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There is no fiber bundle $F\to S^m\to S^q$ when $m>q\ge 3$ and $q$ is odd. A local trivialization makes $F$ a retract of an open subset of $S^m$, so $H^j(F;\mathbb{Q})=0$ for $j>m$; the long exact sequence of homotopy groups of a fibration makes $F$ connected. In the multiplicative rational Serre spectral sequence, only base columns zero and $q$ occur. Since $H^q(S^m)=0$, a fibre class $a\in H^{q-1}(F;\mathbb{Q})$ must satisfy $d_{q}a=u$, the base generator. Its degree is even. If $a^s=0$ is its first vanishing power, the derivation law gives $0=d_q(a^s)=su{a}^{s-1}\ne 0$ on the $E_q$ page, a contradiction. Bounded fibre cohomology ensures such a vanishing power exists. In particular $S^4$ cannot be the total space of a bundle over $S^3$.

The based path space $P_{x_0}X$ consists of continuous paths beginning at $x_0$, with variable endpoint and the compact-open topology. Endpoint evaluation is a Serre fibration with fiber the based loop space. Its total space is contractible by $\gamma (t)\mapsto \gamma (st)$ as $s$ decreases to zero. This differs from a space of paths with both endpoints fixed, which need not be contractible. The long exact sequence of homotopy groups of a fibration gives ${\pi }_i(\Omega X)\cong {\pi }_{i+1}(X)$ for connected $X$ in the usual based ranges.

The SU(5) grand unified theory has simply connected space $G=SU(5)$ and a connected Standard Model subgroup. More precisely that subgroup is $(SU(3)\times SU(2)\times U(1))/{\mathbb{Z}}_6$; the product notation suppresses a finite central quotient. The long exact sequence of homotopy groups of a fibration $H\to G\to G/H$ givesThus stable magnetic monopoles form, whereas this transition does not require walls or strings. The monopole core and mass scales in an SU(5) transition are set by the massive gauge and scalar fields:For order-one couplings, the core size is of order ${\eta }^{-1}$ and the mass is of order $\eta$ up to the often substantial factor $4\pi /g$. These heavy objects are nonrelativistic soon after formation and can survive as stable relics.

Over the Grassmannian ${\mathrm{Gr}}_k({\mathbb{C}}^N)$, the complex tautological bundle on a Grassmannian isIts fibre at $W$ is the k-plane $W$ itself. A neighbourhood of a fixed plane consists of graphs of linear maps $A:W\to W^{\perp }$; $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The inclusions of coordinate spaces give the infinite Grassmannian and its tautological rank-k bundle ${\gamma }_k^{\infty }$.

Use the customary compact Hausdorff base convention for the classification of complex vector bundles by a Grassmannian; equivalently the argument uses finite subordinate partitions of unity. Give a rank-k complex vector bundle $E\to X$ a continuous fiber metric given by Hermitian inner products. Choose a finite trivializing cover with local orthonormal coordinates $t_i:E{|}_{U_i}\to U_i\times {\mathbb{C}}^k$, and a subordinate partition of unity ${\rho }_i$ with support contained in $U_i$. Defineextending each component by zero off its support. This is continuous and satisfies $\Vert J_x(v){\Vert }^2={\sum }_i{\rho }_i(x)\Vert v{\Vert }^2=\Vert v{\Vert }^2$, so it is a fibrewise embedding. The image planes vary continuously: in local frames, their orthogonal projection matrix is $J(J^{\ast }J{)}^{-1}{J}^{\ast }$. Thus they define a map $c_E:X\to {\mathrm{Gr}}_k({\mathbb{C}}^{kr})$, and $v\mapsto (x,J_{x}v)$ identifies $E$ with $c_E^{\ast }{\gamma }_k^{\infty }$. This proves that every bundle is a tautological pullback vector bundle, by the finite-dimensional embedding of a complex vector bundle.

