ of the Serre polynomial generators for mod-two Eilenberg–MacLane cohomology:For an admissible sequence of Steenrod squares, $I=(i_1,\dots ,i_r)$ has $i_j\ge 2i_{j+1}$; its degree increment is $\sum i_j$, and its Steenrod excess is $e(I)=i_1-i_2-\cdots -i_r$. Include the empty sequence. Up to increment three the nonempty possibilities are $(1),(2),(3),(2,1)$; the strict excess bound and possible products explain precisely the small-$n$ exceptions.

The low-degree mod-two cohomology of an Eilenberg–MacLane space isNegative-degree cohomology is zero. For $n=2$, ${\mathrm{Sq}}^{2}u=u^2$ and ${\mathrm{Sq}}^{3}u=0$ by instability, so the product in degree five must not be omitted. For $n=3$, ${\mathrm{Sq}}^{3}u=u^2$ is a product, still independent from ${\mathrm{Sq}}^2{\mathrm{Sq}}^{1}u$. For $n\ge 4$ the listed classes are indecomposable. The Adem relations include ${\mathrm{Sq}}^1{\mathrm{Sq}}^1=0$ and ${\mathrm{Sq}}^1{\mathrm{Sq}}^2={\mathrm{Sq}}^3$, so no further increment-three class comes from reversing the two squares.

Now use the space actually printed in the PDF,the converted TeX dropped the projective-space $P$'s. This stunted real projective space has one cell in dimensions $7,8,9,10$, besides its basepoint, and is $6$-connected. The integral cellular boundary is $2$ in even dimensions and zero in odd dimensions. HenceThe Hurewicz theorem gives ${\pi }_7(Y)=\mathbb{Z}/2$.

For the next two groups, let $K=K(\mathbb{Z}/2,7)$ and choose $f:Y\to K$ representing the generator of $H^7(Y;{\mathbb{F}}_2)$. It induces an isomorphism on ${\pi }_7$, and the target's higher homotopy groups vanish.

We need the low-degree integral homology of a mod-two Eilenberg–MacLane space. Here it isTo justify the torsion orders, the Serre class theorem first makes every positive-degree integral homology group of $K$ a finite 2-group. The universal coefficient theorem for cohomology with ${\mathbb{F}}_2$ and the low-degree dimensions give one cyclic summand in degrees seven, nine and ten and none in degree eight. The mod-two Bockstein homomorphism is ${\mathrm{Sq}}^1$. Its relevant nonzero actions areThe last target is nonzero: $(3,1)$ is admissible with excess two, below seven. It also follows by transgressing the nonzero square $({\mathrm{Sq}}^1{u}_2{)}^2$ through successive path fibrations. Each nonzero Bockstein pairs the mod-two classes associated with a cyclic integral summand of order exactly two; for a summand of order $2^r$ with $r>1$, this first Bockstein would be zero. In degree ten, ${\mathrm{Sq}}^{3}u$ is already the Ext class from $H_9$, so the independent class ${\mathrm{Sq}}^2{\mathrm{Sq}}^{1}u$ detects $H_{10}$. This proves all four integral groups without confusing them with mod-two Betti numbers.

Let $x\in H^1({\mathbb{RP}}^{10};{\mathbb{F}}_2)$ be its usual generator. The quotient classes in degrees $7$ through $10$ identify with $x^7,\dots ,x^{10}$ via the pair's cohomology. The Steenrod squares on real projective space satisfyThussince $\left(\genfrac{}{}{0ex}{}{7}{2}\right)$ and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)$ are odd, while $\left(\genfrac{}{}{0ex}{}{8}{2}\right)$ is even. In degree nine, the nonzero cohomology map detects the map on $H_9$, as $H_8=0$ on both sides. Consequently $f_{\ast }:H_9(Y;\mathbb{Z})\to H_9(K;\mathbb{Z})$ is an isomorphism.

Treat $f$ as a mapping-cylinder pair $(K,Y)$. It is $8$-connected: both spaces are $6$-connected, ${\pi }_7$ is an isomorphism, and ${\pi }_8(K)=0$. The relative homology sequence gives $H_9(K,Y)=0$, since $H_9(Y)\to H_9(K)$ is an isomorphism and $H_8(Y)=0$. The Relative Hurewicz theorem then gives ${\pi }_9(K,Y)=0$, and the relative homotopy sequence identifies this group with ${\pi }_8(Y)$. Thus ${\pi }_8(Y)=0$ and the pair is now $9$-connected.

Next,because $H_{10}(Y)=0$ and the degree-nine map is an isomorphism. Apply the Relative Hurewicz theorem again and use ${\pi }_{10}(K)={\pi }_9(K)=0$:Therefore the homotopy groups of the stunted projective space through degree nine are

A Serre fibration $p:E\to B$ has the homotopy lifting property for every disc: given $h:D^k\times I\to B$ and ${\tilde{h}}_0:D^k\to E$ with $p{\tilde{h}}_0=h(-,0)$, there is a lift $\tilde{h}$ extending ${\tilde{h}}_0$. Equivalently one may test this property on CW complexes.

