[Proof of Theorem \ref{theorem-trace-formula-again}] The proof proceeds in a number of steps. \medskip\noindent Step 1. {\it Let $j : \mathcal{U}\hookrightarrow X$ be an open immersion with complement $Y = X - \mathcal{U}$ and $i : Y \hookrightarrow X$. Then $T''(X, K) = T''(\mathcal{U}, j^{-1} K)+ T''(Y, i^{-1}K)$ and $T'(X, K) = T'(\mathcal{U}, j^{-1} K)+ T'(Y, i^{-1}K)$.} \medskip\noindent This is clear for $T'$. For $T''$ use the exact sequence $$ 0\to j_!j^{-1} K \to K \to i_* i^{-1} K \to 0 $$ to get a filtration on $K$. This gives rise to an object $\widetilde K \in DF(X, \Lambda)$ whose graded pieces are $j_!j^{-1}K$ and $i_*i^{-1}K$, both of which lie in $D_{ctf}(X, \Lambda)$. Then, by filtered derived abstract nonsense (INSERT REFERENCE), $R\Gamma_c(X_{\bar k}, K)\in DF_{perf}(\Lambda)$, and it comes equipped with $\pi_x^*$ in $DF_{perf}(\Lambda)$. By the discussion of traces on filtered complexes (INSERT REFERENCE) we get \begin{eqnarray*} \text{Tr}(\pi_X^* |_{R\Gamma_c(X_{\bar k}, K)}) & = & \text{Tr}(\pi_X^* |_{R\Gamma_c(X_{\bar k}, j_!j^{-1}K)}) + \text{Tr}(\pi_X^* |_{R\Gamma_c(X_{\bar k}, i_*i^{-1}K)}) \\ & = & T''(U, i^{-1}K) + T''(Y, i^{-1}K). \end{eqnarray*} \noindent Step 2. {\it The theorem holds if $\dim X\leq 0$. } \medskip\noindent Indeed, in that case $$ R\Gamma_c(X_{\bar k}, K) = R\Gamma(X_{\bar k}, K) = \Gamma(X_{\bar k}, K) = \bigoplus\nolimits_{\bar x\in X_{\bar k}} K_{\bar x} \leftarrow \pi_X*. $$ Since the fixed points of $\pi_X: X_{\bar k}\to X_{\bar k}$ are exactly the points $\bar x\in X_{\bar k}$ which lie over a $k$-rational point $x\in X(k)$ we get $$ \text{Tr}\big(\pi_X^*|_{R\Gamma_c(X_{\bar k}, K)}\big) = \sum\nolimits_{x\in X(k)}\text{Tr}(\pi_{\bar x}|_{K_{\bar x}}). $$ \medskip\noindent Step 3. {\it It suffices to prove the equality $T'(\mathcal{U}, \mathcal{F}) = T''(\mathcal{U}, \mathcal{F})$ in the case where \begin{itemize} \item $\mathcal{U}$ is a smooth irreducible affine curve over $k$, \item $\mathcal{U}(k) = \emptyset$, \item $K=\mathcal{F}$ is a finite locally constant sheaf of $\Lambda$-modules on $\mathcal{U}$ whose stalk(s) are finite projective $\Lambda$-modules, and \item $\Lambda$ is killed by a power of a prime $\ell$ and $\ell \in k^*$. \end{itemize} } \medskip\noindent Indeed, because of Step 2, we can throw out any finite set of points. But we have only finitely many rational points, so we may assume there are none\footnote{At this point, there should be an evil laugh in the background.}. We may assume that $\mathcal{U}$ is smooth irreducible and affine by passing to irreducible components and throwing away the bad points if necessary. The assumptions of $\mathcal{F}$ come from unwinding the definition of $D_{ctf}(X, \Lambda)$ and those on $\Lambda$ from considering its primary decomposition. \medskip\noindent For the remainder of the proof, we consider the situation $$ \xymatrix{ \mathcal{V} \ar[d]_f \ar[r] & Y \ar[d]^{\bar f} \\ \mathcal{U} \ar[r] & X } $$ where $\mathcal{U}$ is as above, $f$ is a finite \'etale Galois covering, $\mathcal{V}$ is connected and the horizontal arrows are projective completions. Denoting $G=\text{Aut}(\mathcal{V}|\mathcal{U})$, we also assume (as we may) that $f^{-1}\mathcal{F} =\underline M$ is constant, where the module $M = \Gamma(\mathcal{V}, f^{-1}\mathcal{F})$ is a $\Lambda[G]$-module which is finite and projective over $\Lambda$. This corresponds to the trivial monoid extension $$ 1\to G\to \Gamma = G \times \mathbf{N}\to \mathbf{N}\to 1. $$ In that context, using the reductions above, we need to show that $T''(\mathcal{U}, \mathcal{F}) = 0$. \medskip\noindent Step 4. {\it There is a natural action of $G$ on $f_*f^{-1}\mathcal{F}$ and the trace map $f_*f^{-1}\mathcal{F}\to \mathcal{F}$ defines an isomorphism} $$ (f_*f^{-1}\mathcal{F})\otimes_{\Lambda[G]} \Lambda = (f_*f^{-1}\mathcal{F})_G \cong \mathcal{F}. $$ \medskip\noindent To prove this, simply unwind everything at a geometric point. \medskip\noindent Step 5. {\it Let $A = \mathbf{Z}/\ell^n \mathbf{Z}$ with $n\gg 0$. Then $f_*f^{-1}\mathcal{F} \cong (f_*\underline A) \otimes_{\underline A} \underline M$ with diagonal $G$-action.