Dual conventions for the exterior Euler class

The presentation of $K_{\mathrm{v}\mathrm{b}}^0(X)$ is well defined: isomorphism classes of finite locally free modules form a set, since bundles can be encoded by open covers, finite ranks and transition matrices satisfying the overlap identities. Form the free abelian group on those classes and quotient by the subgroup generated by the stated exact-sequence relations. Tensoring an exact sequence of finite locally free modules with another such module preserves exactness, since locally all terms are free and the sequence splits. Consequently tensor product respects the relations in each variable and gives the stated ring multiplication; associativity, commutativity and the identity follow from the corresponding canonical tensor-product isomorphisms.

For each $i$, exterior multiplication gives a canonical pairing ${\mathrm{\Lambda }}^{r-i}E\otimes {\mathrm{\Lambda }}^{i}E⟶\det E\mathrm{.}$ It induces a map ${\mathrm{\Lambda }}^{r-i}E\to \mathcal{H}om({\mathrm{\Lambda }}^{i}E,\det E)$. On a trivializing open set, the increasing wedge indexed by a subset $J$ of size $r-i$ pairs to zero with every increasing $i$-wedge except the one indexed by the complementary subset, with which it pairs to $\pm e_1\wedge \cdots \wedge e_r$. Thus its matrix is a signed permutation matrix, including in characteristic two, and the induced map is an isomorphism. Because the pairing is defined by exterior multiplication, these local isomorphisms agree on overlaps.

Locally, $E^{\vee }$ has the dual basis $e_1^{\vee },\dots ,e_r^{\vee }$. The canonical map ${\mathrm{\Lambda }}^i{E}^{\vee }\to ({\mathrm{\Lambda }}^{i}E{)}^{\vee }$ sends ${\varphi }_1\wedge \cdots \wedge {\varphi }_i$ to the functional $v_1\wedge \cdots \wedge v_i\mapsto \det ({\varphi }_a(v_b))$. It sends each increasing dual wedge to the dual of the corresponding increasing wedge basis vector, hence is an isomorphism. Likewise the evaluation map $\det E\otimes ({\mathrm{\Lambda }}^{i}E{)}^{\vee }\to \mathcal{H}om({\mathrm{\Lambda }}^{i}E,\det E)$ is locally an isomorphism. These maps are canonical and therefore glue. Combining their inverses with step 1.2 gives the asserted canonical isomorphism. The same argument covers $i=0$, $i=r$, and $r=0$, using the empty wedge.

Taking classes in the ring and substituting $j=r-i$ gives $\begin{array}{rl}{\displaystyle {\lambda }_{-1}(E)}& {\displaystyle =\sum _{i=0}^r(-1{)}^i[{\mathrm{\Lambda }}^{i}E]=\sum _{j=0}^r(-1{)}^{r-j}[\det E][{\mathrm{\Lambda }}^j{E}^{\vee }]}\\ {\displaystyle }& {\displaystyle =(-1{)}^r[\det E]{\lambda }_{-1}(E^{\vee })\mathrm{.}}\end{array}$ The dual line $(\det E{)}^{\vee }$ has the property that the evaluation map $\det E\otimes (\det E{)}^{\vee }\to {\mathcal{O}}_X$ is an isomorphism locally and hence globally. Thus $[(\det E{)}^{\vee }]$ is the inverse of $[\det E]$. Multiplication by the unit $(-1{)}^r[\det E]$ is bijective, proving simultaneous vanishing. This establishes all claims.