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For odd $r\ge 1$ and $W_g={\#}_{i=1}^g(S^r\times S^r)$, the integral cohomology has one copy of $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^{2g}$ in degree $r$, and zero elsewhere. Choose the top orientation class $\omega$ and degree-$r$ classes ${\alpha }_i,{\beta }_i$ from the two factors of summand $i$. Their nonzero positive-degree basis products areAll ${\alpha }_i{\alpha }_j$ and ${\beta }_i{\beta }_j$ vanish, and $\omega$ annihilates positive-degree classes. The Künneth theorem computes each sphere product; the Mayer–Vietoris sequence computes the connected sum groups, and its degree-one pinch maps determine the products. Thus its middle-degree Poincare duality pairing is a symplectic vector space after changing coefficients to a field of characteristic different from two. The case $r=1$ is the usual cohomology ring of a closed oriented surface.

A degree of a map between oriented manifolds equal to one makes the pullback on degree-one cohomology injective by the Poincare duality pairing. Thus a degree-one map of closed oriented surfaces requires $2h\le 2g$. Conversely, collapse the extra handles in ${\Sigma }_h\#{\Sigma }_{g-h}$ onto a point. The retained oriented disc shows that this map has degree one.

For a closed oriented manifold of dimension $4r+2$, Poincare duality pairs all rational Betti numbers except the middle one, making their total contribution to the Euler characteristic even. The middle cohomology degree is odd, so its nondegenerate Poincare duality pairing is an alternating bilinear form and has even dimension. Hence the Euler characteristic is even. Nonorientable manifolds can have odd Euler characteristic, as Real projective space of even dimension demonstrates.

Let $W_g$ be a connected sum of oriented manifolds consisting of $g$ copies of $S^r\times S^r$ with odd $r\ge 1$. Suppose an orientation-preserving smooth involution $f$ has finitely many fixed points, all nondegenerate with positive determinant $\det (I-D{f}_x)$. The Lefschetz-Hopf fixed-point theorem and the cohomology ring of a connected sum of odd-dimensional sphere products giveThis middle group has dimension $2g$ and its Poincare duality pairing is alternating. The induced map preserves the pairing and squares to the identity. By eigenspaces of a symplectic involution, its trace is $2g-4b$ for an integer $b$. Therefore $\#\mathrm{Fix}(f)\equiv 2-2g(\mathrm{mod}4)$.

A one-sided section loop and a two-sided fiber loop in the reflection mapping torus have mod-two self-intersections one and zero and meet once. Their intersection form is $Q$. In the evaluation-dual cohomology basis, the Poincare duality pairing has matrix $Q^{-1}$, giving $t^2=0$ and $x^2=tx\ne 0$.

Here closed oriented surfaces are connected, as in the usual single-genus convention. Let $f:X\to Y$ have nonzero mapping degree $d$. Work over $\mathbb{Q}$, so multiplication by $d$ is invertible and cannot vanish.

The Poincare duality pairing on $H^1(Y;\mathbb{Q})$ is nondegenerate. Thus for every $0\ne a\in H^1(Y;\mathbb{Q})$ there is $b\in H^1(Y;\mathbb{Q})$ with $⟨a⌣b,[Y]⟩\ne 0$. Naturality of the cup product and evaluation on the fundamental class giveIn particular $f^{\ast }a\ne 0$, so $f^{\ast }:H^1(Y;\mathbb{Q})\to H^1(X;\mathbb{Q})$ is injective. The cohomology ring of a closed oriented surface has $\dim H^1({\Sigma }_g;\mathbb{Q})=2g$. Comparing dimensions now givesThis proves genus monotonicity under nonzero-degree surface maps by rational cohomology injectivity of a nonzero-degree map, without requiring $f$ to be a covering map or a smooth map.

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.

Let the integer degree of a map between oriented manifolds be $d\ne 0$, so $f_{\ast }[X]=d[Y]$. For any nonzero $\alpha \in H^k(Y;\mathbb{Q})$, nondegeneracy of the Poincare duality pairing supplies $\gamma \in H^{n-k}(Y;\mathbb{Q})$ with $⟨\alpha ⌣\gamma ,[Y]⟩\ne 0$. Naturality of the cup product and its evaluation givesThus $f^{\ast }\alpha \ne 0$, proving injectivity of pullback on rational cohomology. By the universal coefficient theorem for cohomology, these finite-dimensional rational cohomology groups are the duals of the corresponding homology groups, and the dual of $f_{\ast }$ is $f^{\ast }$. Injectivity of the dual map therefore provesOne can see the surjection constructively. Given $u\in H_k(Y;\mathbb{Q})$, choose its Poincare dual $\eta \in H^{n-k}(Y;\mathbb{Q})$ with $\eta \cap [Y]=u$. Then $v=d^{-1}(f^{\ast }\eta \cap [X])$ satisfies $f_{\ast }v=d^{-1}(\eta \cap f_{\ast }[X])=u$ by naturality of the cap product. Division by $d$ is the reason rational coefficients suffice.

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.

The integral Poincare duality theorem says that a closed connected oriented $n$-manifold has a fundamental class $[M]\in H_n(M;\mathbb{Z})$ for which cap product gives $H^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$. The general Poincare duality with the orientation local system uses $[M]\in H_n(M;{\mathcal{O}}_M)$ and givesfor a local coefficient system $\mathcal{A}$. Over ${\mathbb{F}}_2$ it holds without orientability. For constant integral coefficients on a nonorientable connected closed manifold, $H_n=H^0(M;{\mathcal{O}}_M)=0$, while $H^n=H_0(M;{\mathcal{O}}_M)=\mathbb{Z}/2$: the invariants of sign reversal in $\mathbb{Z}$ are zero and its coinvariants are $\mathbb{Z}/(a\sim -a)$.

