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For positive unequal dimensions $p,q$, let $u,v$ be the pullbacks of the integral orientation classes of the factors. The Künneth theorem gives the free abelian basis $1,u,v,uv$, with degrees $0,p,q,p+q$. The products satisfy $u^2=v^2=0$ and $vu=(-1{)}^{pq}uv$, by naturality and graded commutativity of the cup product. In particular, for $p=2,q=4$, the cohomology ring is $\mathbb{Z}[u,v]/(u^2,v^2)$, with degrees two and four.

For an oriented rank-$r$ vector bundle, let $U\in H^r(D(E),S(E);\mathbb{Z})$ be its Thom class. Forgetting relative supports sends $U$ to ${\pi }^{\ast }e(E)$, where $e(E)$ is the Euler class. Compatibility of relative and mixed cup products givesFor odd $r$, graded commutativity of the cup product makes $2U^2=0$. The Thom isomorphism theorem is injective, so $2e(E)=0$. This explains why an odd-rank Euler class can have only two-torsion.

Let $E$ be an oriented real vector bundle of positive even rank $r$, with zero Euler class, and suppose the integral cohomology of $B$ is torsion-free. The Gysin sequence gives $0\to H^j(B)\to H^j(S(E))\to H^{j-r+1}(B)\to 0$. Choose an odd-degree class $a\in H^{r-1}(S(E))$ with ${\pi }_{!}a=1$. The projection formula for sphere bundle integration identifies the cohomology additively with ${\pi }^{\ast }{H}^{\ast }(B)\oplus {\pi }^{\ast }{H}^{\ast }(B)a$. It is torsion-free, so graded commutativity of the cup product gives $a^2=0$ from $2a^2=0$. These facts establish the displayed ring isomorphism, not just a decomposition of groups.

Write $\dim X=4r+2=2m$, with $m=2r+1$ odd. Rational Poincare duality gives $b_k=b_{2m-k}$. Pairing every term outside the middle yieldsThe Poincare duality pairing on $H^m(X;\mathbb{Q})$ is $B(\alpha ,\gamma )=⟨\alpha ⌣\gamma ,[X]⟩$. It is nondegenerate, and graded commutativity of the cup product gives $B(\alpha ,\gamma )=-B(\gamma ,\alpha )$ because $m$ is odd. In particular $B(\alpha ,\alpha )=0$ over $\mathbb{Q}$. Part (b) therefore makes $b_m$ even, soFor the failure without orientability, ${\mathbb{RP}}^2$ is a closed nonorientable surface with one cell in each dimension zero, one and two; hence $\chi =1-1+1=1$. More generally, ${\mathbb{RP}}^{4r+2}$ is nonorientable and has Euler characteristic one. Thus the orientation hypothesis in the Euler characteristic parity in dimensions congruent to two modulo four is essential.

Work with integral cohomology. For a frame $v=(v_1,\dots ,v_k)$ in the complex Stiefel manifold, defineThese fibres form the complement bundle on a complex Stiefel manifold, a complex vector subbundle of the trivial rank-$n$ bundle. Indeedis a smooth orthogonal projector of constant complex rank $n-k$; locally projecting suitable fixed basis vectors and applying Gram-Schmidt process provides bundle frames. The underlying real vector bundle has rank $r=2(n-k)$ and is oriented by its complex structure: a complex change of frame has real determinant equal to the positive number $|{\det }_{\mathbb{C}}A{|}^2$.

A point of its sphere bundle is $(v_1,\dots ,v_k,w)$ with $\Vert w\Vert =1$ and $w$ orthogonal to the frame. Appending $w$ gives a diffeomorphismwhose inverse forgets the last vector while retaining it as the unit vector in the complement. Thus the asserted homotopy equivalence is actually an identification of sphere bundles, for $0\le k<n$.

Start the induction at $V_0({\mathbb{C}}^n)=\{\mathrm{point}\}$, with cohomology the exterior algebra on no generators. SupposeThis is a free abelian group in every degree, and for $k\ge 1$ its smallest positive degree is $2n-2k+1=r+1$. Hence $H^r(V_k;\mathbb{Z})=0$. This vanishing holds at $k=0$ as well, since the base is a point and $r=2n>0$. The Euler class $e(E)$ belongs to this zero group and is therefore zero.

