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For a based space $(X,x_0)$ and $k\ge 1$, the homotopy group ${\pi }_k(X,x_0)$ is the set of based homotopy classes of maps $(I^k,\partial I^k)\to (X,x_0)$, equivalently based maps $S^k\to X$. Concatenating along one coordinate defines its group operation; reversing that coordinate gives the inverse. The group is abelian for $k\ge 2$. The symbol ${\pi }_0(X)$ records path components and is generally only a pointed set. An Eilenberg–MacLane space $K(G,n)$ is a connected CW space with ${\pi }_n=G$ and all other positive homotopy groups zero; for $n>1$ the group $G$ must be abelian.

Projection onto coordinates givesa map to a product is exactly a pair of maps, and a based homotopy to the product is exactly a pair of based homotopies. These identifications respect concatenation. Thus $K(G,n)\times K(H,n)$ has exactly the homotopy groups required of $K(G\times H,n)$. Taking CW models and using the allowed uniqueness of Eilenberg–MacLane spaces gives

Let $p$ be prime. Take $S^{\infty }$ in complex coordinates, with $C_p$ acting by scalar multiplication by a primitive $p$th root of unity. This is a free action. Every map from a finite-dimensional sphere lands in a finite stage and becomes null-homotopic in a sufficiently higher sphere, so $S^{\infty }$ is weakly contractible; the Whitehead theorem for CW complexes makes it contractible. Therefore the infinite lens space $S^{\infty }/C_p$ has contractible universal cover, fundamental group $C_p$, and no higher homotopy: it is a model of $K(C_p,1)$.

Here is the group cohomology computation, including the ring structure. For the cyclic group generated by $g$, the standard free resolution alternates the maps $g-1$ and $N=1+g+\cdots +g^{p-1}$. Applying ${\mathrm{Hom}}_{\mathbb{Z}[C_p]}(-,{\mathbb{F}}_p)$ makes both maps zero, so each cohomology degree is one-dimensional. The period-two identity chain map represents a class $b$ in degree two, and cup multiplication by this periodicity class is an isomorphism from each degree to the degree two higher. For odd $p$, a nonzero degree-one class $a$ has $a^2=0$ by graded commutativity, while $b^j$ and $a{b}^j$ are nonzero by periodicity. For $p=2$ the model is homotopy equivalent to ${\mathbb{RP}}^{\infty }$, whose ring is the infinite version of that computed in question 2. Consequently the mod-p cohomology ring of a cyclic group isThe standard results used are the contractible-cover description of a $K(G,1)$, its identification of singular cohomology with group cohomology, and the cyclic periodic resolution with cup-periodicity just stated.

The product equivalence and the assumed Künneth theorem over the field giveThe tensor product is graded: for odd $p$, $a_1{a}_2=-a_2{a}_1$ and each $a_i^2=0$. The dimensions in degree $d$ are $d+1$: for $p=2$ count $t_1^i{t}_2^{d-i}$; for odd $p$ the generating series is $(1+t{)}^2/(1-t^2{)}^2=(1-t{)}^{-2}$.

If an $n$-dimensional CW complex has exactly one cell in dimension $n+1$, then $C^{n+1}(X;{\mathbb{F}}_p)={\mathbb{F}}_p$. Cellular cohomology is the quotient of the kernel of its outgoing coboundary by the image of the incoming one, a quotient of a subspace of this one-dimensional vector space. ThusThis one-cell bound on next-degree cohomology is valid even with infinitely many higher cells; the printed finite-per-dimension hypothesis is more than it needs.

Now a free finite-group action makes $S^n\to M=S^n/G$ a covering. For $n>1$ the sphere is simply connected, so ${\pi }_1(M)=G$, ${\pi }_i(M)=0$ for $1<i<n$, and ${\pi }_n(M)=\mathbb{Z}$. The covering also identifies every higher homotopy group with that of $S^n$. Use the homotopy-group effect of attaching higher cells: attaching $(i+1)$-cells, $i\ge 2$, preserves homotopy groups below $i$ and quotients ${\pi }_i$ by the subgroup generated by the attaching classes and their fundamental-group translates. Attach a single $(n+1)$-cell along a generator of ${\pi }_n(M)$; its translates do not require additional cells since the original generator already generates the underlying group. This kills ${\pi }_n$ without changing ${\pi }_1$ or the lower groups.

