[proofplan] The proof applies the additivity of trace to two canonical short exact sequences of finite-dimensional $\mathbb{Q}$-vector spaces that fit between the chain groups, cycles, boundaries, and homology. From the short exact sequence $0 \to B_i \to Z_i \to H_i \to 0$ (with the induced maps), additivity gives $\operatorname{tr}(f_i^Z) = \operatorname{tr}(f_i^B) + \operatorname{tr}(f_i^H)$. From the short exact sequence $0 \to Z_i \to C_i \xrightarrow{d_i} B_{i-1} \to 0$, additivity gives $\operatorname{tr}(f_i^C) = \operatorname{tr}(f_i^Z) + \operatorname{tr}(f_{i-1}^B)$. Substituting the first into the second expresses $\operatorname{tr}(f_i^C)$ in terms of $\operatorname{tr}(f_i^H)$ and two boundary traces $\operatorname{tr}(f_i^B), \operatorname{tr}(f_{i-1}^B)$; these boundary traces appear at consecutive indices and therefore telescope under alternating summation, leaving $\sum (-1)^i \operatorname{tr}(f_i^C) = \sum (-1)^i \operatorname{tr}(f_i^H)$. [/proofplan] [step:Set up the chain-level notation and the induced maps on $Z_i$, $B_i$, $H_i$] Fix a finite simplicial complex $K$ (so every $C_i(K; \mathbb{Q})$ is finite-dimensional and only finitely many are nonzero; this is necessary for the traces below to make sense and for the alternating sums to converge). Write \begin{align*} C_i &:= C_i(K; \mathbb{Q}), & Z_i &:= \ker(d_i: C_i \to C_{i-1}), \\ B_i &:= \operatorname{image}(d_{i+1}: C_{i+1} \to C_i), & H_i &:= Z_i / B_i. \end{align*} Each of $C_i, Z_i, B_i, H_i$ is a finite-dimensional $\mathbb{Q}$-vector space; $B_i \subseteq Z_i \subseteq C_i$. Because $f_\bullet$ is a chain map, $f_i \circ d_{i+1} = d_{i+1} \circ f_{i+1}$ and $d_i \circ f_i = f_{i-1} \circ d_i$. From these identities: \begin{itemize} \item $f_i(Z_i) \subseteq Z_i$: if $x \in Z_i$, then $d_i(f_i(x)) = f_{i-1}(d_i(x)) = f_{i-1}(0) = 0$, so $f_i(x) \in Z_i$. \item $f_i(B_i) \subseteq B_i$: if $x = d_{i+1}(y) \in B_i$ for some $y \in C_{i+1}$, then $f_i(x) = f_i(d_{i+1}(y)) = d_{i+1}(f_{i+1}(y)) \in B_i$. \end{itemize} Therefore $f_i$ restricts to maps $f_i^Z := f_i|_{Z_i}: Z_i \to Z_i$ and $f_i^B := f_i|_{B_i}: B_i \to B_i$, and descends to the induced map $f_i^H: H_i \to H_i$ on the quotient $H_i = Z_i/B_i$. We also write $f_i^C := f_i: C_i \to C_i$ for emphasis. [/step] [step:Apply trace additivity to $0 \to B_i \to Z_i \to H_i \to 0$ to relate $\operatorname{tr}(f_i^Z), \operatorname{tr}(f_i^B), \operatorname{tr}(f_i^H)$] Consider the short exact sequence of $\mathbb{Q}$-vector spaces \begin{align*} 0 \to B_i \hookrightarrow Z_i \twoheadrightarrow H_i \to 0, \end{align*} where the first arrow is the inclusion and the second is the quotient map. This sequence is intertwined with the self-maps $f_i^B, f_i^Z, f_i^H$ (i.e. the diagram commutes), by construction of $f_i^B$ and $f_i^H$. Apply the [Additivity of Trace](/theorems/1957) with $V = Z_i$, $W = B_i$, and $A = f_i^Z$. We must verify the hypotheses: $Z_i$ is a finite-dimensional vector space (checked above), $B_i \leq Z_i$ is a subspace, $f_i^Z(B_i) \subseteq B_i$ (checked in the previous step), the restriction to $B_i$ is $f_i^B$, and the induced map on $Z_i / B_i = H_i$ is $f_i^H$. The conclusion is \begin{align*} \operatorname{tr}(f_i^Z) = \operatorname{tr}(f_i^B) + \operatorname{tr}(f_i^H). \end{align*} [/step] [step:Apply trace additivity to $0 \to Z_i \to C_i \xrightarrow{d_i} B_{i-1} \to 0$ to relate $\operatorname{tr}(f_i^C), \operatorname{tr}(f_i^Z), \operatorname{tr}(f_{i-1}^B)$] The first isomorphism theorem for the map $d_i: C_i \to C_{i-1}$ identifies $C_i / \ker(d_i) \cong \operatorname{image}(d_i)$, i.e.\ $C_i / Z_i \cong B_{i-1}$. This gives the short exact sequence \begin{align*} 0 \to Z_i \hookrightarrow C_i \xrightarrow{\bar{d}_i} B_{i-1} \to 0, \end{align*} where $\bar{d}_i$ is the surjection induced by $d_i$. We check that $f_\bullet$ intertwines this sequence with the self-maps $f_i^Z, f_i^C, f_{i-1}^B$. The left square commutes because $f_i(Z_i) \subseteq Z_i$ (Step 1), so $f_i^C$ restricts to $f_i^Z$ on $Z_i$. For the right square, we must verify $\bar{d}_i \circ f_i^C = f_{i-1}^B \circ \bar{d}_i$, i.e.