For this oriented rank-$r$ vector bundle, the Gysin sequence of a sphere bundle has segmentBecause $e(E)=0$, this becomesIn particular $H^{\ast }(V_{k+1};\mathbb{Z})$ is torsion-free. At $j=r-1$, choose a class $a\in H^{r-1}(V_{k+1};\mathbb{Z})$ with ${\pi }_{!}a=1$. Its restriction to the sphere fibre is an integral generator. The projection formula for sphere bundle integration, with the conventional fibre-integration sign, isIt can be obtained from the relative cohomology of the disk/sphere pair and the Thom isomorphism theorem. Consequently every cohomology class of $V_{k+1}$ is uniquely expressible as ${\pi }^{\ast }c+{\pi }^{\ast }b⌣a$: subtract the second term using its fibre integral, then use exactness to identify the remaining term; injectivity of ${\pi }^{\ast }$ and the projection formula prove uniqueness. This establishes an additive splitting by the old monomials and the old monomials multiplied by $a$.

To establish the ring, $|a|=r-1$ is odd, so graded commutativity of the cup product gives $a^2=-a^2$, and hence $2a^2=0$. The torsion-freeness just proved now gives $a^2=0$. Graded commutativity also gives the required anticommutation with each old odd generator, and the old relations persist under the injective ring map ${\pi }^{\ast }$. The additive splitting proves that there are no additional relations. This is exactly Gysin ring splitting for an odd-dimensional sphere bundle, and supplies a new exterior generator of degreeInduction therefore proves the full ring calculationOver $\mathbb{Z}$, graded commutativity by itself only kills twice an odd square; torsion-freeness is the essential extra step in this integral cohomology of a complex Stiefel manifold proof.

Finally an ordered orthonormal $n$-frame is a unitary matrix, so $V_n({\mathbb{C}}^n)=U(n)$. Each exterior generator may be included once or omitted from a basis monomial. The rank of degree $j$ is therefore the number of subsets of the generator degrees $1,3,\dots ,2n-1$ summing to $j$. Multiplication of their two-term generating factors givesThe highest exponent is $1+3+\cdots +(2n-1)=n^2$, agreeing with the real dimension of the unitary group, and the total rank is $2^n$.

Choose orientations on the two sphere factors whose product is the specified orientation of $X$. Give the four factors of $Y=(S^2{)}^4$ the corresponding orientations in the order $1,2,3,4$. The Künneth theorem has no Tor functor terms here and givesLet $u_i\in H^2(Y;\mathbb{Z})$ be the pullback of the positively normalized cohomology class from factor $i$. The cohomology ring iswith $⟨u_1{u}_2{u}_3{u}_4,[Y]⟩=1$. Its generators commute by graded commutativity of the cup product.

Fix a point in each sphere. For each pair $i<j$, let $S_{ij}$ be the four-dimensional submanifold in which factors $i,j$ vary and the remaining factors equal their fixed points. Orient $S_{ij}$ by the ordered factors $i,j$. The six homology classesgive a coordinate sphere intersection basis by the homology cross product description in the Künneth theorem. Moving a fixed point in one of the other two factors produces a disjoint homologous representative of each $S_{ij}$. Hence every basis element has zero self-intersection number.

More precisely, the intersection pairing isFor complementary pairs the representatives intersect transversely at one point. The sign is positive because swapping two-dimensional factors contributes $(-1{)}^{2\cdot 2}=1$. For pairs with a common index, a factor fixed in both representatives can be displaced to make them disjoint. In the paired order $(12,34),(13,24),(14,23)$, the intersection form is three copies of $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$.

For the diagonal map $\delta :X\to Y$, write $a,b\in H^2(X;\mathbb{Z})$ for the two sphere generators. ThenThe coefficient of $[S_{ij}]$ in $[\Delta ]$ is $⟨u_i{u}_j,[\Delta ]⟩$. Pullback and the preceding cup product relations show that these coefficients are one for $12,14,23,34$ and zero for $13,24$. ThereforeThere are exactly two complementary pairings among these four terms, each counted twice in the square. ThusThis computes both the explicit homology class and its intersection pairing, including all orientation signs.

Use the standard orientations on the three spheres and the complex orientation on Complex projective space. Let $h\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ satisfy $⟨h^3,[{\mathbb{CP}}^3]⟩=1$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^4)$. The Künneth theorem givesWrite the induced map on cohomology as $f^{\ast }h=pa+qb+rc$ with integers $p,q,r$. Naturality and graded commutativity of the cup product giveThe mapping degree is therefore $d=6pqr$. This is the factorial degree obstruction for products of two-spheres. Every positive value is a positive multiple of six, so the least possible positive degree is at least $6$.