We next prove the two independence assertions that turn this construction into a bijection. First, homotopic maps have isomorphic pullback vector bundles. Pulling back along a homotopy gives a bundle over compact Hausdorff $X\times I$; embed that bundle in a finite trivial vector bundle by the construction just given. Its image has a continuous family of rank-k orthogonal projection matrices $P(x,t)$. Compactness gives a subdivision $0=t_0<\cdots <t_s=1$ withRestriction of $P(x,t_{j+1})$ to the image at $t_j$ is injective: if it killed $v$, then $v=(P(x,t_j)-P(x,t_{j+1}))v$, impossible by the strict norm bound unless $v=0$. Equal ranks make it an isomorphism, with continuous local inverses. Composing these bundle maps identifies the endpoint vector bundles. The fact that close orthogonal projections identify their image bundles therefore makes the pullback assignment well-defined on homotopy classes.

Conversely, suppose two classifying maps give isomorphic vector bundles. The infinite Grassmannian has its standard weak CW topology; the compact-subset property of CW complexes puts the compact images of the two maps in finite stages. Their tautological pullback vector bundles therefore give two finite-dimensional embeddings $J_0:E\to {\mathbb{C}}^{N_0}$ and $J_1:E\to {\mathbb{C}}^{N_1}$ of the same bundle, after using the vector bundle isomorphism. In orthogonal coordinate blocks interpolate byAt least one coefficient is nonzero, so every $J_t$ is injective and its image planes give a continuous Grassmannian homotopy. One endpoint uses the first block and the other the second block. Moving the latter coordinate copy to the standard one is a fixed unitary change of coordinates, connected to the identity by a path in the unitary group. Such a path exists by diagonalizing the unitary matrix and continuously varying its eigenvalue phases. Hence the original two maps are homotopic in the infinite Grassmannian. This also proves independence of all embedding and metric choices. Together the two directions establishFor arbitrary non-Hausdorff compact spaces without the required numerability, that usual bundle-classification statement needs additional hypotheses; the construction above specifies precisely the conventional setting being used.

Write $B={\mathrm{Gr}}_k({\mathbb{C}}^{\infty })$. For the assumed fiber bundle $U_k\to \nu \to B$ with contractible total space, the long exact sequence of homotopy groups of a fibration has segmentFor $q\ge 2$ this gives group isomorphisms. For $q=1$, the relevant pointed-set part and connectedness of $U_k$ give ${\pi }_1(B)=0$, corresponding to the single component ${\pi }_0(U_k)$. Thus the claimed formula includes its endpoint:Finally the determinant gives a fiber bundle $S{U}_2\to U_2\to S^1$. Since ${\pi }_3(S^1)={\pi }_4(S^1)=0$, its exact sequence yields ${\pi }_3(U_2)\cong {\pi }_3(S{U}_2)$. The given identification $S{U}_2\cong S^3$ and the degree calculation ${\pi }_3(S^3)=\mathbb{Z}$ therefore giveThe equality of free and based homotopy classes uses the simple connectedness just proved. HenceThe classification of rank-two complex vector bundles over the four-sphere is a bijection of isomorphism classes, labelled by the integer clutching function homotopy class after choosing a generator. It is not an assertion that Whitney sum gives a group operation on bundles of fixed rank two.

Sending a matrix to its last column gives the special orthogonal sphere fibrationThe action is transitive and the stabilizer of that column is $SO(n-1)$. Local orthonormal frames, obtained by Gram-Schmidt process near a chosen unit vector, give local sections and hence local bundle trivializations. This allows the long exact sequence of homotopy groups of a fibration to be applied.

At the bottom of the induction, $SO(1)$ is a point and $SO(2)$ is the rotation circle $S^1$. For $SO(3)$, identify ${\mathbb{R}}^3$ with imaginary quaternions. A unit quaternion acts by $v\mapsto qv{q}^{-1}$, preserving length and orientation. Every rotation through angle $\theta$ about a unit imaginary axis $u$ comes from $q=\cos (\theta /2)+u\sin (\theta /2)$. The kernel is precisely $\{1,-1\}$, since a quaternion commuting with all imaginary quaternions is real. Thus this gives the twofold covering space $S^3\to SO(3)$. Its simply connected total space is the universal cover, yielding ${\pi }_{1}SO(3)=\mathbb{Z}/2$. Coverings induce isomorphisms on higher homotopy groups, so ${\pi }_{2}SO(3)={\pi }_2{S}^3=0$ and ${\pi }_{3}SO(3)={\pi }_3{S}^3=\mathbb{Z}$.

