Naturality and uniqueness of Thom classes

Equip $f^{\ast }\xi$ with the pulled-back metric $f^{\ast }h$, defined by $\mathrm{\parallel }(b^{\mathrm{\prime }},v){\mathrm{\parallel }}_{f^{\ast }h}=\mathrm{\parallel }v{\mathrm{\parallel }}_h$. The canonical bundle map of [F2] preserves this norm exactly, so the defining inequalities and equalities restrict it to a continuous map of pairs $(D(f^{\ast }\xi ),S(f^{\ast }\xi ))⟶(D(\xi ),S(\xi ))\mathrm{.}$ Pull $u_{\xi }$ back along this pair map. On the fiber over $b^{\mathrm{\prime }}$, functoriality identifies its restriction with the restriction of $u_{\xi }$ on the fiber over $f(b^{\mathrm{\prime }})$. Because the pullback orientation was specified to preserve that generator, the pulled-back class is normalized. Uniqueness in [F1] gives $f^{\ast }{u}_{\xi }=u_{f^{\ast }\xi }$.

If $u$ and $v$ are any normalized Thom classes for one supplied orientation, the uniqueness clause of [F1] gives $u=v$. Equivalently, the Thom isomorphism writes $u-v=a⌣u_{\xi }$ with $a\in H^0(B;R)$, and fiber normalization forces $a$ to vanish on every component.

Over $\mathbb{Z}$, replacing every orientation generator $o_b$ by $-o_b$ makes $-u_{\xi }$ restrict to the new generator on every fiber. It is therefore normalized for the reversed orientation, and step 1.2 makes it that orientation's unique Thom class.

For the empty base the unique class pulls back to itself; in rank zero the unit orientation reverses to $-1$ and the same calculation applies. Point bases, identity maps, zero classes, and both pullback-square composites are literal instances of step 1.1. In characteristic two the two signs coincide, but the asserted reversal clause is integral. AC is used exactly through [A1] in [F1], and pulling back the supplied class makes no selection.