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For an oriented real vector bundle $\pi :E\to X$ of rank $d$, the Thom isomorphism theorem gives a unique Thom class $u_E\in H^d(D(E),S(E);\mathbb{Z})$ restricting to the chosen generator in each fibre, and isomorphismsEquivalently use the pair $(E,E\setminus X)$, identifying $X$ with the zero section. After forgetting relative supports, pull the Thom class back along the zero section $s$ to define the Euler class:In this expression the relative class is first mapped to absolute cohomology.

A compact embedded submanifold equipped with a coorientation $Y\subset X$ has an oriented normal bundle ${\nu }_Y$ of rank $d$. A tubular neighborhood and excision identify its Thom class with a class $u_Y\in H^d(X,X\setminus Y)$. Its image in $H^d(X)$ is the cohomology class of a cooriented submanifold ${\epsilon }_Y$. In particular it restricts to zero on $X\setminus Y$. If the ambient manifold is oriented, orient $Y$ by the normal-first convention. The class is its Poincare dual, characterized byfor the appropriate complementary degree. This is the stated integration identity in cohomological notation.

First prove the Lefschetz fixed-point theorem for an oriented closed n-manifold $X$. Orient its diagonal by the identification $\Delta \cong X$, and choose the normal orientation so this agrees with the normal-first convention in the product. This produces the integral class ${\epsilon }_{\Delta }$. To compute the part that contributes to the trace, tensor with $\mathbb{Q}$: the trace on a finitely generated integral cohomology group means its trace on the free quotient, equivalently on rational cohomology; torsion contributes nothing.

Choose a basis $e_{p,j}$ of $H^p(X;\mathbb{Q})$ and its Poincare duality pairing dual basis $e_{p,j}^{\vee }\in H^{n-p}(X;\mathbb{Q})$, normalized by $⟨e_{p,i}⌣e_{p,j}^{\vee },[X]⟩={\delta }_{ij}$. The rational image of the cohomology class of the diagonal isIndeed, pairing this expression against $\alpha \times \beta$ with degrees $p,n-p$ picks out the p-summands. The cross-product multiplication contributes $(-1{)}^{p^2}$ and moving $\alpha$ past $e_{p,j}^{\vee }$ contributes $(-1{)}^{p(n-p)}$. Together with $(-1{)}^{np}$ their product is $+1$. The result is $⟨\alpha ⌣\beta ,[X]⟩$, exactly the diagonal's defining identity. The rational Künneth theorem and nondegeneracy of the duality pairing determine the class. An integral torsion component, if present, is not determined by this rational expression and does not affect any evaluation below.

For the graph map ${\Gamma }_f=(1,f):X\to X\times X$, write $f^{\ast }{e}_{p,j}={\sum }_i{A}_{ij}{e}_{p,i}$. Pulling back the expression and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph lies in $(X\times X)\setminus \Delta$. The relative Thom class restricts to zero there, so ${\Gamma }_f^{\ast }{\epsilon }_{\Delta }=0$ and $L(f)=0$. The contrapositive proves the required fixed-point assertion, without requiring smoothness of $f$ or transversality of its graph.

The printed hypothesis does not require $X$ to be orientable. For a nonorientable $X$, the diagonal's normal bundle is $TX$, so an ordinary integral coorientation cannot be assumed. We extend the proved theorem as follows, rather than apply the untwisted diagonal computation incorrectly. We first construct a smooth embedding in Euclidean space using finitely many charts ${\chi }_i:U_i\to {\mathbb{R}}^n$ and smooth functions ${\rho }_i$ supported in those charts, with at least one ${\rho }_i(x)>0$ at every point. The map $\Psi (x)=({\rho }_i(x),{\rho }_i(x){\chi }_i(x){)}_i$, with zero extensions, is injective: a positive first coordinate identifies a chart in which its remaining coordinates recover $x$. Its differential is injective as well, since $d{\rho }_i(v)=0$ and $d({\rho }_i{\chi }_i)(v)=0$ force $d{\chi }_i(v)=0$ in a chart with ${\rho }_i>0$. Compactness makes this injective immersion a homeomorphism onto its image. This finite-chart Euclidean embedding of a compact smooth manifold, enlarged by zero coordinates if necessary, has positive codimension. A compact tubular disc neighbourhood $W$ is an oriented N-manifold with boundary, inheriting its orientation from the ambient Euclidean space. Its double of a manifold $M$ is closed and oriented. Projection of each copy of $W$ to $X$ agrees on their common boundary, giving $r:M\to X$; inclusion into the first copy gives $i:X\to M$ with $ri=1_X$.