For $n\ge 4$, both ${\pi }_1{S}^{n-1}$ and ${\pi }_2{S}^{n-1}$ vanish. The Serre fibration sequence therefore identifies ${\pi }_{1}SO(n-1)$ with ${\pi }_{1}SO(n)$ and propagates vanishing of ${\pi }_2$. Path-connectedness also propagates from connected fiber and base. Consequently the full low-degree answer isHere ${\pi }_0$ records the single connected component.

For the remaining third group use $SO(3)\to SO(4)\to S^3$. The relevant exact segment, using the supplied fourth homotopy group, isThe first homomorphism is zero since a torsion group cannot map nontrivially to $\mathbb{Z}$. Thus it reduces to a short exact sequence $0\to \mathbb{Z}\to {\pi }_{3}SO(4)\to \mathbb{Z}\to 0$. These groups are abelian and the final free group has a lift of its generator, so the sequence splits. ThereforeThese are the low homotopy groups of special orthogonal groups.

Take the Hopf map $\eta :S^3\to {\mathbb{CP}}^1\cong S^2$, given by $\eta (z_0,z_1)=[z_0:z_1]$. The domain has positive-dimensional homology only in degree three, while $H_3(S^2;\mathbb{Z})=0$. In degree two the source group is zero, and all remaining positive-degree source groups are zero too. Hence the induced map on homology is zero in every positive degree.

For nontriviality, use the Hopf fibration $S^1\to S^3\stackrel{\eta }{\to }{S}^2$. It is a fiber bundle: phase multiplication gives each fibre, and choosing a normalized representative on either standard projective chart gives a local trivialization. The theorem invoked is the long exact sequence of homotopy groups of a fibration. Since the universal cover of $S^1$ is contractible, ${\pi }_2(S^1)={\pi }_3(S^1)=0$. Its exact sequence consequently gives an isomorphismThe identity class of $S^3$ is nonzero, since an identity map cannot be null-homotopic while inducing the identity on $H_3(S^3;\mathbb{Z})$. Its image under ${\eta }_{\ast }$ is the class of $\eta$, so that class is nonzero. The Hopf map is not homotopic to a constant despite inducing zero on all positive-degree homology.

False. Take $M=S^4$ and $N=S^3$. Both are closed oriented simply connected manifolds and $4>3$. A cohomological obstruction rules out even a topological fiber bundle.

Suppose a fiber bundle $F\to S^4\to S^3$ existed. The long exact sequence of homotopy groups of a fibration, including its component terms, makes $F$ path-connected. A local trivialization over a small open ball in $S^3$ identifies its preimage, an open subset of $S^4$, with $F\times B^3$. Projection onto $F\times \{0\}$ gives a retraction. An open smooth four-manifold has the homotopy type of a CW complex of dimension at most four, so its rational cohomology vanishes in degrees above four. A retract's cohomology injects into that of the ambient space. Therefore

Use the multiplicative cohomological Serre spectral sequence with rational coefficients. Since $S^3$ is simply connected, there is no nontrivial coefficient action, and only columns zero and three occur on $E_2$. Thus $E_3=E_2$, and the only possibly nonzero differential is $d_3$. Let $u$ be the degree-three base generator. It cannot survive, because the sequence converges to $H^{\ast }(S^4;\mathbb{Q})$ and that ring has zero degree-three part. It has no outgoing differential and its only possible incoming differential is from $E_3^{0,2}$. Hence there is $a\in H^2(F;\mathbb{Q})$ withThe class $a$ is nonzero and, by the degree bound above, has a first vanishing power $a^r=0$ for some $r\ge 2$. Multiplicativity says that $d_3$ is a graded derivation. Since $|a|=2$ is even,But $E_3^{3,2r-2}=H^3(S^3;\mathbb{Q})\otimes H^{2r-2}(F;\mathbb{Q})$, in which $u{a}^{r-1}\ne 0$ by minimality of $r$; also $r$ is nonzero over $\mathbb{Q}$. This is a contradiction. Thus no such fiber bundle exists. This is the special case of a sphere cannot fibre over a lower-dimensional odd sphere required here.