To attain it, identify each oriented sphere with ${\mathbb{CP}}^1$. Send three projective linear forms to the coefficients of their product:whereNo product of nonzero linear forms is the zero polynomial. Rescaling any factor rescales all coefficients by the same nonzero scalar. The map is therefore well-defined and is a holomorphic map on the products of projective charts.

Choose a homogeneous cubic with three distinct roots, all in an affine chart. Its preimages are exactly the $3!=6$ orderings of its three projective linear factors. In root coordinates, the derivative of the elementary-symmetric coefficient map has determinant equal, up to sign, to the Vandermonde determinant, the product of the three pairwise root differences. This is nonzero for distinct roots, so all six preimages are regular. A nonsingular complex-linear derivative preserves the underlying real orientation, since its real determinant is the squared absolute value of its complex determinant. Each preimage consequently contributes $+1$ to the mapping degree. HenceThis polynomial multiplication map to complex projective space provides the requested example without assuming that a prescribed cohomology homomorphism is realizable by a continuous map.

Work first with coefficients ${\mathbb{F}}_2$. For a rank-$r$ real vector bundle $E\to B$ over a CW complex, choose a fibre metric and its disk bundle $D(E)$ and sphere bundle $S(E)$. The mod-two Thom class is the unique class $U\in H^r(D(E),S(E);{\mathbb{F}}_2)$ restricting to the nonzero generator on every fibre pair. The Thom isomorphism theorem asserts thatNo orientation of the vector bundle is needed with these coefficients.

Let $j:H^{\ast }(D(E),S(E))\to H^{\ast }(D(E))$ forget the relative condition, and let $s:B\to D(E)$ be the zero section. Set $e_2(E)=s^{\ast }j(U)$, the top Stiefel–Whitney class $w_r(E)$. Since $D(E)$ retracts onto the zero section, $j(U)={\pi }^{\ast }{e}_2(E)$. Thus the relative-to-absolute map corresponds under the Thom isomorphism theorem to multiplication by $e_2(E)$. Substituting these identifications into the long exact sequence in relative cohomology of $(D(E),S(E))$ gives the unoriented Gysin sequencewhere $p$ is the sphere bundle projection.

Apply this to the real tautological line bundle ${\gamma }_n\to {\mathbb{RP}}^n$. Its sphere bundle is $S^n$, with projection the antipodal double cover, and put $t=e_2({\gamma }_n)\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$. Suppose $n\ge 1$. The map on $H^0$ from the connected base to the connected sphere is an isomorphism, so exactness shows that multiplication by $t$ is injective from $H^0$ to $H^1$. Since both groups are one-dimensional, $t$ is a generator.

For $1<j<n$, the adjacent sphere cohomology groups in the Gysin sequence vanish, so multiplication by $t$ is an isomorphism from degree $j-1$ to degree $j$. If $n>1$, the top portion isThe last nonzero arrow is surjective between one-dimensional groups, hence an isomorphism; the preceding arrow is zero, so multiplication by $t$ is again an isomorphism. For $n=1$ the earlier $H^0$ argument already gives the top multiplication. Thus $1,t,\dots ,t^n$ are the nonzero generators in their respective degrees, while $t^{n+1}=0$ by dimension. We obtainFor $n=0$ this says simply $H^{\ast }({\mathbb{RP}}^0;{\mathbb{F}}_2)={\mathbb{F}}_2$, with $t=0$. This computes the mod-two cohomology ring of real projective space, including its multiplication rather than only its additive groups.

Now use integral coefficients and an oriented rank-$n$ vector bundle. Its orientation selects an integral Thom class $U$. Define its Euler class byAgain $j(U)={\pi }^{\ast }e(E)$. The cup product of two relative classes in $H^{\ast }(D(E),S(E))$ agrees with the mixed relative/absolute product after forgetting the relative condition on either factor. ConsequentlyThis is the cup square of a Thom class. If $n$ is odd, graded commutativity of the cup product gives $U\cup U=-U\cup U$, hence $2U^2=0$. Since $U\cup {\pi }^{\ast }(2e(E))$ is, up to the graded sign, the Thom isomorphism image of $2e(E)$, its vanishing impliesThus the Euler class of an oriented odd-rank vector bundle is two-torsion; the conclusion is integral and does not require the base cohomology to be torsion-free.

