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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 orthogonal group acts transitively on unoriented lines in ${\mathbb{R}}^n$. The stabilizer of the line spanned by the first coordinate isThus $G/H={\mathbb{RP}}^{n-1}$; using only $O(n-1)$ would stabilize a vector and give a sphere instead. For the universal contractible free $G$-space $EG$, the associated bundle $EG/H\to EG/G$ gives the Serre fibration $G/H\to BH\to BG$. Its total space identifies up to homotopy with $BO(1)\times BO(n-1)$.

If $n$ is odd, $n-1$ is even and the rational cohomology of ${\mathbb{RP}}^{n-1}$ is just $\mathbb{Q}$ in degree zero. This follows from the integral projective cellular differentials, alternating zero and multiplication by two, with the even-dimensional top class killed rationally. The cohomological Serre spectral sequence therefore has only the rowwith all other rows zero. The local system in degree zero is trivial because the fiber is connected. There can be no differential or extension, and the multiplicative edge map $H^{\ast }(BO(n);\mathbb{Q})\to H^{\ast }(BH;\mathbb{Q})$ is an isomorphism of rings.

Also $BO(1)={\mathbb{RP}}^{\infty }$ has rational cohomology $\mathbb{Q}$ in degree zero only, by the same cellular calculation. The Künneth theorem identifies $H^{\ast }(BH;\mathbb{Q})$ with $H^{\ast }(BO(n-1);\mathbb{Q})$. The inclusion fixing the first coordinate factors through $H$, setting its $O(1)$ coordinate to one; hence the resulting isomorphism is the actual restriction along this inclusion:This includes $n=1$, with $BO(0)$ a point, and proves rational odd-rank stabilization of orthogonal classifying spaces.

For even $n$, the fiber is odd-dimensional real projective space and has a further rational class in degree $n-1$. The spectral sequence has a possible additional row, now with determinant-sign local coefficients: an orientation-reversing orthogonal matrix acts with degree $-1$ on the covering sphere and hence on the odd-dimensional projective space. Therefore the one-row collapse argument is unavailable; this top row must not be treated as automatically untwisted.

There is an actual failure of injectivity in every even rank. Write $n=2m$. The universal class $p_m\in H^{4m}(BO(2m);\mathbb{Q})$ is nonzero: pull the universal bundle back to the sum of the real underlying bundles of $m$ complex lines over $({\mathbb{CP}}^{\infty }{)}^m$. If their first Chern classes are $x_j$, the Pontryagin formula $p_m(E)=(-1{)}^m{c}_{2m}(E\otimes \mathbb{C})$ gives $p_m={\prod }_j{x}_j^2\ne 0$. On $BO(2m-1)$ the restricted bundle is ${\epsilon }^1\oplus {\gamma }_{2m-1}$; its top complex Chern class $c_{2m}$ vanishes by the rank bound and the trivial summand. Thus $p_m$ restricts to zero. The natural restriction is not injective for even $n$, as the extra fiber cohomology already warns.

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

Choose a contractible free $G$-space $EG$, with $BG=EG/G=K(G,1)$ the classifying space of a discrete group. The Borel construction gives the fibrationBecause the action on the sphere is free, the map $E\to S^{n-1}/G$ has contractible fibre $EG$ and is a homotopy equivalence. The quotient is a closed $(n-1)$-dimensional manifold, so $H^k(E;\mathbb{F})=0$ for $k>n-1$.

The hypothesis on integral sphere cohomology makes the $G$-action on $H^{n-1}(S^{n-1};\mathbb{F})$ trivial as well. Thus the Serre spectral sequence has just two rows, both copies of $H^{\ast }(BG;\mathbb{F})$, in fibre degrees zero and $n-1$. If $u$ is the fibre generator, the only possible differential is $d_n$, and its transgression definesThe multiplicative differential rule gives $d_n(au)=(-1{)}^{|a|}a⌣\Delta$. Up to the graded sign, this is multiplication by $\Delta$. For $i>0$, the terms $E_{\infty }^{i,n-1}$ and $E_{\infty }^{i+n,0}$ must both vanish because their total degrees exceed $n-1$. The first vanishing says the relevant multiplication map has zero kernel; the second says it has zero cokernel. ThereforeThis is periodic group cohomology from a free sphere action. Degree zero was excluded for a reason: $E_{\infty }^{0,n-1}$ can contribute to the quotient's top cohomology.