Let $D_{+}^n$ and $D_{-}^n$ be the two closed hemispheres of the sphere. Use $\gamma$ as the clutching function between the trivial bundles on these hemispheres. The clutching construction definesThe projection to $S^n$ is locally a vector bundle. In a collar of the equator, extend the transition function by keeping it constant in the normal coordinate; this gives explicit vector bundle trivializations across the seam. Since the transition matrices lie in the special orthogonal group, the standard fiber orientations and Euclidean inner products agree on overlaps. Thus this is an oriented rank-$k$ vector bundle with a fiber metric.

For $k\ge 2$, choose a base point $u_0$ on the equator. A constant change of the negative hemisphere frame replaces $\gamma$ by $\gamma (u_0{)}^{-1}\gamma$, so we may suppose $\gamma (u_0)=I$. With $e_k$ the last standard unit vector, define the section obstruction for a clutched vector bundleThe crucial point is that a nowhere-zero section can be normalized using the fiber metric. It is then represented by continuous maps ${\sigma }_{\pm }:D_{\pm }^n\to S^{k-1}$ satisfyingon the equator.

If $f_{\gamma }=0$, the extension-null-homotopy criterion for a sphere extends $u\mapsto \gamma (u)e_k$ to $D_{-}^n$. Take this extension as ${\sigma }_{-}$ and the constant map $e_k$ as ${\sigma }_{+}$. They glue to a nowhere-zero section.

Conversely, suppose such a section exists. Restricting a contraction of $D_{+}^n$ shows that ${\sigma }_{+}{|}_{S^{n-1}}$ is homotopic to a constant. The target sphere is path connected, so the constant can be moved to $e_k$. Multiplying this homotopy pointwise by $\gamma (u)$ gives a homotopy from $\gamma {\sigma }_{+}$ to $\gamma e_k$. The first of these maps equals ${\sigma }_{-}{|}_{S^{n-1}}$ and extends over $D_{-}^n$, hence is null-homotopic. So $\gamma e_k$ is null-homotopic too. A freely null-homotopic based sphere map represents zero in its homotopy group; any change of base point sends the zero class to zero. We have proved both directions:For $n=1$, the same argument reads as the extension criterion for a two-point boundary and uses the trivial set ${\pi }_0(S^{k-1})$ when $k\ge 2$. For $k=1$, the special orthogonal group $SO(1)$ is trivial, so the vector bundle is already a trivial vector bundle; the corresponding constant map to $S^0$ has zero obstruction. These cases require no higher homotopy group calculation.

If $k>n$, then $n-1<k-1$ and the permitted vanishing of lower homotopy groups of a sphere makes $f_{\gamma }=0$. Normalize the resulting nowhere-zero section to a unit section $\sigma$, and setLocal orthonormal frames show that these orthogonal complements form a rank-$(k-1)$ vector bundle. The explicit vector bundle isomorphismhas inverse $w\mapsto (w-⟨w,\sigma ⟩\sigma ,⟨w,\sigma ⟩)$. Orient $V^{\prime }$ so that a positive frame followed by $\sigma$ is positive in $V_{\gamma }$. This is splitting a trivial line from a nowhere-zero section and proves the oriented splittingFinally consider the special orthogonal sphere fibrationThe fiber over $e_{s+1}$ consists exactly of $\mathrm{diag}(B,1)$ with $B\in SO(s)$. Near any unit vector, the Gram-Schmidt process continuously completes it to an oriented orthonormal frame, giving local sections of $p$ and local product fiber bundle charts. In particular $p$ has the homotopy lifting property.