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

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

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

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

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

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

For a closed oriented $d$-dimensional manifold $M$ and a commutative coefficient ring $R$, Poincare duality says that cap product with the fundamental class is an isomorphismOver a field it equivalently gives a nondegenerate Poincare duality pairing $(a,b)\mapsto ⟨a⌣b,[M]⟩$ between complementary cohomological degrees.

For the six-manifold take rational coefficients and write $b_j={\dim }_{\mathbb{Q}}{H}^j(X;\mathbb{Q})$. Poincare duality gives $b_j=b_{6-j}$, so the Euler characteristic isThe middle-degree Poincare duality pairing on $H^3(X;\mathbb{Q})$ is skew-symmetric by graded commutativity of the cup product, since $(-1{)}^{3\cdot 3}=-1$. It is a nondegenerate alternating bilinear form, so its dimension $b_3$ is even. For example, a nonsingular skew-symmetric matrix of odd size would have $\det A=\det (-A)=-\det A$, impossible over $\mathbb{Q}$. Therefore $\boxed{\chi (X)\in 2\mathbb{Z}}$.

To realize every even integer, use $\chi (S^6)=2$, $\chi ({\mathbb{CP}}^3)=4$ and $\chi (S^3\times S^3)=0$. All three are closed connected orientable six-manifolds. For the connected sum of oriented manifolds, removing a ball from each summand and gluing their boundary spheres givesin dimension six: each removed open ball decreases the Euler characteristic by one, while the gluing sphere $S^5$ has Euler characteristic zero. If $k=2\ell$, putomitting zero copies. The Euler characteristic is $2+2r-2s=k$. This supplies a closed connected orientable example for every $k\in 2\mathbb{Z}$.

Without orientability, evenness need not hold. The Real projective space ${\mathbb{RP}}^6$ is a closed six-manifold with one cell in each dimension $0,\dots ,6$, soIt is nonorientable because the antipodal deck transformation on $S^6$ has degree $(-1{)}^7=-1$ and reverses orientation. Thus it gives the required counterexample.

Give the closed orientable surface its standard CW complex structure: one zero-cell, $2g$ one-cells, and one two-cell attached by the product of $g$ commutators. The cellular boundary of the two-cell is zero, since every edge occurs once with each orientation in that word; the one-cell boundaries are also zero. HenceHere we use the cellular homology theorem, identifying cellular and singular homology, and the universal coefficient theorem for cohomology: its exact sequence has terms $\mathrm{Ext}(H_{q-1},\mathbb{Z})$ and $\mathrm{Hom}(H_q,\mathbb{Z})$. All the homology groups here are free, so the Ext terms vanish.

The ring structure comes from Poincare duality and algebraic intersection number of curves on an oriented surface. For a closed oriented surface, cap product with its fundamental class identifies degree-one cohomology with degree-one homology; evaluating the cup product of two such classes equals the signed intersection number of their dual one-cycles. Choose the usual $g$ pairs of handle curves, each pair meeting positively once, and distinct pairs disjoint. Their dual classes can accordingly be named $a_1,b_1,\dots ,a_g,b_g$ so that, for the positive orientation class $\omega$,These formulas include squares. More generally, graded commutativity of the cup product kills every degree-one square here because $H^2$ is torsion-free. The unit generates $H^0$, and products involving $\omega$ and any positive-degree class vanish for dimensional reasons. These additive groups and multiplication rules completely describe the cohomology ring of a closed oriented surface, including $g=0$, when there are no degree-one generators. The intersection pairing is integral and unimodular, rather than merely nondegenerate over a field.

Let $m=\dim M$ and $q=m-\dim Y$. Here a closed submanifold means compact and without boundary, which is the condition needed for a compactly supported dual class. The orientations of $M$ and $Y$ orient the normal bundle ${\nu }_Y$. Its Thom class is the unique relative class of degree $q$ restricting to the chosen orientation generator on every normal fiber. A tubular neighborhood identifies this class with one near $Y$ in $M$. Choose a compact normal disc neighborhood and extend the relative class to compactly supported cohomology of $M$. Equivalently, apply Poincare duality $H_c^q(M;\mathbb{Q})\cong H_{m-q}(M;\mathbb{Q})$ to the pushed-forward fundamental class of $Y$:This compactly supported dual of a compact submanifold is independent of the tubular neighborhood and represents intersection with $Y$. Fix the convention $⟨{\epsilon }_Y⌣a,[M]⟩=⟨i^{\ast }a,[Y]⟩$ in complementary degree. Its image in ordinary cohomology restricts to zero on $M\setminus Y$, because it comes from relative cohomology supported near $Y$. If a submanifold is only closed as a subset but is noncompact, its usual dual lies in ordinary cohomology instead; compact support is not asserted in that situation.