Successively in each higher degree attach cells representing generators of the remaining homotopy group. In the union every higher homotopy group vanishes, since a map and its relevant homotopy have compact domain and are handled at a finite stage. The result is a $K(G,1)$ with exactly one $(n+1)$-cell. No finiteness assumption on the number of later cells is needed for the preceding cochain bound. If desired, one can retain finitely many per degree using the Serre finiteness theorem for homotopy groups: a simply connected CW complex of finite type has finitely generated homotopy groups. At each finite stage the universal cover is simply connected and finite type, since $G$ is finite; choosing finitely many generators then preserves finite type inductively. The first stage needs just the stated ${\pi }_n(S^n)=\mathbb{Z}$.

If $G=C_p\times C_p$, the resulting $K(G,1)$ would have $\dim H^{n+1}\le 1$ by its single cell, but its already computed cohomology has dimension $n+2>1$ in that degree. This contradiction provesThis is the free sphere actions exclude elementary abelian subgroups of rank two obstruction.

Apply the splitting principle for complex vector bundles simultaneously to $E$ and $F$. This can be implemented by successive projective bundle pullbacks; the projective bundle theorem makes their composite pullback injective on integral cohomology. After pullback write $E=L_1\oplus L_2$ and $F=M_1\oplus M_2$, with Chern roots $x_1,x_2$ and $y_1,y_2$. LetSince the First Chern class of a tensor product of complex line bundles is the sum of their first Chern classes, the four roots of $E\otimes F$ are $x_i+y_j$, $1\le i,j\le 2$.

Compute their six pairwise products integrally. The two pairs within a fixed $x_i$ row contributeThe four pairs between the two rows contributeTheir sum is the second elementary symmetric polynomial of the tensor roots. Injectivity of the pullback returns the resulting identity to $X$, givingProducts here are cup products in integral cohomology. This second Chern class of a tensor product of rank-two bundles formula uses no division by two and therefore holds without assuming the integral cohomology of $X$ is torsion-free.

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.

For these linear systems of divisors, take a complex K3 surface $X$ and a nef big divisor $D$, so $D^2>0$. The exceptional terminology refers to the complete system $|D|$, rather than to an arbitrary subsystem.

A monogonal linear system on a K3 surface has the formwhere $|E|$ is an elliptic pencil on a K3 surface and $\Gamma$ is its section. Its intersection with an elliptic fibre is one, and its fixed part is $\Gamma$. For a concrete existence example, take the smooth Fermat quartic surface $X\subseteq {\mathbb{P}}^3$ and two skew lines on it. Such lines can be written aswith ${\zeta }^4=({\zeta }^{\prime }{)}^4={\eta }^4=({\eta }^{\prime }{)}^4=-1$, $\zeta \ne {\zeta }^{\prime }$ and $\eta \ne {\eta }^{\prime }$. The equations show that they are disjoint, and the four partial derivatives show that the quartic is smooth. The adjunction formula gives $K_X=0$, and the hypersurface exact sequence of sheaves $0\to {\mathcal{O}}_{{\mathbb{P}}^3}(-4)\to {\mathcal{O}}_{{\mathbb{P}}^3}\to {\mathcal{O}}_X\to 0$, together with the intermediate cohomology vanishing of these twists, gives $H^1(X,{\mathcal{O}}_X)=0$. Thus it is a K3 surface, and each line has square $-2$. If $H$ is the hyperplane class, the planes through ${\Gamma }_0$ cut out ${\Gamma }_0$ plus a residual cubic pencil $|E|$, $E=H-{\Gamma }_0$. This pencil has no base point off ${\Gamma }_0$. On ${\Gamma }_0$, its two local residual equations are the two normal derivatives of the quartic equation; they cannot vanish simultaneously because $X$ is smooth. Thus it is basepoint-free, withThe degree-one map of $\Gamma$ to the pencil base makes it a section and forces connected fibres. Bertini smoothness theorem and adjunction formula then make a general fibre a smooth genus-one curve.

Now choose $D=mE+\Gamma$. For every irreducible curve other than $\Gamma$, both $E$ and the effective curve $\Gamma$ have nonnegative intersection with it. Also $D\Gamma =m-2\ge 0$, so $D$ is nef, and $D^2=2m-2>0$. On a general elliptic fibre, the restriction of ${\mathcal{O}}_X(D)$ is the degree-one bundle of the section point. Its unique section vanishes at that point. Every global section of ${\mathcal{O}}_X(D)$ therefore vanishes along a dense subset of $\Gamma$, hence along $\Gamma$ itself. Removing this fixed divisor gives exactly $|mE|$. This verifies the monogonal example directly; its general divisor is a section plus $m$ fibres.

