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Let $\gamma$ be the real tautological line bundle over ${\mathbb{RP}}^2$. A homogeneous polynomial of degree $a$ defines a section of the real line bundleIndeed, on a projective line $\ell \subset {\mathbb{R}}^3$ its restriction satisfies $F(rv)=r^{a}F(v)$, exactly the transformation law for a linear functional on ${\ell }^{\otimes a}$. Its zero set is well defined without treating an odd-degree polynomial as a global real-valued function on projective space.

Put $u=w_1(\gamma )\in H^1({\mathbb{RP}}^2;{\mathbb{F}}_2)$. This is the nonzero generator: the tautological line changes sign along the generating projective loop. Dualization preserves this sign, while tensoring multiplies transition signs, soFor generic coefficients both polynomial sections are transverse to zero and their zero curves meet transversely. This genericity follows from parametric transversality: at each projective point some coordinate is nonzero, and varying the coefficient of its $a$th power varies the section's value freely; the two polynomial coefficient sets vary the two values independently. The common zero set is consequently a zero-dimensional submanifold of a compact space, and is finite.

We justify the needed parity of zeros of a real line-bundle section. Along a closed loop transverse to the zero curve, trivialize the pulled-back line bundle on the cut interval. At a simple zero the real component function changes sign. The product of these sign changes equals the sign of the bundle's transition around the loop, which is the evaluation of its first Stiefel–Whitney class. Therefore the intersection parity of the zero curve with every loop equals the evaluation of $w_1$ on that loop. This says that its mod-two Poincare dual is $w_1$, so

The mod-two Poincare duality intersection rule givesThere are no orientation signs in these coefficients, so every transverse intersection contributes one. Two distinct projective lines meet in exactly one point; both have dual class $u$ by the preceding degree-one calculation. Hence $⟨u^2,[{\mathbb{RP}}^2{]}_2⟩=1$, andThis proves the parity of real projective plane-curve intersections.

Equality with $ab$ need not hold. Choose $a=1$, $b=2$, with the line and conicOn the line, the conic equation becomes $x^2+y^2=0$, which has no nonzero real solution. Thus $\boxed{|A\cap B|=0\ne 2=ab}$. Both real curves are smooth and nonempty, and they are disjoint, so their real intersection is transverse vacuously. Even over the complex numbers they meet in two distinct transverse points $[1,i,0]$ and $[1,-i,0]$; the example is not a tangency or shared-component exception. On the unit sphere $F^2+g^2$ has a positive minimum, so disjointness also persists under sufficiently small real coefficient perturbations, so it is compatible with the generic real situation.

Let $m=\dim X$ be odd. Use mod-two Poincare duality, which applies even when the closed manifold is nonorientable. It gives $b_k^{(2)}=b_{m-k}^{(2)}$, where $b_k^{(2)}={\dim }_{{\mathbb{F}}_2}{H}_k(X;{\mathbb{F}}_2)$. The Euler characteristic can be computed with any field of coefficients: for a finite cell model, the alternating sum of chain dimensions equals the alternating sum of homology dimensions, and the former counts cells independently of the field.

Consequently $\chi (X)={\sum }_{k=0}^m(-1{)}^k{b}_k^{(2)}$. Pair each term with index $m-k$. The paired Betti numbers are equal and their signs opposite because $m$ is odd; there is no unpaired middle index. ThereforeThis argument applies componentwise if the manifold is disconnected and does not require an orientation.

All homology and cohomology in this solution use ${\mathbb{F}}_2$ coefficients, so no orientation hypothesis is necessary. The Poincare-Lefschetz duality degrees are complementary:The starred notation in the question must be understood with this degree reversal.

Turn the handle decomposition upside down. An absolute $k$-handle becomes a relative $(n-k)$-handle based on $\partial W$. The absolute cellular boundary coefficient counts intersections of an attaching sphere with a belt sphere. In the reversed handle decomposition, the same intersections are counted in the opposite order. Over ${\mathbb{F}}_2$ the two incidence matrices are transposes, with no sign ambiguity. Thus the absolute chain complex is the complementary-degree dual of the relative chain complex. Taking homology and using the universal coefficient theorem for cohomology over a field proves the displayed isomorphism. This is mod-two handle duality.

