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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 $m\ge 0$, the cohomology ring isandThe generator $a$ reduces to the square of the degree-one generator in the mod-two cohomology ring of real projective space; $b$ is the integral orientation class in odd top dimension.

For $n\ge 2$, a nonsurjective self-map of Real projective space ${\mathbb{RP}}^n$ has zero mod-two degree of a map between closed manifolds. In the mod-two cohomology ring of real projective space, the nonzero degree-one class $a$ has $a^n\ne 0$. If pullback fixed $a$, it would fix $a^n$, contradicting zero degree. Thus the map is trivial on the fundamental group and lifts to $S^n$. A point missed by the original map has both its inverse images missed by the lift. The lift therefore takes values in a punctured sphere, which is contractible, and is null-homotopic. Composing its contraction with the covering projection proves the conclusion.

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.

Work first over ${\mathbb{F}}_2$. The mod-two cohomology ring of real projective space is $H^{\ast }({\mathbb{RP}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2[a]/(a^{n+1})$, with $|a|=1$. A map missing a point has zero mod-two degree of a map between closed manifolds, so $f^{\ast }(a^n)=0$. The group in degree one has only two elements, so $f^{\ast }a$ is either zero or $a$. The latter would imply $f^{\ast }(a^n)=(f^{\ast }a{)}^n=a^n\ne 0$, a contradiction. Therefore $f^{\ast }a=0$, and multiplicativity makes $f^{\ast }$ zero in every positive cohomological degree. Duality of homology and cohomology over a field consequently makes $f_{\ast }$ zero in every positive degree with mod-two coefficients.

For integral coefficients, cellular homology of real projective space gives $H_k\cong \mathbb{Z}/2$ in odd degrees $0<k<n$, zero in even degrees $0<k<n$, and a top group $\mathbb{Z}$ if $n$ is odd and zero if $n$ is even. For each torsion group, the natural reduction map to mod-two homology is injective by the universal coefficient theorem for homology; naturality and the preceding vanishing imply that its integral induced map is zero. When $n$ is odd, the integral top-degree map is also zero because the nonsurjective map has integer degree of a map between oriented manifolds zero. All groups above dimension $n$ vanish. Hence $f_{\ast }$ is zero on all positive integral homology groups as well.

For $n\ge 2$, ${\pi }_1({\mathbb{RP}}^n)=\mathbb{Z}/2$, and its abelianization is already $H_1({\mathbb{RP}}^n;\mathbb{Z})$. The vanishing just proved therefore implies $f_{\ast }{\pi }_1=0$. By the lifting criterion for a covering space, $f$ lifts through the double covering space $q:S^n\to {\mathbb{RP}}^n$ to a map $\tilde{f}:{\mathbb{RP}}^n\to S^n$ with $q\tilde{f}=f$.

Choose a point $y$ missed by $f$. Both points of $q^{-1}(y)$ are missed by $\tilde{f}$. In particular the lift has image in $S^n\setminus \{z\}\cong {\mathbb{R}}^n$, for either such point $z$. This punctured sphere is contractible, so contraction there gives a homotopy from $\tilde{f}$ to a constant map into $S^n$. Composing with $q$ contracts $f$. ThusThis proves the nonsurjective self-map of real projective space is null-homotopic conclusion, and also shows that all positive-degree induced maps vanish with any coefficients.

For a rank-$r$ real vector bundle with a fiber metric, the Thom isomorphism theorem says that if the bundle is oriented over the coefficient ring $R$, there is a Thom class $u\in H^r(D(E),S(E);R)$ restricting to the chosen generator on every fiber pair, andis an isomorphism. For $R={\mathbb{F}}_2$, every real vector bundle has the required orientation. The disk bundle retracts to $B$, so apply the pair long exact sequence and the Thom isomorphism to get the Gysin sequence of a sphere bundleThe first map is multiplication by the Euler class of a vector bundle, the pullback to the zero section of the absolute image of $u$. The connecting map followed by inverse Thom isomorphism gives ${\pi }_{!}$.