For a map $g:E\to B$, its mapping-path fibration replacement isPath spaces have the compact-open topology, with the usual compactly generated convention. The inclusion $s(e)=(e,{\mathrm{const}}_{g(e)})$ satisfies $ps=g$. Forgetting the path is a homotopy inverse to $s$: shorten $\gamma$ to $u\mapsto \gamma (tu)$, from its constant initial path at $t=0$ to the original path at $t=1$. The equivalence here means a commuting triangle over the same base whose map of total spaces is a homotopy equivalence; it does not assert a fibrewise homotopy inverse over the base. Appending the successive segments of a moving base path gives the homotopy lifting property for $p$, even for arbitrary parameter spaces, so $p$ is in particular a Serre fibration. Its ordinary fibre over $b_0$ is the homotopy fibre

For a based Serre fibration, with fibre inclusion $j:F\hookrightarrow E$, the long exact sequence of homotopy groups of a fibration isThe maps $j_{\ast },p_{\ast }$ compose based sphere representatives with $j,p$; the maps on path components do the same with points. To define $\partial$, represent a class by $(I^k,\partial I^k)\to (B,b_0)$ and lift the cube starting at $e_0$ on its bottom face, keeping its vertical sides at $e_0$. The opposite face then gives $(I^{k-1},\partial I^{k-1})\to (F,e_0)$. Relative lifting for a CW pair gives this construction; its based homotopy class is independent of choices. At $k=1$ it records the component of the endpoint of the lifted loop. The last terms are pointed sets rather than groups in general.

Now write $W=S^1\vee S^n$ and let $f:W\to S^n$ be the collapse map. Inclusion of the $n$-sphere is a section, so $f_{\ast }:{\pi }_k(W)\to {\pi }_k(S^n)$ is surjective in every positive degree. Thus all connecting maps in its homotopy fibre sequence vanish. The fibre is path-connected, andThe section also gives the first assertion when $n=2$, by killing the connecting map from ${\pi }_2(S^2)$.

The universal cover of $W$ consists of a line with an $n$-sphere attached at each integer. It is homotopy equivalent to the wedge sum ${\bigvee }_{k\in \mathbb{Z}}{S}^n$. This cover is $(n-1)$-connected, so the Hurewicz theorem identifies its $n$th homotopy group with its $n$th homology. Choose a generator $t$ of the fundamental group whose action shifts the sphere indexed by $k$ to that indexed by $k+1$. The resulting homotopy groups as modules over the fundamental group giveEvery lifted sphere maps to $S^n$ with degree one. Consequently $f_{\ast }$ is the augmentation $\epsilon (\sum a_k{t}^k)=\sum a_k$, with the trivial $R$-action on $\mathbb{Z}$. The inclusion of the fibre respects the fundamental group actions, soThe augmentation ideal of the infinite cyclic group is generated by $t-1$: multiply a Laurent polynomial by a power of $t$ and use divisibility by $t-1$ when its value at one vanishes. Multiplication by $t-1$ is injective, giving the last isomorphism. This is a free module of rank one, whose underlying abelian group has countably infinite rank; its ${\pi }_1$-action is not trivial.

Choose the Hopf map $\eta$ and orientations so its Hopf invariant is one. The cellular cohomology of $Y_d$ is $\mathbb{Z}$ in degrees $0,2,4$ and zero otherwise. Let $x,z$ be generators in degrees two and four. Precomposing $\eta$ with a degree-$d$ self-map of $S^3$ represents $d\eta$. The induced map of mapping cones is the identity on the two-cell and has degree $d$ on the four-cell. Since the Hopf mapping cone has cup square equal to its top generator, naturality of the cup product gives $x^2=dz$. Equivalently, precomposition scales the Hopf invariant by degree. HenceThe coefficient is $d$, not $d^2$: multiplication in ${\pi }_3(S^2)$ is being used, rather than postcomposition by a degree-$d$ map of the target sphere.

For $X=X_{d,f}$, the only nonzero cellular boundary is $C_4\to C_3$, multiplication by $f$. Thus its cellular cohomology givesLet $x$ again generate $H^2$, and let $z$ be the class of the four-cell cochain. Collapsing the three-sphere gives a map $X\to Y_d$ which pulls the degree-two generator back to $x$ and the degree-four generator back to $z$. Hence $x^2=dz$, now interpreted modulo $f$. Every other product of positive-degree classes vanishes by dimension. This is the cohomology ring of a Hopf attachment with a sphere summand. More explicitly, if $f\ne 0$,If $f=0$, there is additionally a generator $y$ of degree three, with $xy=y^2=yz=0$. The displayed description of the groups uses $\mathbb{Z}/0\mathbb{Z}=\mathbb{Z}$, so the zero case is included. In particular, a nonzero $f$ kills the degree-three cohomology but can leave torsion in degree four.

