Index map

Remarks

The class does not depend on the choice of $n$, $u$ in its class, or the lift $v$, and $\partial$ is a group homomorphism into K0 of $I$. A lift $v$ exists because $\operatorname{diag}(u, u^*)$ lies in the connected component of $1$ in the unitary group, and unitaries in that component lift along surjections. If $u$ lifts to a partial isometry $w \in M_n(A)$, then $\partial [u] = [1_n - w^* w] - [1_n - w w^*]$. Example: for $0 \to \mathcal{K} \to B(H) \to B(H)/\mathcal{K} \to 0$ and the image $u$ of a Fredholm partial isometry $w$, this gives $\partial[u] = \dim \ker w - \dim \ker w^*$ under $K_0(\mathcal{K}) \cong \Z$. Non-example of a nonzero value: if $u$ itself lifts to a unitary $w \in M_n(A)$, take $v = \operatorname{diag}(w, w^*)$; then $v \operatorname{diag}(1_n, 0) v^* = \operatorname{diag}(1_n, 0)$ and $\partial [u] = 0$. When $A$ is not unital, one applies the definition to the unitized sequence $0 \to I \to \widetilde{A} \to \widetilde{B} \to 0$.