Set $F=ifr$. A fixed point of $F$ must lie in $i(X)$, and then is exactly the image of a fixed point of $f$. Moreover $F^{\ast }=r^{\ast }{f}^{\ast }{i}^{\ast }$, and cyclicity of trace for maps between finite-dimensional rational cohomology groups givesin each degree, because $i^{\ast }{r}^{\ast }=1$. Thus $L(F)=L(f)$. The oriented theorem for $M$ now gives a fixed point of $F$, hence of $f$, whenever $L(f)\ne 0$. This oriented manifold retract for the Lefschetz theorem proves the full printed result. The double construction also works componentwise if necessary.

Put $B=\partial M$ and let $v$ be the cone vertex. Specify the convention away from the Witt case: first take intersection cohomology to mean the hypercohomology of the intersection complex $I{C}_{\bar{m}}$, normalized in degree zero on the regular stratum. The lower middle perversity at the vertex is $\bar{m}(2m+1)=m-1$. The standard local cone calculation for this convention saysand restriction to $U\simeq B$ is the identity below the cutoff. Here this can also be read directly from the construction $I{C}_{\bar{m}}={\tau }_{\le m-1}R{j}_{\ast }\mathbb{Q}$, with $j:\hat{M}\setminus \{v\}\hookrightarrow \hat{M}$: punctured conical neighbourhoods retract onto $B$, and the truncation discards the degrees above $m-1$.

Define relative intersection cohomology as the hypercohomology of the homotopy fibre of restriction, namely $R\Gamma (\hat{M},\hat{U};IC)=\mathrm{Cone}(R\Gamma (\hat{M};IC)\to R\Gamma (\hat{U};IC))[-1]$. Restricting to the pair $(M,U)$ gives a map to the corresponding ordinary relative cohomology, since $IC{|}_M={\mathbb{Q}}_M$. The induced map of exact triangles gives the requested commutative diagram:Both rows are long exact cohomology sequences of a pair. Commutativity includes the connecting homomorphisms, because they arise from the same map of restriction cones, with a common sign convention. The relative vertical maps are isomorphisms: excision removes the common conical neighbourhood, leaving precisely the manifold pair outside it, where $IC$ is constant. Also the collar retracts onto $B$, so $H^{\ast }(M,U)\cong H^{\ast }(M,B)$.

The only possible nonmanifold stratum is $v$, with link $B$ of dimension $2m$. Since intersection homology of a manifold is ordinary homology, the exact necessary and sufficient condition isThe equivalence with the defining $H_m(B;\mathbb{Q})=0$ follows from the universal coefficient theorem for cohomology over $\mathbb{Q}$. In the Witt case the cone restriction is an isomorphism through degree $m$, since both degree-$m$ groups are zero. Comparing the two exact rows, or applying the Five lemma to each five-term segment, gives $I{H}^i(\hat{M})\cong H^i(M)$ for $i\le m$. For $i>m$, both $I{H}^{i-1}(\hat{U})$ and $I{H}^i(\hat{U})$ vanish, so the top row identifies the absolute group with the relative one. HenceWithout the Witt hypothesis, comparison still gives $I{H}_{\bar{m}}^i(\hat{M})\cong H^i(M)$ for $i<m$. At degree $m$ the upper row readsThe lower row identifies the quotient by $\mathrm{im}\delta$ with the image of the relative-to-absolute map. In higher degrees both cone terms vanish. Thus the full lower-middle Deligne answer isFor clarity, if $I{H}^i$ is instead defined as the linear dual of the lower-middle intersection chains of question 1, it corresponds to the complementary upper middle perversity Deligne sheaf. Its cone group retains $H^m(B)$ as well. The same diagram then givesEquivalently the exceptional group is $H^{m+1}(M,B)/\mathrm{im}(H^m(B)\stackrel{\delta }{\to }{H}^{m+1}(M,B))$. These conventions agree under the stated Witt condition, explaining why that condition removes the ambiguity.