} \medskip\noindent Step 6. {\it There is a canonical isomorphism $(f_*\underline A \otimes_{\underline A} \underline M) \otimes_{\Lambda[G]} \underline \Lambda \cong \mathcal{F}$. } \medskip\noindent In fact, this is a derived tensor product, because of the projectivity assumption on $\mathcal{F}$. \medskip\noindent Step 7. {\it There is a canonical isomorphism $$ R\Gamma_c(\mathcal{U}_{\bar k}, \mathcal{F}) = (R\Gamma_c(\mathcal{U}_{\bar k}, f_*A)\otimes_A^\mathbf{L} M)\otimes_{\Lambda[G]}^\mathbf{L} \Lambda, $$ compatible with the action of $\pi^*_\mathcal{U}$. } \medskip\noindent This comes from the universal coefficient theorem, i.e., the fact that $R\Gamma_c$ commutes with $\otimes^\mathbf{L}$, and the flatness of $\mathcal{F}$ as a $\Lambda$-module. \medskip\noindent We have \begin{eqnarray*} \text{Tr}( \pi_\mathcal{U}^* |_{R\Gamma_c(\mathcal{U}_{\bar k}, \mathcal{F})}) & = & {\sum_{g \in G}}' \text{Tr}_{\Lambda}^{Z_g} \left( (g, \pi_\mathcal{U}^*) |_{R\Gamma_c(\mathcal{U}_{\bar k}, f_*A)\otimes_A^\mathbf{L} M} \right) \\ & = & {\sum_{g\in G}}' \text{Tr}_A^{Z_g} ( (g, \pi_\mathcal{U}^*) |_{R\Gamma_c(\mathcal{U}_{\bar k}, f_*A)} ) \cdot \text{Tr}_\Lambda(g|_M) \end{eqnarray*} where $\Gamma$ acts on $R\Gamma_c(\mathcal{U}_{\bar k}, \mathcal{F})$ by $G$ and $(e, 1)$ acts via $\pi_\mathcal{U}^*$. So the monoidal extension is given by $\Gamma = G \times \mathbf{N} \to \mathbf{N}$, $\gamma \mapsto 1$. The first equality follows from Lemma \ref{lemma-trivial-trace} and the second from Lemma \ref{lemma-weak-trace}. \medskip\noindent Step 8. {\it It suffices to show that $\text{Tr}_A^{Z_g}((g, \pi_\mathcal{U}^*) |_{R\Gamma_c(\mathcal{U}_{\bar k}, f_*A)}) \in A$ maps to zero in $\Lambda$. } \medskip\noindent Recall that \begin{eqnarray*} \# Z_g \cdot \text{Tr}_A^{Z_g}((g, \pi_\mathcal{U}^*) |_{R\Gamma_c(\mathcal{U}_{\bar k}, f_*A)}) & = & \text{Tr}_A((g, \pi_\mathcal{U}^*) |_{R\Gamma_c(\mathcal{U}_{\bar k}, f_*A)})\\ & = & \text{Tr}_A((g^{-1}\pi_\mathcal{V})^* |_{R\Gamma_c(\mathcal{V}_{\bar k}, A)}). \end{eqnarray*} The first equality is Lemma \ref{lemma-gamma-z-gamma-trace}, the second is the Leray spectral sequence, using the finiteness of $f$ and the fact that we are only taking traces over $A$. Now since $A=\mathbf{Z}/\ell^n\mathbf{Z}$ with $n \gg 0$ and $\# Z_g = \ell^a$ for some (fixed) $a$, it suffices to show the following result. \medskip\noindent Step 9. {\it We have $\text{Tr}_A((g^{-1}\pi_\mathcal{V})^* |_{R\Gamma_c(\mathcal{V}, A)}) = 0$ in $A$.} \medskip\noindent By additivity again, we have \begin{eqnarray*} & \text{Tr}_A((g^{-1}\pi_\mathcal{V})^* |_{R\Gamma_c(\mathcal{V}_{\bar k} A)}) + \text{Tr}_A((g^{-1}\pi_\mathcal{V})^* |_{R\Gamma_c(Y-\mathcal {V})_{\bar k}, A)}) \\ & = \text{Tr}_A((g^{-1}\pi_Y)^* |_{R\Gamma(Y_{\bar k}, A)}) \end{eqnarray*} The latter trace is the number of fixed points of $g^{-1}\pi_Y$ on $Y$, by Weil's trace formula Theorem \ref{theorem-weil-trace-formula}. Moreover, by the 0-dimensional case already proven in step 2, $$ \text{Tr}_A((g^{-1}\pi_\mathcal{V})^*|_{R\Gamma_c(Y-\mathcal{V})_{\bar k}, A)}) $$ is the number of fixed points of $g^{-1}\pi_Y$ on $(Y-\mathcal{V})_{\bar k}$. Therefore, $$ \text{Tr}_A((g^{-1}\pi_\mathcal{V})^* |_{R\Gamma_c(\mathcal{V}_{\bar k}, A)}) $$ is the number of fixed points of $g^{-1}\pi_Y$ on $\mathcal{V}_{\bar k}$. But there are no such points: if $\bar y\in Y_{\bar k}$ is fixed under $g^{-1}\pi_Y$, then $\bar f(\bar y) \in X_{\bar k}$ is fixed under $\pi_X$. But $\mathcal{U}$ has no $k$-rational point, so we must have $\bar f(\bar y)\in (X-\mathcal{U})_{\bar k}$ and so $\bar y\notin \mathcal{V}_{\bar k}$, a contradiction. This finishes the proof.