There are no one-cells, and attaching cells of dimension at least three does not change ${\pi }_1$. Thus $X$ is simply connected. The Hurewicz theorem in degree two gives

To compute ${\pi }_3$, choose a map $g:X\to {\mathbb{CP}}^{\infty }=K(\mathbb{Z},2)$ representing $x$. This Eilenberg–MacLane space can be realized by the classifying space of $S^1$. On $S^2$ the map is the standard inclusion; on $S^3$ it is constant, and it extends over the four-cell since ${\pi }_3(K(\mathbb{Z},2))=0$. Its homotopy fibre $T$ is the pullback of the universal circle bundle $S^{\infty }\to {\mathbb{CP}}^{\infty }$. The map on ${\pi }_2$ is an isomorphism, so the long exact sequence of homotopy groups of a fibration gives ${\pi }_1(T)={\pi }_2(T)=0$ and ${\pi }_3(T)\cong {\pi }_3(X)$. Therefore the Hurewicz theorem identifies the latter with $H_3(T;\mathbb{Z})$.

Over $W=S^2\vee S^3$, the circle-bundle total space isThe first summand is the Hopf fibration total space over $S^2$, and the second is the trivial bundle over $S^3$. Their intersection is the fibre over the wedge point. The Mayer–Vietoris sequence gives $H_3(P)=\mathbb{Z}A\oplus \mathbb{Z}B$, with generators from the two three-spheres. These are the lifts of ${\eta }_1$ and $i_2$. The circle bundle over $W$ also has a 2-connected total space, so lifting and the Hurewicz theorem identify $d{\eta }_1+f{i}_2$ with $dA+fB$.

Over the attached four-disc the bundle is trivial. The resulting relative pair has the homology of $(S^1\times D^4,S^1\times S^3)$, so $H_4(T,P)=\mathbb{Z}$ and $H_3(T,P)=0$. Its boundary map sends a generator to the lift of the attaching map, namely $dA+fB$. The long exact sequence in relative homology now givesApplying Smith normal form to this one relation proves the third homotopy group of a Hopf attachment with a sphere summand:Since $d>0$, the greatest common divisor is positive even when $f=0$. This derivation uses the required map to an Eilenberg–MacLane space and determines the extension, rather than merely the orders of its pieces.

Because $S^n$ is simply connected for $n\ge 2$, the homological Serre spectral sequence has constant coefficients and only two nonzero columns:Its only possible nonzero differential isUsing the orientation generator of $H_n(S^n)$ to identify both columns with $H_{\ast }(F)$ defines the Wang homomorphismThe kernel and cokernel descriptions of the two surviving columns splice with the filtration of $H_{\ast }(E)$ to give the Wang sequence over a sphere

Let $F_m$ be the homotopy fiber of a degree-$m$ map $f_m:S^n\to S^n$. Apply this sequence to the fibrationAt the bottom, the map between the two copies of $H_n(S^n)$ is multiplication by $m$. It follows thatThe first torsion group can also be seen from ${\pi }_{n-1}(F_m)\cong \mathbb{Z}/m$ and the Hurewicz theorem; the Wang map then propagates it periodically.

The universal coefficient theorem for cohomology givesEvery product of two positive-degree classes is zero. For $n>2$, this follows immediately because the sum of two degrees of the form $k(n-1)+1$ is not of that form. For $n=2$, degree counting does not suffice, since $H^j(F_m)\cong \mathbb{Z}/m$ for every $j\ge 2$. Use instead the multiplicative cohomological Serre spectral sequence forIts transgression in degree one is multiplication by $m$. Every positive integral cohomology class that survives lies in filtration two, while the product of two such classes lies in filtration four; the base has dimension two, so filtration four is zero. Thus the reduced cohomology is a square-zero ideal, andas a graded ring, with zero multiplication on the second summand.

Products of Eilenberg–MacLane spaces satisfywhere the last isomorphism uses the Chinese remainder theorem. This space is $2$-connected, so the Hurewicz theorem identifiesThe assumed surjection on $H_3$ is an isomorphism because the group is finite. Hence $f$ is an isomorphism on ${\pi }_3$; all other homotopy groups of the source and target vanish. Thus $f$ is a weak homotopy equivalence, and the Whitehead theorem for Kan complexes makes it a homotopy equivalence.

The loop-space shift of homotopy groups givesThus $\Omega S^4$ is $2$-connected and its first nonzero homotopy group isThe Hurewicz theorem now givesand every positive integral homology group below degree $3$ vanishes. The smallest requested degree is therefore $i=3$.

If an orientable three-manifold has nonzero ${\pi }_2$, it contains an embedded essential two-sphere. Consequently a closed orientable irreducible three-manifold with infinite fundamental group is aspherical: its simply connected noncompact universal cover has zero second homotopy and top ordinary homology, hence is acyclic and contractible by the Hurewicz theorem and Whitehead theorem.