Apply this over ${\mathbb{F}}_2$ to the tautological real line bundle $\gamma$ on ${\mathbb{RP}}^n$. Its unit sphere bundle is $S^n$, with projection sending a unit vector to its line. Put $a=e_2(\gamma )=w_1(\gamma )\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$, the mod-two Euler class of a real line bundle. For $n\ge 1$ the degree-zero pullback $H^0({\mathbb{RP}}^n)\to H^0(S^n)$ is an isomorphism. Exactness therefore makes the subsequent transfer zero, and multiplication by $a$ from $H^0$ into $H^1$ injective. In intermediate degrees the sphere cohomology vanishes, making each successive multiplication by $a$ an isomorphism. At degree $n$, the map from $H^{n-1}$ into $H^n$ is injective; the transfer $H^n(S^n)\to H^n({\mathbb{RP}}^n)$ is surjective because $H^{n+1}({\mathbb{RP}}^n)=0$. Since $H^n(S^n)={\mathbb{F}}_2$, that top group is one-dimensional too, and $a^n$ is its nonzero generator. The same endpoint reasoning covers $n=1$. This gives the Gysin proof of mod-two projective-space cohomology and the mod-two cohomology ring of real projective space:For $n=0$ this says simply $H^{\ast }({\mathbb{RP}}^0)={\mathbb{F}}_2$.

For a self-map of ${\mathbb{RP}}^4$, the induced map has $f^{\ast }a=\lambda a$ with $\lambda \in {\mathbb{F}}_2$, so $f^{\ast }{a}^j={\lambda }^j{a}^j$. The mod-two alternating trace isfor either value of $\lambda$. The alternating trace on homology equals that on the finite cellular chain complex; consequently this is the reduction modulo two of the integral Lefschetz number. That integer is nonzero, and the Lefschetz fixed-point theorem forces a fixed point. Thus every self-map of the real projective four-space has a fixed point, a case of the Lefschetz fixed-point property of even-dimensional real projective space.

For the odd-dimensional example, let $J(x_1,x_2,x_3,x_4)=(-x_2,x_1,-x_4,x_3)$, so $J^2=-I$. Its projectivization isIt is well-defined and continuous on ${\mathbb{RP}}^3$. A fixed projective point would have $Jv=tv$ for a nonzero real vector and real $t$, forcing $-v=J^{2}v=t^{2}v$ and therefore $t^2=-1$, impossible. This is a fixed-point-free complex-structure map on odd-dimensional real projective space.

Use cohomology with coefficients in ${\mathbb{F}}_2={\mathbb{Z}}_2$. The cellular homology of real projective space has one cell in degrees zero, one and two, and its boundary coefficients become zero modulo two. Hence there is a one-dimensional cohomology group in each of those degrees. Let $x\in H^1(P;{\mathbb{F}}_2)$ be the Poincare dual of a projective line. Two distinct projective lines intersect transversely in one point, so $⟨x^2,[P{]}_2⟩=1$. ThusThis is the mod-two cohomology ring of real projective space in dimension two. The Künneth theorem over a field makes the cross product an isomorphism of graded rings. Writing $x,y$ for the factor pullbacks givesThere are no additional signs over ${\mathbb{F}}_2$.

The obstruction to a graph avoiding the diagonal is the mod-two diagonal class of the real projective plane. Its dual class $D\in H^2(P\times P;{\mathbb{F}}_2)$ is determined bywhere $i:P\to P\times P$ is the diagonal embedding. Test $z=x^2,xy,y^2$. Each diagonal pullback is $x^2$, with evaluation one. Since $x^2{y}^2$ is the product's top generator, comparison of coefficients givesIf the submanifold were the graph of a smooth map $f:P\to P$, its parametrization would be $j=(\mathrm{id},f)$. For some $\epsilon \in {\mathbb{F}}_2$, $f^{\ast }x=\epsilon x$, and thereforeOn the other hand, the Poincare dual of the diagonal comes from its relative Thom class in $H^2(P\times P,(P\times P)\setminus \Delta ;{\mathbb{F}}_2)$. It restricts to zero on the complement. A graph disjoint from the diagonal would factor through that complement and force $j^{\ast }D=0$, a contradiction. Such a submanifold cannot be the graph of a smooth map $P\to P$. The relative Thom-class argument also applies to continuous self-maps.