Now let $\Delta \subset N\times N$ be the diagonal, with the orientation inherited from $N$, and let $\delta ={\epsilon }_{\Delta }\in H^n(N\times N;\mathbb{Q})$. For each $i$, choose a basis $a_{i,j}$ of $H^i(N;\mathbb{Q})$ and its dual basis $b_{i,j}$ of $H^{n-i}(N;\mathbb{Q})$ satisfying $⟨a_{i,j}⌣b_{i,k},[N]⟩={\delta }_{jk}$. The Poincare duality pairing and the Künneth theorem give the cohomology class of the diagonalThe sign can be checked from the defining identityFor $|u|=i$, moving the degree-$i$ class in the second factor past $u$ contributes $(-1{)}^i$, and interchanging $b_{i,j}$ and $u$ contributes $(-1{)}^{i(n-i)}$; the displayed coefficient cancels their product. Equivalently its exponent is $ni$ modulo two.

The cohomological pushforward between closed oriented manifolds is characterized byfor complementary degrees. This follows immediately from its definition by Poincare duality and naturality of evaluation on pushed-forward homology. In degree $i$, the matrix trace of $f^{!}{g}^{\ast }$ on $H^i(N;\mathbb{Q})$ is consequentlyFor finite-dimensional maps between two possibly different vector spaces, $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. Thus this matrix trace equals that of $g^{\ast }{f}^{!}$ on $H^i(M;\mathbb{Q})$. Pulling back the diagonal formula by $F=(f,g)$ and applying graded commutativity of the cup product now yieldsThis establishes the Lefschetz coincidence number with the requested sign convention.

If $f(m)\ne g(m)$ for every $m$, the map $(f,g)$ factors through $(N\times N)\setminus \Delta$. The diagonal class restricts to zero there, so its pullback is zero and $L(f,g)=0$. Taking the contrapositive proves the Lefschetz coincidence theorem:No transversality or isolated-coincidence assumption is needed. This cohomological argument also applies to continuous maps, which covers the last request.

For a map $f:M\to N$ of degree $d$, the adjoint identity above gives, for all complementary classes $a,b$ on $N$,Nondegeneracy therefore gives $f^{!}{f}^{\ast }=d\mathrm{id}$. Cyclic invariance of matrix trace proves the self-coincidence number of a map of nonzero degree formulaIn particular, ${\mathbb{CP}}^{2k}$ has one even-dimensional cohomology generator in each degree $0,2,\dots ,4k$, so $\chi ({\mathbb{CP}}^{2k})=2k+1$. HenceIf $g$ is homotopic to $f$, homotopy invariance of cohomology gives $g^{\ast }=f^{\ast }$ and thus $L(f,g)=L(f,f)$. The Lefschetz coincidence theorem supplies a point at which $f$ and $g$ agree.

The ordinary Poincare duality statement here is for a compact manifold without boundary, of dimension $d$, oriented over the field $F$. Its fundamental class $[M{]}_F$ induces isomorphismsfor every $q$. For a manifold with boundary the appropriate statement is Poincare-Lefschetz duality with relative groups; the ordinary pairing need not be nondegenerate. Over a field, the universal coefficient theorem for cohomology identifies $H^{d-q}(M;F)$ with the full dual of $H_{d-q}(M;F)$. The cap-cup evaluation identity and duality therefore makea perfect pairing. Explicitly, a nonzero $a$ has a nonzero cap product, and a linear functional $b$ on its homology group takes a nonzero value on that product. The same argument in the other variable proves nonsingularity. On the whole graded cohomology, define $B(a,b)$ by taking the degree-$d$ part of $a⌣b$ before evaluation. For a nonzero component $a_q$ choose $b$ homogeneous of degree $d-q$ to pair nontrivially with it. All other components contribute zero in degree $d$. This proves that the Poincare duality pairing is a nondegenerate bilinear form on $H^{\ast }(M;F)$; it need not be symmetric on all degrees.

Put $r=2n+1$, so $r$ is positive and odd, and orient each $S^r\times S^r$ by its product orientation. The Künneth theorem gives integral cohomology $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^2$ in degree $r$, and zero elsewhere. Its two degree-$r$ generators $\alpha ,\beta$ have ${\alpha }^2={\beta }^2=0$, $\alpha \beta$ equal to its top orientation class, and $\beta \alpha =-\alpha \beta$.

