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Suppose a continuous map $g:S^n\times S^n\to S^n$ changes sign on negating either input. Its quotient $\bar{g}:{\mathbb{RP}}^n\times {\mathbb{RP}}^n\to {\mathbb{RP}}^n$ pulls back the real tautological line bundle to the tensor product of vector bundles $p_1^{\ast }{\gamma }_n\otimes p_2^{\ast }{\gamma }_n$. The fiber map sends $(sx)\otimes (ty)$ to $stg(x,y)$. The First Stiefel–Whitney class of a tensor product of real line bundles consequently gives ${\bar{g}}^{\ast }a=a_1+a_2$. The Künneth theorem and the mod-two cohomology ring of real projective space implyEvery interior binomial coefficient in row $n+1$ must be even. By binomial coefficients with even interior terms, $n=2^k-1$ for some $k\ge 0$. This is a necessary condition, not an assertion of existence in every such dimension.

For an odd map between spheres $f:S^n\to S^m$ and its quotient $\bar{f}$, there is an isomorphism ${\gamma }_n\cong {\bar{f}}^{\ast }{\gamma }_m$ of real tautological line bundles. With $x$ a unit vector, send $tx$ to $tf(x)$; replacing $x$ by $-x$ and $t$ by $-t$ gives the same vector. This proves both linearity on fibers and well-definedness. Therefore ${\bar{f}}^{\ast }{w}_1({\gamma }_m)=w_1({\gamma }_n)$. The mod-two cohomology ring of real projective space then implies $n\le m$, since a vanishing $(m+1)$st power must pull back to a vanishing power.

A degree-$a$ homogeneous real polynomial defines a section of the $a$th tensor power of the dual real tautological line bundle over ${\mathbb{RP}}^2$. Its first Stiefel–Whitney class is $au$, where $u$ generates $H^1({\mathbb{RP}}^2;{\mathbb{F}}_2)$. By parity of zeros of a real line-bundle section, a generic zero curve has mod-two dual class $au$. Two transverse zero curves of degrees $a,b$ therefore have intersection parity $⟨ab{u}^2,[{\mathbb{RP}}^2{]}_2⟩=ab$, since two distinct projective lines intersect once. For example, the line $z=0$ and the conic $x^2+y^2-z^2=0$ are disjoint in the real projective plane, although their degree product is two.

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.

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.

The real tautological line bundle has total space $E=\{(\ell ,v):\ell \in {\mathbb{RP}}^n,v\in \ell \}$ and projection $(\ell ,v)\mapsto \ell$. On the open chart $U_i=\{[x]:x_i\ne 0\}$, the vector $s_i([x])=x/x_i$ is a continuous nowhere-zero section of a vector bundle, independent of the representative of $[x]$. The explicit local trivialization iswith inverse $([x],v)\mapsto ([x],v_i)$. Both maps are continuous and linear on each fiber. The charts cover the base, proving the existence of the required local trivializations.

Let $E\to B$ be a rank-$r$ vector bundle oriented over a commutative ring $R$, with $r\ge 1$. An $R$-orientation is a coherent choice of generator of each fiber's top relative cohomology, or equivalently a Thom class $u_E$. The Thom isomorphism theorem identifies $H^{q-r}(B;R)$ with $H^q(D(E),S(E);R)$ by $a\mapsto {\pi }^{\ast }a⌣u_E$. The disk bundle retracts to $B$, and the relative-to-absolute map becomes multiplication by the Euler class $e(E)$. The pair's long exact sequence therefore becomes the Gysin sequence of a sphere bundle:With real coefficients, the bundle must be oriented in the usual sense. For the computation here use $R={\mathbb{F}}_2$, over which every real vector bundle is oriented, including the nonorientable real tautological line bundle when $n\ge 1$.