Work with coefficients ${\mathbb{F}}_2$. For $n\ge 1$, the mod-two cohomology ring of real projective space and the Künneth theorem givewhere all three generators have degree one. Any induced map on cohomology has $f^{\ast }z=\alpha x+\beta y$, with $\alpha ,\beta \in {\mathbb{F}}_2$. By naturality of the cup product, the only possible term in top degree isAll other terms vanish because their exponent of $x$ or $y$ exceeds $n$. Butis even for every $n\ge 1$. Thus the displayed top-degree induced map on cohomology is zero for every $f$.

Both top-degree homology groups are one-dimensional over ${\mathbb{F}}_2$, and their dual cohomology groups pair with them nondegenerately. Naturality of this evaluation makes $f_{\ast }$ dual to $f^{\ast }$, so $f_{\ast }$ is also zero and cannot be an isomorphism. This is the mod-two degree obstruction for equal projective factors. ConsequentlyIf zero is included among the allowed values, ${\mathbb{RP}}^0$ is a point and the unique map is an isomorphism on $H_0$. In that convention, the only value is $n=0$.

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.

Cells and attaching maps. Regard ${\mathbb{RP}}^k$ as the lines in ${\mathbb{R}}^{k+1}$ and include ${\mathbb{RP}}^{k-1}$ as the lines whose last coordinate is zero. Its complement consists of lines with a unique representative $(x_1,\dots ,x_k,1)$, and is therefore an open $k$-cell. Inductively this makes Real projective space a CW complex with one cell in each dimension from zero to $n$.

A characteristic map is obtained from the northern closed hemisphere of $S^k$: send a unit vector to the line it spans. Its interior maps homeomorphically onto the open cell, while its equator has the antipodal identification. Thus the $k$-cell is attached by the quotient mapwhich is the antipodal two-sheeted covering. This includes the two endpoints of the one-cell attaching to the zero-cell.

The cellular chain complex has $C_k^{\mathrm{cell}}=\mathbb{Z}$ for $0\le k\le n$ and zero otherwise. To compute its differential, follow the attaching map by collapse of the $(k-2)$-skeleton. The resulting map to ${\mathbb{RP}}^{k-1}/{\mathbb{RP}}^{k-2}\cong S^{k-1}$ has two local contributions. They differ by the mapping degree $(-1{)}^k$ of the antipodal map on $S^{k-1}$. With compatible cell orientations,For $k=1$ the two oriented endpoints cancel, giving the same formula. Consecutive differentials compose to zero, as required.

The mod-two cup products. Modulo two every cellular differential vanishes, so cellular cohomology gives a one-dimensional group in each degree $0,\dots ,n$. Let $a\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$ be the Poincare dual of a projective hyperplane. This class is nonzero: a projective line transverse to that hyperplane meets it once. Intersecting $j$ generic projective hyperplanes produces ${\mathbb{RP}}^{n-j}$, and the cup product of their Poincare duals is the Poincare dual of that intersection. In particular, $a^n$ evaluates to one on the mod-two fundamental class. Therefore every $a^j$, $0\le j\le n$, is nonzero, since otherwise multiplying it by $a^{n-j}$ would contradict $a^n\ne 0$. Dimension makes $a^{n+1}=0$. The mod-two cohomology ring of real projective space isFor $n=0$ this simply means the cohomology of a point.