To obtain a perfect pairing from the assumed closed case, form the double of a manifold $D(W)$ from two copies of $W$. The fold map $D(W)\to W$ is a retraction, so the inclusion of either copy induces an injection on homology. If $a\ne 0$ in $H_k(W)$, its image in $H_k(D(W))$ is nonzero. Closed-manifold mod-two Poincare duality supplies a class $z\in H_{n-k}(D(W))$ with intersection $1$ against that image. Cut a representative of $z$ along the boundary and retain its part in the first copy; it defines $b\in H_{n-k}(W,\partial W)$. Equivalently, apply the quotient map $D(W)\to D(W)/W_{-}\cong W/\partial W$ and excision. Intersections with a representative of $a$ in the interior are unchanged, so $a\cdot b=1$. This proves nondegeneracy in the first variable. Mod-two handle duality gives equal finite dimensions for the two spaces, hence nondegeneracy in the second variable as well.

For the three-dimensional conclusion, setThe long exact sequence in relative homology says $L=\mathrm{im}\partial$, where $\partial :H_2(W,\partial W)\to V$. The closed-surface intersection pairing on $V$ is nondegenerate. A collar calculation gives the adjoint identityIndeed, push $x$ slightly into the collar: intersections with the relative surface representing $b$ correspond to its boundary intersections with $x$. By the absolute-relative perfect pairing already proved, $x$ is orthogonal to every $\partial b$ precisely when $i_{\ast }x=0$. Therefore $L^{\perp }=L$.

For any finite-dimensional space with a perfect pairing, $\dim L+\dim L^{\perp }=\dim V$. Hence the half-lives-half-dies theorem givesBut $H_1({\mathbb{RP}}^2;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ has dimension one. It cannot be the entire boundary of a compact three-manifold, because the displayed dimension would be $1/2$. This excludes nonorientable three-manifolds as well as orientable ones.

A closed connected three-manifold has a handle decomposition with one zero-handle and one three-handle. To arrange this, cancel the zero-handles along a spanning tree of connecting one-handles; perform the corresponding construction in the dual handle decomposition to consolidate the three-handles. This does not require the three-manifold to be orientable.

If there are $r$ one-handles and $s$ two-handles, the Euler characteristic isMod-two Poincare duality pairs the Betti numbers of a closed odd-dimensional manifold, giving $\chi (Y)=0$. Thus $r=s$. The one-handles give $r$ generators of a group of the fundamental group, the attaching circles of the two-handles give $r$ relators, and the three-handle does not change the fundamental group. Therefore ${\pi }_1(Y)$ admits a balanced presentation.

Now letbe any finite group presentation. Start with a four-dimensional zero-handle and $r$ one-handles. Its fundamental group is the free group on the $x_i$. Represent the finitely many relators by disjoint embedded circles in its three-dimensional boundary; a small perturbation makes the circles disjoint. Choose framings of an embedded sphere and attach $s$ two-handles. The resulting compact connected oriented four-manifold $W$ has ${\pi }_1(W)\cong G$ by the Seifert-van Kampen theorem.

The map ${\pi }_1(\partial W)\to {\pi }_1(W)$ is surjective. One way to see this is to turn the handle decomposition upside down: the relative handles based on $\partial W$ have indices $2,3,4$, and so add no fundamental group generators. Also $\partial W$ is connected, since surgery on embedded circles in a connected three-manifold preserves connectedness.

Take the double of a manifoldIt is a closed connected smooth oriented four-manifold. The Seifert-van Kampen theorem givesBoth maps from ${\pi }_1(\partial W)$ are the same surjection, under the natural identification of the two copies. The pushout identifies the two copies of every element of $G$ and imposes no new relations: the fold homomorphism is inverse to either inclusion. ThusThis is four-manifold realization of finitely presented groups.