Naturality, orientation sign, and Whitney product for Euler classes

Naturality. Choose the metric $h$ used for $D(E),S(E)$ and equip $f^{\ast }E$ with the pulled-back metric $h^{\mathrm{\prime }}$ defined by $\mathrm{\parallel }(b^{\mathrm{\prime }},v){\mathrm{\parallel }}_{h^{\mathrm{\prime }}}=\mathrm{\parallel }v{\mathrm{\parallel }}_h$. The canonical bundle map $\mathrm{\Phi }:f^{\ast }E\to E$, $(b^{\mathrm{\prime }},v)\mapsto v$, of [F4] preserves this norm exactly. By the disk-sphere definitions [F9], it therefore restricts to a continuous map of pairs $\mathrm{\Phi }:(D(f^{\ast }E),S(f^{\ast }E))\to (D(E),S(E))$ over $f$. By [F2], ${\mathrm{\Phi }}^{\ast }{u}_E$ is the normalized Thom class $u_{f^{\ast }E}$ for the pulled-back orientation. By [F4], the square formed by $\mathrm{\Phi }$ and the two zero sections commutes. Naturality of the pair sequence [F5] therefore gives a commuting square $f^{\ast }{s}^{\ast }{j}^{\ast }(u_E)=s^{\mathrm{\prime }\ast }{j}^{\mathrm{\prime }\ast }({\mathrm{\Phi }}^{\ast }{u}_E),$ where $j^{\mathrm{\prime }\ast }$ is the relative-to-absolute map for $f^{\ast }E$. Substituting ${\mathrm{\Phi }}^{\ast }{u}_E=u_{f^{\ast }E}$ and applying [F1] gives $e(f^{\ast }E)=s^{\mathrm{\prime }\ast }{j}^{\mathrm{\prime }\ast }{u}_{f^{\ast }E}=f^{\ast }{s}^{\ast }{j}^{\ast }{u}_E=f^{\ast }e(E)$.

Orientation sign. Suppose $R=\mathbb{Z}$, $m>0$, and $E$ carries the reversed orientation $-o$. By [F2] the normalized Thom class of the reversed orientation is $-u_E$. Substituting into the defining composite of [F1] and using linearity of $j^{\ast }$ and $s^{\ast }$ gives $e(E,-o)=s^{\ast }{j}^{\ast }(-u_E)=-s^{\ast }{j}^{\ast }(u_E)=-e(E,o)$. In characteristic two the two orientations give the same class, and the statement's integral clause is the one asserted.

Whitney product. Equip $E$ and $F$ with metrics and put $P=(D(E){\times }_{B}D(F),(S(E){\times }_{B}D(F))\cup (D(E){\times }_{B}S(F)))\mathrm{.}$ This is the disk-sphere pair for the maximum norm on $E\oplus F$, not the disk-sphere pair for the usual sum metric. The construction in [F3] uses the base-preserving radial homeomorphism from this maximum-norm pair to the sum-metric pair, fixes the zero section, and identifies the normalized Thom class of the ordered sum with ${\mathrm{pr}}_E^{\ast }{u}_E⌣{\mathrm{pr}}_F^{\ast }{u}_F\in H^{m+n}(P;R)\mathrm{.}$ Let $z:B\to P$ be $z(b)=(0_b,0_b)$ and let $j_P^{\ast }$ be the relative-to-absolute map for $P$. Naturality of the relative product and its compatibility with the relative-to-absolute maps in [F8] give $\begin{array}{rl}{\displaystyle e(E\oplus F)}& {\displaystyle =z^{\ast }{j}_P^{\ast }({\mathrm{pr}}_E^{\ast }{u}_E⌣{\mathrm{pr}}_F^{\ast }{u}_F)}\\ {\displaystyle }& {\displaystyle =(s_E^{\ast }{j}_E^{\ast }{u}_E)⌣(s_F^{\ast }{j}_F^{\ast }{u}_F)=e(E)e(F)\mathrm{.}}\end{array}$ For $m=0$ or $n=0$ the corresponding bundle is zero, its Thom class and Euler class are the unit by [F1], and the product formula reduces to the rank-zero unit.

Koszul sign. Interchanging the two ordered summands is the bundle isomorphism $\sigma :E\oplus F\to F\oplus E$ over the identity. On each fiber it is the block swap, which multiplies the ordered product orientation generator by $(-1{)}^{mn}$ by [F7]. The Thom swap formula [F3], followed by the relative-to-absolute map and the zero section, therefore gives $e(E\oplus F)=(-1{)}^{mn}e(F\oplus E)\mathrm{.}$ Applying step 1.3 in the two displayed orders yields $e(E)e(F)=(-1{)}^{mn}e(F)e(E)$, exactly as also required by graded commutativity [F6]. No unsigned equality of the two products is used.

Boundary cases. Rank zero has the stipulated unit orientation, so fiber normalization gives $u=1$, $s$ and $j$ are identities, so $e(0_B)=1$ and the product formula reads $e(0\oplus F)=1\cdot e(F)=e(F)$, the unit convention matching [F1] on these inputs. No reversed rank-zero orientation is an input to clause 2; an arbitrary cohomological generator in degree zero need not be the unit. For two rank-one bundles, $mn=1$ and the block swap reverses the ordered orientation, so step 2.1 gives the sign $-1$ exactly. The empty base carries the unique zero class on both sides, and the zero ring has its unit equal to its zero element, so the displayed identities hold. Pullback along the identity and along composites are the two ends of the naturality square of step 1.1. AC is used only through the Thom-class suppliers [F2] and [F3].