A hyperelliptic linear system on a K3 surface is a basepoint-free nef big system whose morphism is generically two-to-one. Equivalently its general smooth curve is hyperelliptic. Let $\pi :X\to {\mathbb{P}}^2$ be a double cover branched over a smooth sextic, and put $D={\pi }^{\ast }\mathcal{O}(1)$. The double-cover construction givesThus $X$ is a K3 surface. Further,since the second summand contributes $H^0(\mathcal{O}(-2))=0$. The map given by $|D|$ is precisely the double cover. A general member is the double cover of a line branched at six points, so Riemann-Hurwitz formula gives genus two. This is the hyperelliptic example.

A trigonal linear system on a K3 surface is a basepoint-free nonhyperelliptic nef big system whose general smooth curve has gonality three. For an example, take the same smooth quartic with a line ${\Gamma }_0$ and the hyperplane system $|H|$. Its general member is a smooth plane quartic of genus three; adjunction formula identifies its plane embedding with its canonical map, so it is not hyperelliptic. The elliptic pencil $E=H-{\Gamma }_0$ hasIts restriction to a general hyperplane curve is a basepoint-free degree-three pencil. The curve is neither rational nor hyperelliptic, so its gonality is exactly three. This verifies the trigonal example.

Thus the three examples exhibit respectively a fixed section with elliptic fibres, a double-plane map, and an embedded quartic with trigonal hyperplane sections. The last example also illustrates why low-genus trigonal sections should not be confused with a blanket numerical criterion for every higher-genus model.

Here a smooth family is understood as a smooth proper family of compact complex manifolds: $f:\mathcal{X}\to S$ is a proper holomorphic submersion, with compact Kähler fibres. Properness and compactness are essential for the usual statement. Indeed, the projection $({\mathbb{C}}_t\times {\mathbb{C}}_z)\setminus \{(0,0)\}\to {\mathbb{C}}_t$ is a holomorphic submersion with Kähler fibres, but its fibres are $\mathbb{C}$ for $t\ne 0$ and ${\mathbb{C}}^{\ast }$ for $t=0$. Their first Betti numbers are zero and one, so their cohomology cannot give a fixed-rank variation. This supplies a counterexample if “smooth family” is read without the usual proper compact-fibre hypothesis. The Ehresmann fibration theorem gives local differentiable trivializations, sois a local system with fibre $H^k(X_s,\mathbb{Z})/\text{torsion}$. Its associated flat bundle has the Gauss–Manin connection, given by differentiating cohomology classes in topologically constant frames.

The Kähler Hodge decomposition givesThe Betti numbers are locally constant by differentiable triviality. The dimensions of $H^q(X_s,{\Omega }_{X_s}^p)$ are upper semicontinuous, and their sums for fixed $p+q=k$ equal the fixed Betti number. Near any fixed fibre, every summand can only decrease, while their sum cannot change; hence none decreases. Thus the Hodge numbers are locally constant. The relative de Rham complex and the Grauert base change theorem then makethe fibres of holomorphic subbundles $F^p\mathcal{H}$. More intrinsically these are the images of the hypercohomology of the truncated relative de Rham complex ${\Omega }_{\mathcal{X}/S}^{\ge p}$ in the full relative de Rham cohomology. Kähler degeneration of the spectral sequence ensures that these images have the required ranks.

For the transversality proof, let $A^2$ denote the ideal in the absolute holomorphic de Rham complex generated by two pulled-back base one-forms. Keeping zero or one base differential gives a short exact sequence of cochain complexes, with the usual shifted differential in the left-hand complexThe connecting morphism in degree $k$ is the Gauss–Manin connectionTo compute it, lift a relative de Rham cocycle to the middle complex, apply the absolute differential, and retain the terms with one base differential. A cocycle in filtration degree $p$ can be represented using absolute forms of degree at least $p$. In a term containing one base differential, removing that factor leaves a relative form of degree at least $p-1$. The connecting construction therefore lands in $F^{p-1}$:This is the filtration shift behind Griffiths transversality. It explains the inequality rather than merely naming a theorem. The unfiltered construction agrees with topological parallel transport by the de Rham comparison, and its square is zero, as required for the flat connection.