Take the solid torus $M=S^1\times D^2$, whose boundary is the two-dimensional torus $B=T^2$. Its ordinary cohomology is $H^0(M)=H^1(M)=\mathbb{Q}$, zero otherwise; Poincare duality for a manifold with boundary gives $H^2(M,B)=H^3(M,B)=\mathbb{Q}$ and all other relative groups zero. The lower-middle Deligne formula givesThus degrees one and two have different dimensions, so there can be no three-dimensional Poincare duality. In the dual-lower-chain convention the dimensions are $(1,1,0,1)$, which likewise fail duality. Both computations exhibit the obstruction $H^1(T^2)={\mathbb{Q}}^2\ne 0$.

All cohomology below has rational coefficients. First describe the incidence space geometrically. The image of $g$ is $\mathbb{P}(W)\cong {\mathbb{CP}}^1$, inside $\mathrm{Gr}(1,4)\cong {\mathbb{CP}}^3$. For a line $U\subset W$, choosing a plane $V$ containing $U$ is equivalent to choosing the line $V/U$ in ${\mathbb{C}}^4/U$. If $\mathcal{L}$ is the tautological line bundle on $\mathbb{P}(W)$, thenThis is the projective bundle of a rank-three quotient bundle, with fibre ${\mathbb{CP}}^2$. Local frames of the quotient bundle give local product charts. In particular $F_W$ is an oriented real six-dimensional manifold.

For $V\in S_W\setminus \{W\}$, the intersection $V\cap W$ is exactly a line, so $f^{-1}(V)$ is a single point and the inverse map is $V\mapsto (V\cap W,V)$. The intersection line varies continuously, indeed holomorphically, wherever its dimension is constant: in local Grassmannian charts it is the kernel of a matrix of constant rank, whose kernel bases can be chosen locally. Thus $f$ is an isomorphism over the open stratum. At the remaining point,Set $K=R{f}_{\ast }{\mathbb{Q}}_{F_W}$, a derived direct image of sheaves. The proper base change for sheaves theorem identifies its derived stalks with the cohomology complexes of the fibres. Consequently $K$ is the constant rational sheaf in degree zero on $S_W\setminus \{W\}$, andThe cohomology sheaves are therefore finite-rank local systems on the two strata, with no other nonzero degrees. This proves that $K$ is a constructible complex of sheaves and verifies normalization. The open stratum has real dimension six, so the isolated point has codimension six. Its middle perversity is $\bar{m}(6)=2$, and the intersection-complex support and cosupport axioms require stalk vanishing in degrees above two. The fibre calculation verifies the support axiom, as well as vanishing in negative degrees.

For an oriented real six-dimensional manifold, Verdier duality gives $D_{F_W}{\mathbb{Q}}_{F_W}\cong {\mathbb{Q}}_{F_W}[6]$. For a proper map, Verdier duality commutes with derived direct image. Applying these statements here yieldsThe duality identity $i_W^{!}{D}_{S_W}K\cong D_{\{W\}}{i}_W^{\ast }K$ converts this intoHere $i_W^{!}K$ is the costalk, and dualization of rational complexes reverses degrees. Thus its only nonzero groups are $\mathbb{Q}$ in degrees four and six. In particular they vanish for $a\le 3$, which is precisely the cosupport axiom at this isolated six-dimensional singularity. The uniqueness characterization by constructibility, normalization, support and cosupport now identifies $K$ with the unshifted intersection complex of $S_W$. Taking hypercohomology, and using the composition of derived global sections with $R{f}_{\ast }$, givesFinally apply the fibration result allowed in the question to $g:F_W\to \mathbb{P}(W)$. Its base is the simply connected complex projective line, its fibre is the Complex projective plane, and the projective-bundle charts above verify the fibration hypothesis. Hence, as graded rational vector spaces,The resulting Betti numbers and Poincaré polynomial areThis uses $H^{\ast }({\mathbb{CP}}^r)=\mathbb{Q}[h]/(h^{r+1})$ with $\mathrm{deg}h=2$. No assertion that an arbitrary projective fibration has the product cohomology ring is needed.

