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 \begin{itemize} \item there exist $a, b\in \mathbf{Z}$ independent of $n$ such that $K_n^i = 0$ for all $i\notin[a, b]$, and \item the rank of $K_n^i$ is independent of $n$. \end{itemize} Therefore, the module $K_\infty^i = \lim_n \{K_n^i, \varphi_n^i\}$ 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 $H^i\left(K_\infty^\bullet\right) = \lim_n H^i\left(K_n^\bullet\right)$ (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 {\it 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 $\pi_\infty^\bullet = \lim_n\pi_n^\bullet$ of $K_\infty^\bullet$. Then part {\it 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 \ref{theorem-C}.