The product with integral coefficients. The final product's coefficients are unstated; take $\mathbb{Z}$ as the default. Dualizing the cellular chain complex above giveswith all unlisted groups zero. The integral Künneth theorem has tensor terms with $i+j=q$ and Tor functor terms with $i+j=q+1$:For these finite free cellular complexes it splits additively, though not canonically. Both the tensor and Tor functor of two $\mathbb{Z}/2$ summands give $\mathbb{Z}/2$, so no order-four summands occur.

For an efficient count, write $F_A(t)$ for the free-rank polynomial and $T_A(t)$ for the number of $\mathbb{Z}/2$ summands in each degree. The integral Künneth torsion polynomial rule isThe last two factors record respectively the tensor contribution in summed degree and the Tor functor contribution one degree lower. The three factors haveThe first two give $F_{23}=1+t^3$ and $T_{23}=2t^2+t^3+t^4+t^5$. Multiplying by the third givesConsequently the integral cohomology of a product of finite real projective spaces in this case isIf the intended coefficients were instead ${\mathbb{F}}_2$, the Künneth theorem over a field gives the dimension polynomialThus the mod-two groups in degrees zero through nine are respectivelyand all other degrees vanish.

The standard CW complex structure on Real projective space has one cell in each dimension from zero to three. Its integral cellular boundary is multiplication by $2$ in even positive degrees and zero in odd degrees. Thus the integral cellular cochain complex for ${\mathbb{RP}}^3$ isin degrees $0,1,2,3$. With coefficients ${\mathbb{F}}_2$, all its differentials vanish, so every one of these four cohomology groups is one-dimensional.

The lift-and-divide construction of the Bockstein homomorphism turns the integral differential $2$ into $1$ modulo $2$. Therefore $\beta :H^1\to H^2$ is an isomorphism, while the maps from degrees $0,2,3$ are zero. The Bockstein cohomology is consequentlyFor comparison, in the mod-two cohomology ring of real projective space ${\mathbb{F}}_2[t]/(t^4)$, $|t|=1$, this says $\beta (t)=t^2$, $\beta (t^2)=0$ and $\beta (t^3)=0$. The last two formulas also follow from the Bockstein derivation rule and the truncation $t^4=0$.

Put $F={\mathbb{F}}_2$. Under the usual bundle convention of a paracompact base, choose a fiber metric and let $D(E),S(E)$ be the disk bundle and sphere bundle of the rank-$n$ real vector bundle. A mod-two Thom class is a relative cohomology classwhose restriction to every fiber pair $(D^n,S^{n-1})$ is the nonzero generator. Equivalently it is a class in $H^n(E,E\setminus B;F)$, where $B$ is identified with the zero section. No orientation choices are needed over $F$.

The Thom isomorphism theorem asserts that this class exists and that, for every $r$, the mapis an isomorphism. The disk bundle retracts to the zero section $s:B\to D(E)$, so $H^r(D(E);F)\cong H^r(B;F)$. Define the mod-two Euler class $e_2(E)=s^{\ast }u\in H^n(B;F)$, where the pullback includes the passage from relative to absolute cohomology. Under the Thom identification, the map from relative to absolute cohomology sends $a$ to $a⌣e_2(E)$. Substituting into the pair's long exact sequence in cohomology gives the Gysin sequence of a sphere bundle:Here ${\pi }_{!}$ is the connecting homomorphism followed by ${\Phi }^{-1}$. Thus the sequence and the cup-product map have been derived from the Thom theorem, not just stated. The Thom/Gysin constructions are also discussed in Hatcher's Vector Bundles and K-Theory, §3.2.

The original PDF specifies the tautological bundle on ${\mathbb{RP}}^n$; the occurrences of ${\mathbb{R}}^n$ in this part of the TeX transcription have lost the projective-space symbol. Let $L\to {\mathbb{RP}}^n$ be this real line bundle. Its sphere bundle consists of pairs $(\ell ,v)$ with $v$ a unit vector in the line $\ell$, so $S(L)\cong S^n$, with projection the antipodal double covering. Write $t=e_2(L)$.