Consequently the fibre cohomology gives a pure variation of Hodge structure. To claim a polarized integral variation one needs the additional compatible flat polarization. In a smooth projective family, a relative ample line bundle supplies a flat integral Kähler class; its primitive pieces are polarized by the Hodge–Riemann bilinear relations, and the full cohomology is polarized using the appropriate primitive-piece signs. For a general compact Kähler family, the unpolarized variation above is automatic, but a flat integral polarization is not supplied merely by knowing that each fibre is Kähler. This distinction is particularly relevant to unrestricted K3 families.

Work in the homotopy category of simply connected CW complexes. This is one of the permitted assumptions on the fundamental group; it makes every coefficient system below untwisted. A rational space is a simply connected space whose homotopy groups are vector spaces over $\mathbb{Q}$. A rationalization of a topological space is a map $r:X\to X_{\mathbb{Q}}$ to such a space which inducesIts purpose includes a universal mapping property, which will be described after the construction.

An Eilenberg–MacLane space $K(A,n)$ has exactly one nonzero homotopy group, ${\pi }_n=A$, where $A$ is abelian when $n\ge 2$. The representability theorem says $[B,K(A,n)]\cong H^n(B;A)$, naturally in $B$ and $A$. More generally, a principal fibration is a homotopy pullback of a path-space fibration $PC\to C$ along a map from its base to $C$; the fibre $\Omega C$ acts on the fibre paths by concatenation. For the Postnikov tower construction take $C=K(A,n+1)$. A principal Eilenberg–MacLane fibration with fibre $K(A,n)$ is the homotopy pullback of the path fibration of $K(A,n+1)$. Equivalently, its total space is the homotopy fibre of a map $k:B\to K(A,n+1)$. It is classified by the cohomology class $[k]\in H^{n+1}(B;A)$; there is no nontrivial action of the base fundamental group on the fibre coefficients in this principal description.

The Postnikov tower theorem supplies spaces $P_{n}X$ and compatible mapswith ${\pi }_i(P_{n}X)={\pi }_i(X)$ for $i\le n$ and zero for $i>n$. For a simply connected $X$, the fibre of $P_{n}X\to P_{n-1}X$ is $K({\pi }_{n}X,n)$, and the fibration is principal. Its Postnikov invariant isrepresented by $P_{n-1}X\to K({\pi }_{n}X,n+1)$, whose homotopy fibre is $P_{n}X$. Thus the homotopy groups alone are not the tower data: the classes $k_n$ specify how each new group is attached. For instance a zero class gives a product with $K({\pi }_{n}X,n)$.

State also the standard rational comparison results used in constructing the tower: $K(A,n)\to K(A\otimes \mathbb{Q},n)$ is a rational homology equivalence; a map between simply connected spaces is a rational homology equivalence if and only if it induces isomorphisms on all rational homotopy groups; and a rational homology equivalence induces isomorphisms in cohomology with any constant rational vector-space coefficients. The last assertion uses the universal coefficient theorem over the field $\mathbb{Q}$. These results are used here without proof, as allowed for this essay.

Construct rational stages inductively, with $P_1^{\mathbb{Q}}=\ast$. Suppose $r_{n-1}:P_{n-1}X\to P_{n-1}^{\mathbb{Q}}$ has already been constructed and is a rational homology equivalence. Put $V_n={\pi }_{n}X\otimes \mathbb{Q}$ and change coefficients in $k_n$ to obtain ${\overline{k}}_n\in H^{n+1}(P_{n-1}X;V_n)$. The comparison result gives an isomorphismLet $k_n^{\mathbb{Q}}$ be its unique inverse image and defineEquality of the pulled-back classes gives a homotopy-commuting square of classifying maps, and hence a map $r_n:P_{n}X\to P_n^{\mathbb{Q}}$ over $r_{n-1}$. On the fibre it is the map induced by ${\pi }_{n}X\to V_n$. The homotopy exact sequences therefore show that $r_n$ tensors all homotopy groups through degree $n$ with $\mathbb{Q}$, while the higher groups of both stages vanish. The rational comparison theorem makes $r_n$ a rational homology equivalence, allowing the induction to continue.

Take the homotopy inverse limit of the rational tower:The convergence theorem for a Postnikov tower gives $X\simeq {\mathrm{holim}}_n{P}_{n}X$ and the compatible stage maps induce $r:X\to X_{\mathbb{Q}}$. To see what convergence says here, fix a homotopy degree $i$. From stage $n\ge i$ onwards the group and its transition maps are constant; the inverse-limit group is ${\pi }_{i}X\otimes \mathbb{Q}$ and the associated derived-limit obstruction is zero. Thus this construction has exactly the claimed rational homotopy groups. Taking a homotopy limit, rather than an arbitrary set-theoretic inverse limit, is part of the construction.