The unit sphere bundle of ${\gamma }_n$ is $S^n$: a unit vector determines its line. Put $a=e_2({\gamma }_n)=w_1({\gamma }_n)\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$. For $n\ge 1$, the pullback $H^0({\mathbb{RP}}^n;{\mathbb{F}}_2)\to H^0(S^n;{\mathbb{F}}_2)$ is an isomorphism, so the next connecting homomorphism is zero and multiplication by $a$ is injective on $H^0$. Since the sphere's cohomology vanishes strictly between degrees zero and $n$, exactness makes multiplication by $a$ an isomorphism $H^{q-1}\to H^q$ for $1\le q<n$, and injective for $q=n$. For $q=1$ these assertions also use the just-noted vanishing of the connecting homomorphism from $H^0(S^n)$.

The standard $n$-dimensional CW complex structure of Real projective space gives $H^q=0$ for $q>n$. Consequently the segment $H^n(S^n;{\mathbb{F}}_2)\to H^n({\mathbb{RP}}^n;{\mathbb{F}}_2)\stackrel{⌣a}{\to }{H}^{n+1}({\mathbb{RP}}^n;{\mathbb{F}}_2)=0$ makes $H^n$ at most one-dimensional. The preceding injectivity makes it exactly one-dimensional, generated by $a^n$. This includes $n=1$; $n=0$ is a point separately. We have derived the multiplication, not just the dimensions:

An odd map between spheres descends to $\bar{f}:{\mathbb{RP}}^n\to {\mathbb{RP}}^m$. The odd maps pull back the real tautological line bundle argument gives ${\gamma }_n\cong {\bar{f}}^{\ast }{\gamma }_m$: with $x$ a unit vector the fiber map is $tx\mapsto tf(x)$, unchanged when $(x,t)$ is replaced by $(-x,-t)$. Naturality of the first Stiefel–Whitney class gives ${\bar{f}}^{\ast }b=a$. If $n>m$, the relation $b^{m+1}=0$ would imply $a^{m+1}=0$ in a ring where this power is nonzero. Hence $\boxed{n\le m}$. The case $n=0$ is immediate, and the same argument rules out $m=0<n$.

For the map with separate oddness, pass to $\bar{g}:{\mathbb{RP}}^n\times {\mathbb{RP}}^n\to {\mathbb{RP}}^n$. Let $a,b$ be the degree-one classes from its two domain factors and $c$ the target class. There is an isomorphismgiven on fibers by $(sx)\otimes (ty)\mapsto stg(x,y)$; flipping either unit representative leaves this map well defined. The First Stiefel–Whitney class of a tensor product of real line bundles gives ${\bar{g}}^{\ast }c=a+b$. By the Künneth theorem,Pulling back $c^{n+1}=0$ shows that $(a+b{)}^{n+1}=0$. The two end monomials already vanish, while every interior monomial $a^j{b}^{n+1-j}$, $1\le j\le n$, is a distinct nonzero basis element. Therefore all interior binomial coefficients in row $N=n+1$ are even. To identify such rows, write $N={\sum }_{i\in I}{2}^i$. In ${\mathbb{F}}_2[t]$ repeated squaring gives $(1+t{)}^N={\prod }_{i\in I}(1+t^{2^i})$. If $I$ has at least two elements, the coefficient of $t^{2^{\min I}}$ is one and its exponent is strictly between zero and $N$, a contradiction. Thus $N$ is a power of two, as in binomial coefficients with even interior terms, andThis is the necessary cohomological obstruction to separately odd sphere multiplication; it does not assert existence in every dimension of this form.

For a real rank-$r$ vector bundle $E\to B$, the Thom isomorphism theorem with ${\mathbb{F}}_2$ coefficients identifies the relative cohomology sequence of its disk bundle and sphere bundle withThe multiplier is the top Stiefel–Whitney class, also called the mod-two Euler class. No orientation of $E$ is required. For the real tautological line bundle, the sphere bundle is the antipodal cover of Real projective space; this sequence proves that all powers of the degree-one generator through the dimension of the base are nonzero.