The additive mod-two groups, obtainable from the one-cell-in-each-dimension cellular chain complex, areFor $n\ge 1$, both the base and $S^n$ are connected, so ${\pi }^{\ast }:H^0({\mathbb{RP}}^n;F)\to H^0(S^n;F)$ is an isomorphism. Exactness makes the following ${\pi }_{!}$ zero and multiplication by $t$ injective from $H^0$ to $H^1$. For $1\le r<n$, the term $H^r(S^n;F)$ vanishes, so the same Gysin sequence makesinjective. Since these groups are one-dimensional, the maps are isomorphisms. Hence $1,t,\dots ,t^n$ are the respective nonzero generators, while $t^{n+1}=0$ by dimension. The mod-two cohomology ring of real projective space is thereforeFor $n=0$ this is simply $F$, with $t=0$.

For integral coefficients, the additive groups areIndeed, the cellular homology of real projective space has boundary multiplication by $2$ in even positive chain degrees and zero in odd degrees, and the integral cochain differential is its transpose. It remains to determine the products; the additive groups alone do not do so.

For $n\ge 2$, the coefficient sequence $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to F\to 0$ has a Bockstein homomorphism $\beta$. Because $H^1({\mathbb{RP}}^n;\mathbb{Z})=0$ and $H^2({\mathbb{RP}}^n;\mathbb{Z})=\mathbb{Z}/2$, exactness makes $a=\beta (t)$ the nonzero integral degree-two class. Reduction $\rho :H^2({\mathbb{RP}}^n;\mathbb{Z})\to H^2({\mathbb{RP}}^n;F)$ is injective: its kernel is the image of multiplication by $2$, which is zero. ThusMore generally, reduction is injective in each positive even degree, and its compatibility with the cup product gives $\rho (a^j)=t^{2j}$. Consequently $a^j$ is the nonzero generator whenever $2j\le n$. Also $2a=0$, and powers beyond the dimension vanish.

If $n=2m$ is even, these powers and the unit account for all groups, so the integral cohomology ring of real projective space isIf $n=2m+1$ is odd, add an integral top-dimensional orientation class $b$ of degree $2m+1$. Its reduction is the nonzero class $t^{2m+1}$. Every product $ab$ and $b^2$ is zero by dimension, and $b$ has infinite additive order. HenceThese formulas include $m=0$: the even space is a point with ring $\mathbb{Z}$, and the odd space is a circle with ring $\mathbb{Z}[b]/(b^2)$, $|b|=1$. The polynomial presentations are interpreted with the displayed grading and products; no extra positive-degree products are left unspecified.

Let $E$ have rank $r\ge 1$, let $p:\mathbb{P}(E)\to X$ be its projectivization of a real vector bundle, and let $\lambda \subset p^{\ast }E$ be its tautological bundle. Put $u=w_1(\lambda )\in H^1(\mathbb{P}(E);{\mathbb{F}}_2)$. The Projective bundle definition of Stiefel–Whitney classes starts from the isomorphismIn particular, $p^{\ast }$ is injective. Expand $u^r$ uniquely in this basis. The coefficients define the Stiefel–Whitney classes through the relationSet $w_0(E)=1$ and $w_i(E)=0$ for $i>r$; for a rank-zero vector bundle set $w(E)=1$. Uniqueness also proves naturality of the Stiefel–Whitney classes under a pullback vector bundle.

To prove the Whitney product formula for Stiefel–Whitney classes, apply the splitting principle for real vector bundles simultaneously to $E$ and $E^{\prime }$. Its injective pullback on mod-two cohomology follows by iterating the Projective bundle definition of Stiefel–Whitney classes. It suffices to compute for a split vector bundle $E={⨁}_{i=1}^r{L}_i$. Put $x_i=w_1(L_i)$.

