Whitney sum formula for Stiefel–Whitney classes

Suppose first $m,n\ge 1$ and put $A=P(E)$, $C=P(F)$ inside $X=P(E\oplus F)$, and $U_2=X\setminus A$, $U_1=X\setminus C$. Both $A$ and $C$ are closed subbundles and they are disjoint, since a line contained in both $E_b$ and $F_b$ would lie in $E_b\cap F_b=0$. In a chart $U\subseteq B$ write a line as $[v:w]$ with $v\in {\mathbb{R}}^m$, $w\in {\mathbb{R}}^n$, not both zero; then $A{\mathrm{\mid }}_U=\{[v:0]\}$, $C{\mathrm{\mid }}_U=\{[0:w]\}$. The formula $H_t[v:w]=[v:(1-t)w]$ for $0\le t\le 1$ is well defined and independent of the local chart because it is induced by the canonical linear map $E\oplus F\to E\oplus F$, $(v,w)\mapsto (v,(1-t)w)$ on the complement of $C$; it is continuous there, fixes $A$ pointwise, and satisfies $H_0=\mathrm{id}$, $H_1(X\setminus C)\subseteq A$. Hence $U_1=X\setminus C$ deformation retracts onto $A$, and symmetrically $U_2=X\setminus A$ deformation retracts onto $C$. Both are open, and $U_1\cup U_2=X\setminus (A\cap C)=X$.

The class ${\omega }_E:={\sum }_{i=0}^m{p}^{\ast }{w}_i(E)⌣x^{m-i}$ restricts to zero on $A$, and ${\omega }_F:={\sum }_{j=0}^n{p}^{\ast }{w}_j(F)⌣x^{n-j}$ restricts to zero on $C$. Indeed $A=P(E)\subseteq X$ and [F6] gives $i_E^{\ast }x=x_E$ and $i_E^{\ast }{p}^{\ast }{w}_i(E)=p_E^{\ast }{w}_i(E)$, so the restricted expression is exactly the defining projective-bundle relation of $E$. The argument for $C$ is symmetric.

Each class lifts to the relative group of the corresponding complement. Since ${\omega }_E$ restricts to zero on $A$, exactness of the pair sequence [F2] exhibits ${\omega }_E$ as the image of a class ${\tilde{\omega }}_E\in H^m(X,A;{\mathbb{F}}_2)$. The inclusion of pairs $(X,A)\to (X,U_1)$ is an isomorphism on relative cohomology: by step 1.1 the inclusion $A\hookrightarrow U_1$ is a homotopy equivalence, so in the map of pair sequences [F2] the two vertical maps $H^{\ast }(X)\to H^{\ast }(X)$ and $H^{\ast }(U_1)\to H^{\ast }(A)$ are isomorphisms, and the five lemma (equivalently, the long exact sequences split into commuting exact pieces with two isomorphisms out of three) gives that $H^{\ast }(X,U_1)\to H^{\ast }(X,A)$ is an isomorphism; [F3] supplies the homotopy invariance of the restriction. Thus ${\omega }_E$ has a preimage ${\tilde{\omega }}_E\in H^m(X,U_1;{\mathbb{F}}_2)$. Symmetrically ${\omega }_F$ has a preimage ${\tilde{\omega }}_F\in H^n(X,U_2;{\mathbb{F}}_2)$.

The relative product vanishes. Both $U_1$ and $U_2$ are open in $X$ and open in their union $X$, so [F4] applies to $A=U_1$, $B=U_2$ and defines ${\tilde{\omega }}_E⌣{\tilde{\omega }}_F\in H^{m+n}(X,U_1\cup U_2;{\mathbb{F}}_2)=H^{m+n}(X,X;{\mathbb{F}}_2)=0.$ Under the relative-to-absolute map, which by the naturality clause of [F4] with $A^{\mathrm{\prime }}=B^{\mathrm{\prime }}=\mathrm{\varnothing }$ sends the product to the absolute cup product of the images, this class maps to ${\omega }_E⌣{\omega }_F\in H^{m+n}(X;{\mathbb{F}}_2)$. Hence ${\omega }_E{\omega }_F=0$.

Expand the vanishing product: ${\omega }_E{\omega }_F={\sum }_{k=0}^{m+n}({\sum }_{i+j=k}{w}_i(E)w_j(F))x^{m+n-k}=0,$ with $x^{m+n}$-coefficient $w_0(E)w_0(F)=1$, so this is a monic relation of degree $m+n$ for $x$ on $X=P(E\oplus F)$. By the uniqueness clause of [F5] it is the defining relation of $E\oplus F$, so $w_k(E\oplus F)={\sum }_{i+j=k}{w}_i(E)w_j(F)(0\le k\le m+n),$ and for $k>m+n$ both sides vanish by the rank conventions. Summing gives $w(E\oplus F)=w(E)w(F)$.

Degenerate ranks and trivial summands. If $m=0$ then $E=0_B$, $E\oplus F\cong F$, and $w(E)=1$, so the formula holds; symmetrically for $n=0$. Taking $F={\epsilon }_B^r$ trivial of rank $r$: its classifying map is the constant map, so $w({\epsilon }_B^r)=1$ by naturality [F6] and the rank conventions, and the formula gives $w(E\oplus {\epsilon }_B^r)=w(E)$. For $r=1$ this says the positive classes are unchanged by adding a trivial line. If $m=n=1$ then $X=P(E\oplus F)$ is a ${\mathbb{P}}^1$-bundle, $A$ and $C$ are two disjoint sections, and the argument reduces to the displayed computation with $k\le 2$. The empty base gives zero groups and the zero relation, and the formulas hold in the zero ring with unit $1$.

Axiom audit. The argument uses AC only through the projective-bundle theorem [F5] for $E\oplus F$; the pair sequences, the relative cup product and the deformation retraction of step 1.1 are choice-free, and the only geometry used is the canonical linear homotopy inside each fiber.