For a cochain complex $E$, define the shift of a cochain complex byFor a cochain map $a:E\to F$, put $(\Sigma a{)}^i=a^{i+1}$. This is again a cochain map, so the construction is a functor. The sign ensures the standard cone identities; the shifted differential squares to zero. Iteration gives $E[r{]}^i=E^{i+r}$ and differential $(-1{)}^r{d}_E$, and $H^i(E[r])=H^{i+r}(E)$.

For $\varphi :E\to F$, its algebraic mapping cone isApplying this differential twice gives first component $d_F^{2}f+d_F\varphi e-\varphi d_{E}e=0$, since $\varphi$ is a cochain map, and second component $d_E^{2}e=0$. The mapsare cochain maps: $d_{C}u(f)=(d_{F}f,0)=u(d_{F}f)$ and $v(d_C(f,e))=-d_{E}e=d_{\Sigma E}v(f,e)$. They are maps of complexes of sheaves, because every displayed operation is a sheaf morphism in each degree.

The bounded below derived category of sheaves $D(X)=D^{+}(X)$ can be constructed by first quotienting the category of bounded below cochain complexes by cochain homotopy and then formally inverting all quasi-isomorphisms. A morphism may be represented by a roof $E\stackrel{\sim }{\leftarrow }{E}^{\prime }\to F$. Alternatively, take injective resolutions of sheaves and use homotopy classes of maps between the resulting injective complexes. The cohomology sheaf functors descend to the localization, because they already identify homotopic maps and turn quasi-isomorphisms into isomorphisms. If $E\cong 0$ in $D(X)$, applying these functors gives ${\mathcal{H}}^i(E)=0$ for every $i$. Conversely, if all these sheaves vanish, the unique map $E\to 0$ is a quasi-isomorphism, which becomes invertible in $D(X)$. HenceAn exact triangle in a derived category is a triangle isomorphic to a cone triangleIn particular the last arrow is part of the structure; a mere sequence of three objects with successive composite zero does not suffice.

Suppose $0\to E\stackrel{\varphi }{\to }F\stackrel{\psi }{\to }G\to 0$ is degreewise exact. DefineThen $r(d_C(f,e))=\psi (d_{F}f)+\psi \varphi (e)=d_G\psi (f)=d_{G}r(f,e)$, so $r$ is a cochain map. It is natural under maps of short exact sequences. It is degreewise surjective as a sheaf map, and its kernel can be identified with $E^i\oplus E^{i+1}$ using $(a,e)\mapsto (\varphi (a),e)$. In these coordinates the kernel differential isThere is a contracting homotopy $h^i(a,e)=(0,a)$ in degree $i-1$:Thus the kernel has zero cohomology sheaves. The cohomology-sheaf exact sequence is the sequence of the short exact sequence of complexes $0\to \ker r\to C(\varphi )\to G\to 0$, or equivalently its ordinary complex-cohomology sequence at every stalk. It follows that $r$ induces isomorphisms of all cohomology sheaves, provingUnder this isomorphism the cone triangle becomes the exact trianglewhere $\delta =v\circ r^{-1}$ in the derived category.