For a connected sum of oriented manifolds, excision and the long exact sequence for deleting a ball show that deleting a ball removes the top homology class and leaves all lower positive homology groups unchanged. The boundary sphere represents zero in the punctured manifold: it is the boundary of its relative fundamental chain. In the Mayer–Vietoris sequence for the two punctured pieces joined along the separating sphere, the orientation class of the connected sum maps onto that sphere's class. The intermediate positive groups are consequently the direct sums of the groups of the two original manifolds. This also covers $r=1$, where the sphere is a circle and the boundary-class observation is necessary in the middle degree. Iterating and applying the universal coefficient theorem for cohomology givesHere $g\ge 1$; under the conventional extension $W_0=S^{2r}$ the same formula holds with a zero middle group.

Let $\omega$ be the top cohomological orientation class. The connected-sum pinch map to the wedge of the $g$ sphere products gives degree-one projections to each summand. Pull back the two factor classes from summand $i$ to obtain ${\alpha }_i,{\beta }_i$. Their product is $\omega$, since the projection has degree one. Classes from different wedge summands have zero positive-degree cup products, and graded commutativity of the cup product supplies the reversed sign. Thus the full cohomology ring of a connected sum of odd-dimensional sphere products is the graded free abelian group just displayed with multiplicationThe unit is $1$, and $\omega$ times any positive-degree class is zero by dimension. This describes all products, including ${\omega }^2=0$.

The smooth involution is a diffeomorphism, since it is its own inverse. Its fixed set is closed; discreteness and compactness therefore make it finite. The supplied positivity of $\det (I-D{f}_x)$ makes every fixed point nondegenerate with local index $+1$. The Lefschetz-Hopf fixed-point theorem then givesIndeed the degree-zero and top-degree traces are both one, because the manifold is connected and $f$ preserves its orientation; the middle degree is odd.

On $V=H^r(W_g;\mathbb{R})$, the form $\Omega (a,b)=⟨a⌣b,[W_g]⟩$ is a nondegenerate alternating bilinear form by Poincare duality and the oddness of $r$. Thus it is a symplectic vector space of dimension $2g$. Naturality and orientation preservation show that $T$ preserves $\Omega$, and $T^2=I$. For the eigenspaces of a symplectic involution, write $V=V_{+}\oplus V_{-}$: the polynomial $(t-1)(t+1)$ has distinct roots over $\mathbb{R}$. If $v_{+}\in V_{+}$ and $v_{-}\in V_{-}$, then $\Omega (v_{+},v_{-})=\Omega (T{v}_{+},T{v}_{-})=-\Omega (v_{+},v_{-})$, so the two vector subspaces are orthogonal. Each restricted form is nondegenerate, since a vector annihilating its own vector subspace also annihilates the other and hence all of $V$. Their dimensions are therefore even, say $\dim V_{+}=2a$, $\dim V_{-}=2b$, with $a+b=g$. ConsequentlyThis proves the fixed-point congruence for an involution on an odd-sphere connected sum:

The short exact coefficient sequenceinduces the integral Bockstein homomorphismIf $\rho :H^{i+1}(X;\mathbb{Z})\to H^{i+1}(X;\mathbb{Z}/m)$ is reduction modulo $m$, thenEquivalently, if an integral cochain $a$ lifts a modulo-$m$ cocycle and $\delta a=mb$, then $\tilde{\beta }[a]=[b]$ and $\beta [a]=[b\mathrm{mod}m]$.

For integral lifts $a,c$ of classes $x,y$, the cup product coboundary formula isDividing by $m$ and reducing modulo $m$ proves the Bockstein derivation rule

Now assume that the stated closed five-manifold $M$ exists. Its top integral cohomology makes it connected and orientable. The long exact sequence from a coefficient sequence for multiplication by $p$ shows thatand that reduction $H^3(M;\mathbb{Z})\to H^3(M;\mathbb{Z}/p)$ is an isomorphism. HenceThe same coefficient sequence gives $H^4(M;\mathbb{Z}/p)=0$.

Choose $0\ne x\in H^2(M;\mathbb{Z}/p)$ and put $y=\beta x\ne 0$. By Poincare duality over $\mathbb{Z}/p$, the pairingis nondegenerate. Both factors are one-dimensional, so $x⌣y\ne 0$. But $x⌣x=0$ because $H^4(M;\mathbb{Z}/p)=0$, while the derivation rule and graded commutativity of the cup product giveThis contradicts $p>2$. Therefore no such manifold exists.