On $\mathbb{P}(E)$, the coordinate projections of the tautological inclusion $\lambda \hookrightarrow p^{\ast }E$ give sections of the real line bundles ${\lambda }^{\ast }\otimes p^{\ast }{L}_i$ with no common zero. The mod-two Euler class of a real line bundle is its first Stiefel–Whitney class, and the First Stiefel–Whitney class of a tensor product of real line bundles gives $u+p^{\ast }{x}_i$. Multiplication of Euler classes under a direct sum, and vanishing of the Euler class of a vector bundle with a nowhere-zero section, therefore giveComparing with the unique monic relation defining the Stiefel–Whitney classes shows that $w_j(E)$ is the $j$th elementary symmetric polynomial in the $x_i$. Thus $w(E)={\prod }_i(1+x_i)$. A direct sum joins the two lists, so injectivity of the splitting pullback gives

For the immersion obstruction, use the notation $g=(f,f)$: the printed $f\times f$ here means the map from one copy of Real projective space into the product. Write $x=w_1({\gamma }_{\mathbb{R}})$ and let $z$ be the mod-two reduction of $c_1({\gamma }_{\mathbb{C}}^{\ast })$. The mod-two cohomology ring of real projective space and the cohomology ring of complex projective space giveNontriviality in degree two forces $f^{\ast }z=x^2$. Necessarily $k\ge 1$, since the degree-two cohomology of ${\mathbb{CP}}^0$ vanishes.

The standard tangent bundle identities for Real projective space and Complex projective space areFor the underlying real vector bundle of a complex line bundle, its total Stiefel–Whitney class is $1+c_1\mathrm{mod}2$. HenceIf $g$ is homotopic to an immersion $j$, its normal bundle $\nu$ has rank $4k-(2k+1)=2k-1$, andThe Whitney product formula for Stiefel–Whitney classes and homotopy invariance of cohomology yieldThe division is legitimate in the truncated cohomology ring, since $1+x$ is a unit. The degree-$2k$ component isIt must vanish because $2k>\mathrm{rank}\nu$, while $x^{2k}\ne 0$. Therefore $k$ must be odd.

Let $p:\mathbb{P}(E)\to X$ be the projectivization of a real vector bundle, let $L_E$ be its tautological bundle, and putThe mod-two projective bundle formula says that $H^{\ast }(\mathbb{P}(E);{\mathbb{F}}_2)$ is a free $H^{\ast }(X;{\mathbb{F}}_2)$-module on $1,u,\dots ,u^{d-1}$. The Projective bundle definition of Stiefel–Whitney classes is the unique relation

Apply the splitting principle for real vector bundles. After an injective pullback, writeIf $x_i=w_1(L_i)$, the projective-bundle relation factors assoThe line summands of $E\oplus E^{\prime }$ are the union of the two lists, henceComparing the degree-$k$ components proves the Whitney product formula for Stiefel–Whitney classesInjectivity of the splitting pullback returns the identity to $X$.

For real line bundles, the transition functions take values in $O(1)=\{\pm 1\}$. Tensor product multiplies these signs, while the identification $\{\pm 1\}\cong \mathbb{Z}/2$ turns multiplication into addition. The corresponding degree-one characteristic classes therefore satisfy the First Stiefel–Whitney class of a tensor product of real line bundles formulaEquivalently, this follows from the classification of real line bundles by $H^1(X;{\mathbb{F}}_2)$.

Now take $M={\mathbb{RP}}^n$ and $E=k{\gamma }_{\mathbb{R}}^{1,n+1}$. Sincea line in $E_x$ is the fixed line ${\gamma }_x$ tensored with a line in ${\mathbb{R}}^k$. Thus the projectivization of copies of the real tautological line bundle isLet $x$ and $v$ be the degree-one generators pulled back from the first and second factors. The mod-two cohomology ring of real projective space and the Künneth theorem giveThe tautological line $L_E$ is the tensor product of the two tautological lines, so $w_1(L_E)=x+v$. In the alternative generator $u=w_1(L_E)$, the same ring is