For the final construction write $U=S^2\setminus S^1$ and set $A=i_{!}{\mathbb{Q}}_U$, $B={\mathbb{Q}}_{S^2}$, $C=j_{\ast }{\mathbb{Q}}_{S^1}$, each initially placed in degree zero. The stated sequence $0\to A\to B\to C\to 0$ is exact on stalks: on $U$ its first map is an isomorphism and its last sheaf is zero; on the equator its first sheaf is zero and the second map is an isomorphism. Take its connecting morphismThe source has only ${\mathcal{H}}^0(C)=C$, whereas the target has only ${\mathcal{H}}^{-1}(A[1])=A$. Therefore every induced map ${\mathcal{H}}^i(\mu )$ is zero: in any degree at least one of its source and target is zero. This is a ghost morphism in a derived category.

To prove the derived morphism itself is nonzero, apply derived global sections, giving the hypercohomology sequence. The extension by zero identifies $H^{\ast }(S^2;A)$ with the compactly supported cohomology of $U$, and closed direct image identifies $H^{\ast }(S^2;C)$ with $H^{\ast }(S^1;\mathbb{Q})$. The complement $U$ is two open disks, each homeomorphic to ${\mathbb{R}}^2$, soThe degree-one part of the resulting exact sequence isThus ${\mu }_{\ast }$ is injective, in particular nonzero. With the hemispheres oriented from $S^2$, the last map is $(a,b)\mapsto a+b$, so an appropriately oriented equator generator maps to $(1,-1)$. A zero morphism in the derived category of sheaves would induce a zero map after derived global sections and taking cohomology. Consequently $\mu \ne 0$, although ${\mathcal{H}}^i(\mu )=0$ for every $i$. This establishes explicitly that the collection of cohomology-sheaf functors detects zero objects but is not faithful on morphisms.

For integral singular cochains $a\in C^p(X)$ and $b\in C^q(X)$, define the cup product on an ordered simplex byThe same construction works over any commutative coefficient ring. Expanding the coboundary as the alternating sum of faces gives the Leibniz identityIndeed faces deleting vertices before the split occur in the first sum, faces after the split occur in the second, and the two terms at the common vertex cancel with opposite signs. Thus the product of cocycles is a cocycle. Replacing $a$ by $a+\delta c$ changes the product with a cocycle $b$ by $\delta (c⌣b)$; replacing $b$ by $b+\delta d$ changes the product with a cocycle $a$ by $(-1{)}^p\delta (a⌣d)$. Hence the product is independent of representatives and defines a well-defined product on cohomology classes.

With ${\mathbb{F}}_2={\mathbb{Z}}_2$ coefficients, the real-projective-space cohomology ring isFor $n>m\ge 1$, the pullback of the degree-one generator is either zero or $t$. The second choice would send the zero class $t^{m+1}$ to the nonzero class $t^{m+1}\in H^{m+1}({\mathbb{RP}}^n;{\mathbb{F}}_2)$, contradicting the naturality of the cup product. Thus the generator pulls back to zero, and so do all its positive powers. For $m=0$ the target's reduced cohomology is already zero. Therefore every such map induces zero on reduced mod-two cohomology.

Suppose a continuous $f:S^n\to {\mathbb{R}}^n$ had $f(x)\ne f(-x)$ everywhere. Thenis a continuous odd map to $S^{n-1}$ and descends to $\bar{h}:{\mathbb{RP}}^n\to {\mathbb{RP}}^{n-1}$. For $n\ge 2$, a path from $x$ to $-x$ projects to a loop detected by the degree-one generator of the double cover. Its image lifts to a path from $h(x)$ to $-h(x)$, so is detected by the target's double cover as well. Thus ${\bar{h}}^{\ast }$ is nonzero in degree one, contradicting the preceding calculation. For $n=1$, an odd continuous map from the connected circle to $S^0$ is already impossible. The case $n=0$ is immediate. HenceThis proves the Borsuk-Ulam theorem rather than assuming it.

For a closed cover $A_1,\dots ,A_{n+1}$, assume no member contains an antipodal pair. Apply the just-proved theorem to $F(x)=(d(x,A_1),\dots ,d(x,A_n))$, using any metric on the sphere. Empty sets can be assigned a constant positive coordinate. It gives equal distances at $x,-x$. If either point belongs to an $A_i$ with $i\le n$, zero distance and closedness place both points in it, a contradiction. Hence neither lies in any of the first $n$ sets, and both must lie in $A_{n+1}$, another contradiction. This proves the Lusternik-Schnirelmann-Borsuk theorem for closed sets.

