Proof.
Since \mathbf{Z}/\ell ^ n\mathbf{Z} is a local ring and K_ n is perfect, each K_ n can be represented by a finite complex K_ n^\bullet of finite free \mathbf{Z}/\ell ^ n \mathbf{Z}-modules such that the map K_ n^ p \to K_ n^{p+1} has image contained in \ell K_ n^{p+1}. It is a fact that such a complex is unique up to isomorphism. Moreover \pi _ n can be represented by a morphism of complexes \pi _ n^\bullet : K_ n^\bullet \to K_ n^\bullet (which is unique up to homotopy). By the same token the isomorphism \varphi _ n : K_{n+1} \otimes _{\mathbf{Z}/\ell ^{n+1}\mathbf{Z}}^{\mathbf{L}} \mathbf{Z}/\ell ^ n\mathbf{Z}\to K_ n is represented by a map of complexes
\varphi _ n^\bullet : K_{n+1}^\bullet \otimes _{\mathbf{Z}/\ell ^{n+1}\mathbf{Z}} \mathbf{Z}/\ell ^ n\mathbf{Z} \to K_ n^\bullet .
In fact, \varphi _ n^\bullet is an isomorphism of complexes, thus we see that
there exist a, b\in \mathbf{Z} independent of n such that K_ n^ i = 0 for all i\notin [a, b], and
the rank of K_ n^ i is independent of n.
Therefore, the module K_\infty ^ i = \mathop{\mathrm{lim}}\nolimits _ n \{ K_ n^ i, \varphi _ n^ i\} is a finite free \mathbf{Z}_\ell -module and K_\infty ^\bullet is a finite complex of finite free \mathbf{Z}_\ell -modules. By induction on the number of nonzero terms, one can prove that H^ i\left(K_\infty ^\bullet \right) = \mathop{\mathrm{lim}}\nolimits _ n H^ i\left(K_ n^\bullet \right) (this is not true for unbounded complexes). We conclude that H_\infty ^ i = H^ i\left(K_\infty ^\bullet \right) is a finite \mathbf{Z}_\ell -module. This proves ii. To prove the remainder of the lemma, we need to overcome the possible noncommutativity of the diagrams
\xymatrix{ {K_{n+1}^\bullet } \ar[d]_{\pi _{n+1}^\bullet } \ar[r]^{\varphi _ n^\bullet } & {K_ n^\bullet } \ar[d]^{\pi _ n^\bullet } \\ {K_{n+1}^\bullet } \ar[r]_{\varphi _ n^\bullet } & {K_ n^\bullet .} }
However, this diagram does commute in the derived category, hence it commutes up to homotopy. We inductively replace \pi _ n^\bullet for n\geq 2 by homotopic maps of complexes making these diagrams commute. Namely, if h^ i : K_{n+1}^ i \to K_ n^{i-1} is a homotopy, i.e.,
\pi _ n^\bullet \circ \varphi _ n^\bullet - \varphi _ n^\bullet \circ \pi _{n + 1}^\bullet = dh + hd,
then we choose \tilde h^ i : K_{n+1}^ i\to K_{n+1}^{i-1} lifting h^ i. This is possible because K_{n+1}^ i free and K_{n+1}^{i-1}\to K_ n^{i-1} is surjective. Then replace \pi _ n^\bullet by \tilde\pi _ n^\bullet defined by
\tilde\pi _{n+1}^\bullet = \pi _{n+1}^\bullet + d\tilde h+\tilde hd.
With this choice of \{ \pi _ n^\bullet \} , the above diagrams commute, and the maps fit together to define an endomorphism \pi _\infty ^\bullet = \mathop{\mathrm{lim}}\nolimits _ n\pi _ n^\bullet of K_\infty ^\bullet . Then part i is clear: the elements t_ n = \sum (-1)^ i \text{Tr}\left(\pi _ n^ i |_{K_ n^ i}\right) fit into an element t_\infty of \mathbf{Z}_\ell . Moreover
\begin{align*} t_\infty & = \sum (-1)^ i \text{Tr}_{\mathbf{Z}_\ell }(\pi _\infty ^ i |_{K_\infty ^ i}) \\ & = \sum (-1)^ i \text{Tr}_{\mathbf{Q}_\ell }( \pi _\infty ^ i |_{K_\infty ^ i \otimes _{\mathbf{Z}_\ell }\mathbf{Q}_\ell }) \\ & = \sum (-1)^ i \text{Tr}( \pi _\infty |_{H^ i(K_\infty ^\bullet \otimes \mathbf{Q}_\ell )}) \end{align*}
where the last equality follows from the fact that \mathbf{Q}_\ell is a field, so the complex K_\infty ^\bullet \otimes \mathbf{Q}_\ell is quasi-isomorphic to its cohomology H^ i(K_\infty ^\bullet \otimes \mathbf{Q}_\ell ). The latter is also equal to H^ i(K_\infty ^\bullet )\otimes _{\mathbf{Z}}\mathbf{Q}_\ell = H_\infty ^ i \otimes \mathbf{Q}_\ell , which finishes the proof of the lemma, and also that of Theorem 64.20.5.
\square
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