The rational localization theorem says that this map has the universal propertyIn other words every map from $X$ to a rational space factors through $r$, and the factorization is unique up to homotopy. This concerns homotopy classes, not strict uniqueness of point-set maps. The theorem can be formulated as localization of the simply connected homotopy category at rational homotopy equivalences; rational spaces are its local objects. It also implies uniqueness of rationalization of a topological space: apply the property to two candidate rational spaces to obtain inverse equivalences under $X$. In particular the construction forgets torsion in homotopy while retaining precisely the information needed for maps to rational targets.

The same theory extends to nilpotent spaces, where the fundamental group is nilpotent and its action on higher homotopy groups is nilpotent. That version requires localization of the fundamental group and compatible treatment of its coefficient action. The simply connected assumption chosen here avoids those additional issues and supplies the full construction and mapping property requested.

Let $Z={\mathbb{CP}}^5/{\mathbb{CP}}^2$. Its quotient cell structure has one cell in each of dimensions $0,6,8,10$. In particular $Z$ is five-connected. The relative cohomology sequence, or its cellular cochain complex, givesThe subscripts denote degrees. Under the quotient map back to ${\mathbb{CP}}^5$, the three positive classes pull back to $x^3,x^4,x^5$. All positive products vanish, since their degree is at least twelve. Thus the cohomology ring of a collapsed projective subspace in this case is the square-zero algebra $R$ on these three classes.

Now compute its Sullivan minimal model $(\Lambda V,d)$, using cohomological grading and the convention that even generators generate a polynomial ring and odd generators an exterior algebra. Five-connectivity gives $V^j=0$ for $j\le 5$. Since a minimal differential is decomposable, it cannot have a nonzero value below degree twelve. Matching cohomology through degree ten therefore forceswith no other generators in degrees at most ten. The first products must then be killed by generatorsThese follow by killing respectively the unwanted degree-twelve, degree-fourteen, and two degree-sixteen classes. Subsequent generators kill $bc,c^2$ and the resulting syzygies. For example $av-bu$ is closed of degree nineteen, so it requires a further generator of degree eighteen. Thus the full minimal model is not just the algebra on the first seven generators.

There is nevertheless a simple map from the full model to $R$: send $a,b,c$ to the three displayed classes and every other generator to zero. It is a chain map, since every generator differential is a sum of products of positive-degree elements, and all positive products in $R$ vanish. It induces an isomorphism in cohomology: the entire cohomology of $\Lambda V$ is concentrated in degrees $0,6,8,10$, and its indicated basis maps to the indicated basis of $R$. This proves formality of a topological space and identifies the full minimal model as a minimal resolution of $R$, rather than only matching the initial part of its cohomology.

An explicit description of that full resolution is useful. Take the free graded Lie algebraIts Chevalley–Eilenberg cochain algebra $C^{\ast }(L)$ has one generator of degree $|\ell |+1$ for every homogeneous Lie basis word $\ell$, and its quadratic differential is dual to the bracket, with the usual graded signs. Rescaling the bracket-word duals gives exactly the generators and differentials listed above. Positive Lie degrees make this algebra Sullivan minimal. Its cohomology is $R$: the universal enveloping algebra is $T(p,q,r)$, and the augmentation module has the exact free resolutionExactness is immediate because every nonempty tensor word has a unique last letter. Applying ${\mathrm{Hom}}_{T(p,q,r)}(-,\mathbb{Q})$ gives just the unit and the three dual classes in total degrees $6,8,10$; products of positive classes land in an absent Ext degree two and are zero. Chevalley–Eilenberg cochains of a graded Lie algebra compute this Ext, so this also verifies the full model's cohomology.

The wedge sum $W=S^6\vee S^8\vee S^{10}$ has the same connectivity, cohomology ring, and initial closed minimal generators. The same full-model-to-$R$ argument applies to $W$, so its minimal model is also $C^{\ast }(L)$, by uniqueness of Sullivan minimal models of quasi-isomorphic simply connected rational cochain algebras. The Sullivan minimal-model classification identifies isomorphic minimal models with the same rational homotopy type. ConsequentlyThe explicit infinite minimal resolution above is shared by both spaces.