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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.

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.

The zero section $C=\{t=0\}\subset M$ is a circle. Its normal bundle is the real line bundle obtained fromso its clutching map reverses sign once around $C$. Its mod-two Euler class of a real line bundle, equivalently its first Stiefel–Whitney class, therefore satisfiesIf $M$ were orientable, the splittingand the orientation of the circle would orient ${\nu }_C$, forcing this mod-two Euler number to vanish. This contradiction proves that the Möbius band is nonorientable.

Over Real projective space ${\mathbb{RP}}^n$, the real tautological line bundle isOn the standard chart $U_i=\{[x]:x_i\ne 0\}$, let $s_i([x])=x/x_i$. A local trivialization sends $([x],t)$ to $([x],t{s}_i([x]))$, with inverse $([x],v)\mapsto ([x],v_i)$. These maps prove that this is a real line bundle. Its unit sphere bundle is the antipodal map quotient $S^n\to {\mathbb{RP}}^n$. Its mod-two Euler class of a real line bundle is the degree-one generator of the mod-two cohomology ring of real projective space.