Naturality, normalization, and Whitney sum for Chern classes

Naturality of the class $x$: the projective bundle of $f^{\ast }E$ is canonically $f^{\ast }P(E)$ and the tautological line of $f^{\ast }E$ is the pullback of ${\gamma }_E$, so $x_{f^{\ast }E}=f^{\ast }{x}_E$ by naturality of the Euler class in the complex orientation.

Rank conventions: by [F1] one has $c_0=1$, $c_i=0$ above the rank, $c(0_B)=1$, and on a line $c_1(L)=e(L_{\mathbb{R}})$ with $c_i(L)=0$ for $i\ge 2$; this is assertion 2 and the first part of assertion 4.

If $m=0$ or $n=0$, the Whitney formula follows immediately from $c(0_B)=1$; hence assume $m,n\ge 1$. Let $X=P(E\oplus F)$ and put $U_1=X\setminus P(E)$, $U_2=X\setminus P(F)$; the subsets $P(E),P(F)\subseteq X$ are disjoint and $U_1\cup U_2=X$. The map that sends a point of $U_2$, a line not contained in $F$, to the line spanned by its $E$-component is a deformation retraction of $U_2$ onto $P(E)$, and symmetrically $U_1$ deformation retracts onto $P(F)$.

Applying $f^{\ast }$ to the defining relation of $E$ and using step 1.1 expresses $x_{f^{\ast }E}^m$ as a monic relation with coefficients $f^{\ast }{c}_i(E)$; by uniqueness in [F2] these are the coefficients of $f^{\ast }E$, so $c_i(f^{\ast }E)=f^{\ast }{c}_i(E)$ for all $i$, which is assertion 1.

The classes ${\omega }_1:={\sum }_{j=0}^m(-1{)}^j{c}_j(E)x^{m-j}$ and ${\omega }_2:={\sum }_{j=0}^n(-1{)}^j{c}_j(F)x^{n-j}$, where $x=x_{E\oplus F}$ and $c_j$ of a summand means the pullback of $c_j$ to $X$, satisfy: on $P(E)\subseteq X$ the class $x$ restricts to $x_E$ and ${\omega }_1$ restricts to the defining relation of $E$, hence to $0$. Since $P(E)\hookrightarrow U_2$ is a deformation retract, [F6] shows that the restriction of ${\omega }_1$ to $U_2$ is also zero. The long exact sequence of $(X,U_2)$ therefore gives a relative lift of ${\omega }_1$ in $H^{\ast }(X,U_2)$. Symmetrically ${\omega }_2$ has a lift in $H^{\ast }(X,U_1)$.

Changing coefficients in step 2.1 gives the same identity over $R=\mathbb{Z}$ and over every ${\mathbb{F}}_p$, and the rank cutoff is preserved because $f^{\ast }E$ has the same rank.

The relative cup product [F5] with $A=U_2$, $B=U_1$ maps ${\omega }_1\otimes {\omega }_2$ into $H^{\ast }(X,U_1\cup U_2)=H^{\ast }(X,X)=0$ because $U_1\cup U_2=X$. Hence the product of the two classes is zero in $H^{\ast }(X)$: ${\sum }_{j=0}^{m+n}(-1{)}^j({\sum }_{r+s=j}{c}_r(E)c_s(F))x^{m+n-j}=0$.

Comparison with the defining relation of $E\oplus F$: by [F2] the unique monic relation of $E\oplus F$ is $x^{m+n}+{\sum }_{j=1}^{m+n}(-1{)}^j{c}_j(E\oplus F)x^{m+n-j}=0$. Subtracting the relation of step 3.2 from it gives a polynomial relation of degree less than $m+n$ in $x$, so by the basis property of [F2] all its coefficients vanish. Hence $c_j(E\oplus F)={\sum }_{r+s=j}{c}_r(E)c_s(F)$ for all $j$, which is assertion 3.

Stabilization. For the trivial complex line ${\epsilon }^1$ the identity section is nowhere zero, so $e({\epsilon }_{\mathbb{R}}^1)=0$ by [F7] and $c({\epsilon }^1)=1$ by assertion 2; the trivial bundle ${\epsilon }^r$ is a sum of trivial lines, so assertion 3 gives $c({\epsilon }^r)=1$ and hence $c(E\oplus {\epsilon }^r)=c(E)$ for all $r\ge 0$.

Boundary cases. The cases $m=0$ and $n=0$ were discharged in step 1.3. In the rank-one case $m=1$ the class ${\omega }_1=x-c_1(E)$ and assertion 2 is [F1]. The empty base is excluded by the path-connected hypothesis, and a disconnected base is treated componentwise. AC is used only through [A1] in the bundle, relative-cohomology and Euler suppliers.