Let $\alpha :S^r\to SO(s+1)$ be based and suppose $r<s$. The composite $p\alpha$ is a based null-homotopic map by the given vanishing of ${\pi }_r(S^s)$. Lift a based null-homotopy of $p\alpha$, beginning at $\alpha$, to a homotopy $G:S^r\times [0,1]\to SO(s+1)$. The lift might initially move its base point $u_0$, but $G(u_0,t)$ always fixes $e_{s+1}$. Right multiplication by $G(u_0,t{)}^{-1}$ corrects this: the homotopystill covers the same sphere homotopy, remains based, and starts at $\alpha$. At its endpoint every matrix fixes $e_{s+1}$, so $\tilde{G}(-,1)$ lies in the included $SO(s)$. Thus every homotopy class in the target is represented by one from $SO(s)$:This proves surjectivity of special orthogonal stabilization below the sphere dimension. For $r=0$, this is the corresponding assertion about path components; the positive-rank special orthogonal groups are connected. The lifting argument also explains the zero term ${\pi }_r(S^s)$ in the long exact sequence of homotopy groups of a fibration.

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.

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.

A map $p:E\to B$ is a Serre fibration when it has the homotopy lifting property for every disc: given $g:D^k\to E$ and $H:D^k\times I\to B$ with $pg=H(-,0)$, there is a lift $\tilde{H}:D^k\times I\to E$ satisfying $\tilde{H}(-,0)=g$ and $p\tilde{H}=H$.

Fix $b_0\in B$, $e_0\in F=p^{-1}(b_0)$, and write $i:F\hookrightarrow E$. The long exact sequence of homotopy groups of a fibration isThe first two maps are induced by inclusion and projection. To define $\partial$, represent a class of ${\pi }_k(B)$ by a map of pairs $(D^k,S^{k-1})\to (B,b_0)$, lift it beginning at $e_0$ along radial paths, and restrict the lift to $S^{k-1}$; that restriction lies in $F$. At the bottom, exactness continues through the pointed sets ${\pi }_1(B)\to {\pi }_0(F)\to {\pi }_0(E)\to {\pi }_0(B)$.

Every fiber bundle is a Serre fibration. Pull a bundle back along $H:D^k\times I\to B$; a lift of $H$ is the same as a section of this pullback extending the section over $D^k\times \{0\}$ supplied by $g$. Since $D^k\times I$ is compact, finitely many bundle charts cover it. A Lebesgue-number subdivision of $I$, followed by a finite subdivision of $D^k$, makes each resulting prism lie in one chart. In a trivialization $U\times F\to U$, extend the section across a prism by keeping its $F$-coordinate constant along the interval direction. Proceed prism by prism and time-slab by time-slab; on an already treated face use its prescribed coordinate, and the transition functions ensure agreement on overlaps. The resulting sections glue to the required lift $\tilde{H}$.

Now consider $i:{\mathbb{RP}}^n\hookrightarrow {\mathbb{CP}}^n$. The target is simply connected, so $i_{\ast }$ is zero on ${\pi }_1$. For $k\ge 2$, every based map $S^k\to {\mathbb{RP}}^n$ lifts through the double cover to $S^n$. The compositelifts through the Hopf fibration $S^{2n+1}\to {\mathbb{CP}}^n$ to the real-coordinate inclusion $S^n\hookrightarrow S^{2n+1}$. This inclusion is null-homotopic because ${\pi }_n(S^{2n+1})=0$. Hence $i_{\ast }$ is zero on every ${\pi }_k$, including the cases $n=1$ and $k\ge 2$ where the source groups already vanish.

The last-column map $p(A)=A{e}_{s+1}$ has fiber $SO(s)$, included as $\mathrm{diag}(B,1)$. Local completion of a unit vector by the Gram-Schmidt process gives local fiber bundle charts. Consequently it has the homotopy lifting property and the long exact sequence of homotopy groups of a fibration.