Four closed sets on $S^2$ need not contain an antipodal pair. Take the four unit vertices $v_i$ of a regular tetrahedron centered at zero and define $C_i=\{x:x\cdot v_i\ge x\cdot v_j\text{ for every }j\}$. These closed sets cover the sphere by maximizing the four scalar products. If $x$ and $-x$ belonged to $C_i$, both inequalities would force all four scalar products to be equal. Since $\sum v_i=0$, they would all be zero, and because the vertices span ${\mathbb{R}}^3$, $x$ would be zero. This is impossible on the sphere. Thus these four sets give the antipodal-free closed cover from simplex Voronoi cells counterexample.

For a locally compact space, compactly supported cohomology iswith transition maps induced by enlarging supports. In ${\mathbb{R}}^n$ closed balls are cofinal; their complements retract onto $S^{n-1}$. The relative cohomology long exact sequence and contractibility of ${\mathbb{R}}^n$ giveThis is compactly supported cohomology of Euclidean space; for $n=0$ it is simply the cohomology of a point.

For an oriented $n$-manifold without boundary, Poincare duality in its compact-support form states that capping with the locally finite fundamental class gives $H_c^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$. If $M$ is closed, the ordinary fundamental class $[M]\in H_n(M;\mathbb{Z})$ gives $H^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$ and a perfect complementary-degree pairing over $\mathbb{Q}$.

Choose an oriented embedded closed ball $B$ in a connected closed oriented $M$. Collapse $M\setminus \mathrm{int}B$ to a point, identifying $B/\partial B$ with $S^n$ using its orientation. The quotient map is continuous. By excision, the fundamental class maps to the local oriented generator in $H_n(B,\partial B)$, and then to the generator of $H_n(S^n)$. Hence the collapse map has $\boxed{\mathrm{deg}f=1}$.

The reverse direction is not always possible. More generally a degree-one map $f:N\to M$ between closed oriented manifolds induces an injective map on rational cohomology: if $\alpha \ne 0$, duality supplies $\beta$ with $⟨\alpha ⌣\beta ,[M]⟩\ne 0$, andIn particular there is no degree-one map $S^2\to T^2$, since $H^1(T^2;\mathbb{Q})={\mathbb{Q}}^2$ but $H^1(S^2;\mathbb{Q})=0$. This is a counterexample to an unconditional reverse assertion.

Finally let $p:{\mathbb{CP}}^3\to S^4$ be the given bundle projection. The standard cohomology ring of complex projective space is $H^{\ast }({\mathbb{CP}}^3;\mathbb{Z})=\mathbb{Z}[h]/(h^4)$, $|h|=2$; the integral generator $u$ of $H^4(S^4)$ has $p^{\ast }u=m{h}^2$ for some integer $m$. If a section of a fiber bundle $s$ existed, $p\circ s=1$ would give $s^{\ast }{p}^{\ast }=1$ by functoriality. But $H^2(S^4)=0$, so $s^{\ast }h=0$ and $s^{\ast }(m{h}^2)=0$, contradicting $s^{\ast }{p}^{\ast }u=u\ne 0$. The facts used here are the sphere cohomology groups, the projective-space ring just stated, and multiplicative functoriality of cohomology. Thus the bundle has no section, the cohomological obstruction to a section of the twistor bundle.

Take the double cover of $S^2$ branched at $2g+2$ points, with deck involution $\iota$. One explicit construction cuts the sphere along $g+1$ disjoint arcs pairing the branch points and cross-glues two copies along the cuts. Its Euler characteristic is $2\chi (S^2)-(2g+2)=2-2g$, so it is an oriented closed surface of genus $g$. Locally at a branch point the projection is $z\mapsto z^2$ and the involution is $z\mapsto -z$; thus the involution is nonidentity and orientation-preserving, also when $g=0$.