For $C_p=\mathbb{Z}/p$, use its periodic free resolution with alternating maps $g-1$ and $N=1+g+\cdots +g^{p-1}$. Applying ${\mathrm{Hom}}_{\mathbb{Z}[C_p]}(-,{\mathbb{F}}_p)$ makes every differential zero, so $H^i(B{C}_p;{\mathbb{F}}_p)$ is one-dimensional in every nonnegative degree. The two-step shift of the resolution supplies a nonzero degree-two class $u$ whose cup product gives the periodicity isomorphisms. For odd $p$, let $v$ be a nonzero degree-one class. Graded commutativity gives $v^2=0$, and the shift gives nonzero $v{u}^k$ and $u^k$ in every degree. For $p=2$, the two resolution maps agree over ${\mathbb{F}}_2[C_2]$, giving a one-step shift whose degree-one class has nonzero powers. Hence the cohomology ring of a finite cyclic group over its prime field isHere the exterior factor is an exterior algebra and the polynomial factor is a polynomial ring. For odd $p$, $u$ may be chosen as the Bockstein homomorphism of $v$.

If $G$ contained $C_p\times C_p$, restriction would give that subgroup the same free, cohomologically trivial sphere action. It would therefore also have period $n$ in positive-degree cohomology. But $B(C_p\times C_p)\simeq B{C}_p\times B{C}_p$, and the Künneth theorem givesThese dimensions strictly increase, so the groups in degrees $i$ and $i+n$ cannot be isomorphic. Thus $G$ contains no subgroup isomorphic to $\mathbb{Z}/p\times \mathbb{Z}/p$ for any prime $p$.

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.

The space $K(\mathbb{Z}/n,1)$ is a classifying space $B(\mathbb{Z}/n)$. The periodic resolution of a finite cyclic group givesThus positive odd cohomology vanishes and every positive even group is $\mathbb{Z}/n$.

Let $p$ be odd. If $p\nmid n$, transfer makes multiplication by $n$ both zero and invertible on positive-degree cohomology, soIf $p\mid n$, restriction to the cyclic Sylow $p$-subgroup and transfer giveFor $n=p$, one may take $v=\beta u$, where $\beta$ is the mod-$p$ Bockstein homomorphism.

For the second part put $A=\mathbb{Z}/p$ and $X=K(A,1)\times K(A,1)$. The long exact homotopy sequence of the homotopy fibre of $f:X\to K(A,2)$ givesand ${\pi }_k(F)=0$ for $k>1$. Therefore $F$ is a $K(G,1)$ and $|G|=p^3$.

Regard it as the fibrationWriteand write $w,t$ for the degree-one and degree-two generators of the fibre. The fibration is classified by $u_1{u}_2$, so in its cohomological Serre spectral sequenceThe Kudo transgression theorem and the mod-$p$ Bockstein giveConsequently $H^1(F;{\mathbb{F}}_p)$ has basis $u_1,u_2$, while $H^2(F;{\mathbb{F}}_p)$ has the four surviving classes represented byIf $G$ were abelian, an abelian group of order $p^3$ mapping onto $A^2$ would be either $A^3$ or $\mathbb{Z}/p^2\times A$. The first has three-dimensional $H^1(-;{\mathbb{F}}_p)$; the second has two-dimensional $H^1$ but three-dimensional $H^2$. Both contradict the dimensions just calculated. Hence $G$ is nonabelian.

Let $i:F\hookrightarrow E$. A pairis transgressive pair when, in the long exact sequence of the pair $(E,F)$,where $H^{r+1}(B,b_0)$ is identified with reduced cohomology. In the Serre spectral sequence, this says that $x$ survives to the transgression andunder the edge identifications, modulo the usual earlier-differential indeterminacy.

The Kudo transgression theorem says that if $(x,y)$ is transgressive and $0\le j\le r$, thenis transgressive. To prove it, use relative Steenrod squares. Naturality givesand stability, equivalently compatibility with the suspension isomorphism, makes squares commute with the connecting map:Applying ${\mathrm{Sq}}^j$ to $\delta x={\pi }^{\ast }y$ proves the theorem. The properties used are naturality, stability, additivity, and the instability conditions ${\mathrm{Sq}}^{j}z=0$ for $j>|z|$ and ${\mathrm{Sq}}^{|z|}z=z^2$.

Let $x\in H^1({\mathbb{RP}}^{\infty };{\mathbb{F}}_2)$ be the generator. Instability givesThe Cartan formula says the total square is multiplicative, soComparing components yields the complete formulawith the binomial coefficient reduced modulo two.

