The Stacks project

Proof. We will use Derived Categories, Lemma 13.19.8 without further mention in this proof. Let $P^\bullet$ be a bounded complex of finite projective $\Lambda$-modules and let $\alpha : P^\bullet \to K$ be an isomorphism in $D(\Lambda )$. Then $\alpha ^{-1}\circ f\circ \alpha$ corresponds to a morphism of complexes $f^\bullet : P^\bullet \to P^\bullet$ well defined up to homotopy. Set

Given $P^\bullet$ and $\alpha$, this is independent of the choice of $f^\bullet$. Namely, any other choice is of the form $\tilde{f}^\bullet = f^\bullet + dh +hd$ for some $h^ i : P^ i \to P^{i-1}(i\in \mathbf{Z})$. But

and so $\sum _ i (-1)^ i \text{Tr} ((dh+hd)|_{P^ i})=0$. Furthermore, this is independent of the choice of $(P^\bullet , \alpha )$: suppose $(Q^\bullet , \beta )$ is another choice. The compositions

are representable by morphisms of complexes $\gamma _1^\bullet$ and $\gamma _2^\bullet$ respectively, such that $\gamma _1^\bullet \circ \gamma _2^\bullet$ is homotopic to the identity. Thus, the morphism of complexes $\gamma _2^\bullet \circ f^\bullet \circ \gamma _1^\bullet : Q^\bullet \to Q^\bullet$ represents the morphism $\beta ^{-1}\circ f\circ \beta$ in $D(\Lambda )$. Now

by the fact that $\gamma _1^\bullet \circ \gamma _2^\bullet$ is homotopic to the identity and the independence of the choice of $f^\bullet$ we saw above. $\square$