The invariant rational cohomology of a finite surface action identifies with the cohomology of its quotient. For this action, triangulate the quotient with the branch points as vertices and lift the triangulation. Quotient cochains identify with invariant cochains, and averaging over the two deck transformations makes taking invariants exact over $\mathbb{Q}$. Hence the invariant part of $H^1(S;\mathbb{Q})$ is $H^1(S^2;\mathbb{Q})=0$. Since ${\iota }^2=1$, its eigenvalues are $\pm 1$ and it is therefore $-I$ on the $2g$-dimensional $H^1$. Its actions on $H^0$ and $H^2$ are both the identity. ThusHomotopy invariance and the result just proved show that every map homotopic to this involution has fixed points, for every genus.

Let $\pi :E\to B$ be an oriented real vector bundle of rank $r$ over a paracompact base. A Thom class is a class $U\in H^r(E,E\setminus B;\mathbb{Z})$ restricting on every oriented fiber pair $({\mathbb{R}}^r,{\mathbb{R}}^r\setminus 0)$ to its positive generator. The Thom isomorphism theorem states that this class exists uniquely with those fiber restrictions and thatHere $B$ is identified with the zero section. Equivalently use the disk/sphere bundle pair. The compact-support version is $H_c^q(B;\mathbb{Z})\cong H_c^{q+r}(E;\mathbb{Z})$, with fiberwise compact support in the Thom construction. If the bundle is not oriented, the integral assertion requires the corresponding orientation local system.

For a closed co-oriented embedded submanifold $Y$ of codimension $r$, co-orientation means an orientation of its normal bundle. A tubular neighborhood identifies a neighborhood of $Y$ with that bundle. Its Thom class, transferred by excision, is a class in $H^r(M,M\setminus Y;\mathbb{Z})$. For compact $Y$ this maps into the direct limit defining compactly supported cohomology, giving $\boxed{{\epsilon }_Y\in H_c^r(M;\mathbb{Z})}$. The construction is independent of the tubular neighborhood and uses the chosen co-orientation. In an oriented ambient manifold it is the Poincare dual of the oriented submanifold. Here “closed submanifold” has the usual compact-without-boundary meaning; if it meant merely a closed subset and $Y$ were noncompact, the supported relative class would still exist but need not have compact support.

We next prove the fixed-point implication without imposing an unprinted orientability assumption on $M$. SetUse a finite triangulation of the closed smooth manifold. The elementary Hopf trace identity says that the alternating trace of a chain endomorphism equals its alternating trace on homology. To verify it, split each finite-dimensional chain space as boundaries, a complement representing homology, and a complement mapping isomorphically onto the preceding boundaries. The traces on these two boundary summands cancel in adjacent degrees, leaving exactly the homology traces. The cohomology traces equal the homology traces because the induced maps are dual over $\mathbb{Q}$.

If $f$ has no fixed points, compactness gives $\delta ={\min }_{x}d(x,f(x))>0$. Choose a triangulation $K$ of mesh less than $\delta /4$ and a sufficiently fine subdivision $K^{\prime }$ admitting a simplicial approximation $F:K^{\prime }\to K$ uniformly within $\delta /4$ of $f$. The subdivision chain map $s:C_{\ast }(K)\to C_{\ast }(K^{\prime })$ sends each simplex to simplices inside it. For a vertex $v$ of such a small simplex, $d(v,F(v))>3\delta /4$, so $F(v)$ cannot lie in the original coarse simplex. Consequently $F_{\#}s$ has zero diagonal coefficient on every coarse simplex. Its trace vanishes in every degree, while it induces $f_{\ast }$ on homology. The trace identity yields $L(f)=0$. Taking the contrapositive provesThis is the Lefschetz fixed-point theorem, established here rather than invoked. For oriented $M$ the same identity is the graph-diagonal formula for the Lefschetz number: $L(f)=⟨(1,f{)}^{\ast }{\epsilon }_{\Delta },[M]⟩$, which vanishes when the graph misses the diagonal. Finally, homotopic maps induce the same cohomology maps, so their Lefschetz numbers agree.

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.