= Solution
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 <isomorphisms>
$$
H^q(M;F)\xrightarrow{\ a\mapsto[M]_F\frown a\ }H_{d-q}(M;F)
$$
for 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 make
$$
H^q(M;F)\times H^{d-q}(M;F)\longrightarrow F,\qquad (a,b)\longmapsto\langle a\smile b,[M]_F\rangle
$$
a <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 $a\smile b$ 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 $H^*(M;F)$; 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> gives
$$
\boxed{H^q(W_g;\mathbb Z)\cong\begin{cases}\mathbb Z,&q=0,2r,\\\mathbb Z^{2g},&q=r,\\0,&\text{otherwise}.\end{cases}}
$$
Here $g\geq1$; 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 multiplication
$$
\boxed{\alpha_i\alpha_j=\beta_i\beta_j=0,\qquad \alpha_i\beta_j=\delta_{ij}\omega,\qquad \beta_j\alpha_i=-\delta_{ij}\omega.}
$$
The 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 $\det(I-Df_x)$ makes every fixed point nondegenerate with local index $+1$. The <Lefschetz-Hopf fixed-point theorem> then gives
$$
\#\operatorname{Fix}(f)=L(f)=2-\operatorname{tr}T,\qquad T=f^*:H^r(W_g;\mathbb R)\to H^r(W_g;\mathbb R).
$$
Indeed 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 $\Omega(a,b)=\langle a\smile b,[W_g]\rangle$ 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 $\Omega(v_+,v_-)=\Omega(Tv_+,Tv_-)=-\Omega(v_+,v_-)$, 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 $a+b=g$. Consequently
$$
\operatorname{tr}T=2a-2b=2g-4b,\qquad \#\operatorname{Fix}(f)=2-2g+4b.
$$
This proves the <fixed-point congruence for an involution on an odd-sphere connected sum>:
$$
\boxed{\#\operatorname{Fix}(f)\equiv2-2g\pmod4.}
$$
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