Finally, ${\mathrm{Sq}}^1$ is the Bockstein homomorphism associated withIf a mod-two cocycle representing $y$ is lifted to an integral cochain $A$, write $dA=2B$. Then $B$ modulo two represents ${\mathrm{Sq}}^{1}y$. But $dB=0$, because integral cochains are torsion-free and $2dB=d^{2}A=0$. Thus $B$ itself is a cocycle lift, so its Bockstein vanishes. Thereforefor every space $Y$ and every $y\in H^{\ast }(Y;{\mathbb{F}}_2)$.

Apply the rational Serre spectral sequence to the path-loop fibration of $K(\mathbb{Z},3)$. Its fiber iswhose rational cohomology ring is $\mathbb{Q}[c]$ with $|c|=2$. Since the path space is contractible, $c$ must transgress to a nonzero class $x\in H^3(K(\mathbb{Z},3);\mathbb{Q})$. Multiplicativity givesOver $\mathbb{Q}$ these differentials pair and kill every positive-degree class except $x$, while graded commutativity gives $x^2=0$. Hence

Use the path-loop fibrationand its rational Serre spectral sequence. The total space is contractible, while $H^{\ast }(S^4;\mathbb{Q})$ is $\mathbb{Q}$ in degrees $0$ and $4$ and zero otherwise. The only possible nonzero differential isConvergence to the cohomology of a point first forces $d_4:H^3(\Omega S^4;\mathbb{Q})\to H^4(S^4;\mathbb{Q})$ to be an isomorphism, and then inductively forces an isomorphism from each nonzero vertical group to the group three degrees below it in the other column. Therefore

Write $F=K(\mathbb{Z}/4,1)$, $E=K(\mathbb{Z},2)$, and $B=K(\mathbb{Z},2)$. In total degree at most two, the $E_2$ page of the mod-two Serre spectral sequence hasand $E_2^{1,0}=E_2^{1,1}=0$. A periodic free resolution of the cyclic group gives $H_1(F;\mathbb{Z})=\mathbb{Z}/4$ and $H_2(F;\mathbb{Z})=0$. Hence $H^1(F;{\mathbb{F}}_2)\cong \mathrm{Hom}(\mathbb{Z}/4,{\mathbb{F}}_2)$, while the universal coefficient theorem for cohomology gives $H^2(F;{\mathbb{F}}_2)\cong \mathrm{Ext}(\mathbb{Z}/4,{\mathbb{F}}_2)$; both are one-dimensional.

The edge map $H^2(B;{\mathbb{F}}_2)\to H^2(E;{\mathbb{F}}_2)$ is induced by multiplication by $4$ and is therefore zero modulo two. Consequentlyis an isomorphism. The differential out of $E_2^{0,2}$ is zero, because the total space has a one-dimensional $H^2$ which must survive in filtration zero. Thus for every $r\ge 3$ and $p+q\le 2$,and all other groups in that range vanish.

It follows thatThe surviving filtration-zero class is the restriction of the degree-two class of $E$, so

Suppose a finite group acts freely on $S^{n-1}$ and trivially on its integral cohomology. The Borel construction gives a sphere fibration over $BG$ whose total space is homotopy equivalent to the $(n-1)$-dimensional quotient. Its Serre spectral sequence has just two rows with trivial local coefficients. Transgressing the fibre generator gives $\Delta$; the sole differential is multiplication by it up to sign. For $i>0$, both the source and target terms have total degrees above $n-1$, so their surviving kernel and cokernel must vanish. Hence multiplication by $\Delta$ gives period $n$ in positive-degree cohomology over any field.

The ring is exterior on its degree-$n$ universal class for odd $n$, and polynomial for even $n$. The path-loop Serre spectral sequence alternates these two types by transgression. In the polynomial-fiber case its derivation has nonzero coefficients $k$ on powers, which become invertible over the rationals.

The stabilizer of a line in $O(n)$ is $O(1)\times O(n-1)$, giving the homogeneous space ${\mathbb{RP}}^{n-1}$. Its classifying-space Serre fibration has rationally acyclic fiber for odd $n$, so its cohomological Serre spectral sequence has only its degree-zero row and gives the displayed restriction isomorphism. The factor $BO(1)={\mathbb{RP}}^{\infty }$ is itself rationally acyclic. For even $n$, the fiber has a top rational class, with determinant-sign monodromy, so the one-row argument fails. The top universal Pontryagin class $p_{n/2}$ is nonzero on $BO(n)$ but vanishes on restriction to $BO(n-1)$, witnessing genuine failure of stabilization.