If the three-manifold is orientable, duality and the universal coefficient theorem for cohomology giveFor a nonorientable three-manifold, $H_3=0$ and $H^3=\mathbb{Z}/2$. The top-degree universal coefficient sequence therefore givesSince the homology groups are finitely generated, their cyclic decomposition shows that this Ext group is isomorphic to the torsion subgroup of $H_2$. Thus the torsion is exactly $\mathbb{Z}/2$, not the possibly larger torsion subgroup of $H_1$. Also odd-dimensional closed manifolds have zero Euler characteristic, by pairing complementary mod-two Betti numbers. If $r$ is the free rank of $H_2$, it follows that $0=\chi (M)=1-b+r$, giving the second homology of a closed nonorientable three-manifoldThe inequality is forced by the nonnegative free rank $b-1$.

For the final graded-group construction, one always has $K^0=K^3=K^6=\mathbb{Z}$, since each of these is copied from $H^0$ of a connected three-manifold. Suppose it were the cohomology of a closed six-manifold $N$. The group $K^0$ makes $N$ connected, and $K^6=\mathbb{Z}$ makes it orientable: the nonorientable alternative has top integral cohomology $\mathbb{Z}/2$, as above. Its third rational cohomology would then be one-dimensional. But the middle Poincare duality pairing on $H^3(N;\mathbb{Q})$ is nondegenerate and skew-symmetric by graded commutativity, hence must have even dimension. Equivalently, a generator $u$ would satisfy $u⌣u=-u⌣u$, so its square in the torsion-free top group is zero and the entire rank-one pairing vanishes. This contradicts duality and the even rank of middle cohomology in dimension four k plus two. ThereforeBoth existence questions have negative answers, independently of the other groups of the original three-manifold.

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.

Suppose $F:{\Sigma }_g\to {\Sigma }_h$ has degree one. For degree-one classes $u,v$ on the target, naturality of the cup product and the definition of the degree of a map between oriented manifolds giveIf a nonzero $u$ had $F^{\ast }u=0$, nondegeneracy of the Poincare duality pairing would supply $v$ with nonzero right side, a contradiction. Thus $F^{\ast }$ injects the $2h$-dimensional real degree-one cohomology into the $2g$-dimensional source. Hence $g\ge h$. This is the cohomological injectivity of a degree-one map in this setting.

Conversely, for $g\ge h$, express ${\Sigma }_g$ as ${\Sigma }_h\#{\Sigma }_{g-h}$. Collapse the second punctured summand and the joining circle to a point. The quotient of the retained punctured summand by its boundary is homeomorphic to ${\Sigma }_h$, giving a continuous map to that surface. Its restriction to a small oriented disc away from the collapsing region is an orientation-preserving homeomorphism, and a point in this disc has exactly one preimage. The induced map on local top homology, and hence on the fundamental class, has coefficient $+1$. The map therefore has degree one. For $h=0$, the same construction is the familiar collapse of the complement of a disc to obtain $S^2$.

ConsequentlyThis proves both directions of the degree-one maps between closed oriented surfaces criterion. The case $g=h$ also admits the identity map.

Put $A={\Sigma }_h^{\partial }$. This surface with boundary deformation retracts onto a wedge of $2h$ circles, so $H^1(A;\mathbb{R})$ has dimension $2h$ and $H^2(A;\mathbb{R})=0$. If $r$ were a retraction, ${\iota }^{\ast }{r}^{\ast }=\mathrm{id}$ would make $r^{\ast }$ injective. Its image $W\subset H^1({\Sigma }_g;\mathbb{R})$ would have dimension $2h$.

Every pair of classes $u,v$ in $H^1(A;\mathbb{R})$ has $u⌣v=0$, since $H^2(A;\mathbb{R})=0$. Naturality givesThus $W$ is an isotropic subspace of a symplectic vector space for the nondegenerate skew Poincare duality pairing on the $2g$-dimensional space $H^1({\Sigma }_g;\mathbb{R})$. The stated linear-algebra bound gives $2h\le g$. Hence

The bound is sharp. Double $A$ along its boundary: $D(A)=A{\cup }_{\partial A}A$ is the closed oriented surface of genus $2h$. Identify each copy with $A$ and fold them onto one copy. The two maps agree on the joining circle, so they give a continuous retraction fixing the first copy pointwise. This includes $h=0$, where the double of a disc is a sphere.

More generally, for $g\ge 2h$, add $g-2h$ handles in the interior of the second copy. Pinch those extra handles onto their connecting point, keeping the boundary fixed, and then fold onto $A$. This is still a retraction. In fact the construction works for any embedding in the question: the connected complement has one boundary component and genus $g-h$ by Euler-characteristic additivity, and the classification theorem for surfaces identifies it, relative to that boundary, with a copy of $A$ having $g-2h$ additional handles. Thus the exact existence criterion is $2h\le g$, as expressed by retraction onto a punctured oriented surface.

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:

A once-punctured genus-$h$ closed orientable surface retracts to a wedge of $2h$ circles. A retraction from ${\Sigma }_g$ would inject its degree-one cohomology as an isotropic subspace of a symplectic vector space for the Poincare duality pairing, since the punctured surface has zero degree-two cohomology. The maximal isotropic dimension is $g$, giving $2h\le g$. Equality is realized by folding the double of the punctured surface onto one copy. Additional handles in the other copy can be pinched away while fixing the joining boundary, proving sufficiency for every embedding of such a one-boundary subsurface when $g\ge 2h$.