\ that the diagram \begin{align*} \begin{array}{ccc} C_i & \xrightarrow{f_i^C} & C_i \\ \downarrow{d_i} & & \downarrow{d_i} \\ B_{i-1} & \xrightarrow{f_{i-1}^B} & B_{i-1} \end{array} \end{align*} commutes. This is exactly the chain-map identity $d_i \circ f_i = f_{i-1} \circ d_i$ restricted to the image $B_{i-1} \subseteq C_{i-1}$. So the induced quotient map on $C_i / Z_i \cong B_{i-1}$ is $f_{i-1}^B$. Apply the [Additivity of Trace](/theorems/1957) with $V = C_i$, $W = Z_i$, and $A = f_i^C$. The hypotheses: $C_i$ is finite-dimensional, $Z_i \leq C_i$ is a subspace, $f_i^C(Z_i) \subseteq Z_i$ (checked), the restriction is $f_i^Z$, and the induced map on the quotient $C_i / Z_i \cong B_{i-1}$ is $f_{i-1}^B$. The conclusion is \begin{align*} \operatorname{tr}(f_i^C) = \operatorname{tr}(f_i^Z) + \operatorname{tr}(f_{i-1}^B). \end{align*} [guided] The short exact sequence $0 \to Z_i \to C_i \xrightarrow{d_i} B_{i-1} \to 0$ comes from the first isomorphism theorem applied to $d_i: C_i \to C_{i-1}$: the kernel is $Z_i$ and the image is $B_{i-1}$, so $C_i / Z_i \cong B_{i-1}$ via the map induced by $d_i$. This is the canonical way to "extract" the boundary image as a quotient, and it is the key structural identity that lets us relate the trace on $C_i$ to the traces on $Z_i$ and $B_{i-1}$. Why is the quotient map on $C_i / Z_i$ equal to $f_{i-1}^B$ (with index shifted down to $i-1$)? Because the isomorphism $C_i / Z_i \cong B_{i-1}$ is realised by $\bar{d}_i$, not by the identity; when we push $f_i^C$ through this identification, the chain-map commutativity $d_i \circ f_i = f_{i-1} \circ d_i$ turns the induced quotient map into $f_{i-1}^B$. The index drop from $i$ to $i-1$ is the feature that will produce the telescoping cancellation in the final step. Why apply additivity of trace to this sequence rather than directly to $Z_i \subseteq C_i$ thought of as a subspace? Because the quotient $C_i / Z_i$ needs a named target, and $B_{i-1}$ is the natural one (giving us access to the $B$-traces we have already seen in Step 2). Applying the [Additivity of Trace](/theorems/1957) with $V = C_i$, $W = Z_i$ delivers exactly the trace-equality we want. [/guided] [/step] [step:Substitute to express $\operatorname{tr}(f_i^C)$ in terms of $\operatorname{tr}(f_i^H)$ and two boundary traces] From Step 2, $\operatorname{tr}(f_i^Z) = \operatorname{tr}(f_i^B) + \operatorname{tr}(f_i^H)$. Substitute this into the identity of Step 3: \begin{align*} \operatorname{tr}(f_i^C) = \operatorname{tr}(f_i^Z) + \operatorname{tr}(f_{i-1}^B) = \operatorname{tr}(f_i^B) + \operatorname{tr}(f_i^H) + \operatorname{tr}(f_{i-1}^B). \end{align*} Rearranging, \begin{align*} \operatorname{tr}(f_i^H) = \operatorname{tr}(f_i^C) - \operatorname{tr}(f_i^B) - \operatorname{tr}(f_{i-1}^B). \end{align*} [/step] [step:Sum against the alternating sign and telescope the boundary traces] Multiply both sides by $(-1)^i$ and sum over $i \geq 0$: \begin{align*} \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_i^H) = \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_i^C) - \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_i^B) - \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_{i-1}^B). \end{align*} The last sum, after reindexing $j = i - 1$, is \begin{align*} \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_{i-1}^B) = \sum_{j \geq -1} (-1)^{j+1} \operatorname{tr}(f_j^B) = -\sum_{j \geq -1} (-1)^j \operatorname{tr}(f_j^B). \end{align*} By convention $B_{-1} = 0$ (there are no $-1$-chains contributing to boundaries), so $\operatorname{tr}(f_{-1}^B) = 0$ and the $j = -1$ term vanishes. Thus \begin{align*} \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_{i-1}^B) = -\sum_{j \geq 0} (-1)^j \operatorname{tr}(f_j^B) = -\sum_{i \geq 0} (-1)^i \operatorname{tr}(f_i^B). \end{align*} Substituting back, \begin{align*} \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_i^H) &= \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_i^C) - \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_i^B) - \Bigl(-\sum_{i \geq 0} (-1)^i \operatorname{tr}(f_i^B)\Bigr) \\ &= \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_i^C). \end{align*} The two boundary sums cancel exactly. This is the claimed equality \begin{align*} \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_i: C_i \to C_i) = \sum_{i \geq 0} (-1)^i \operatorname{tr}(f_*: H_i \to H_i). \end{align*} [guided] The telescoping is the crucial mechanism. In the rearranged identity \begin{align*} \operatorname{tr}(f_i^H) = \operatorname{tr}(f_i^C) - \operatorname{tr}(f_i^B) - \operatorname{tr}(f_{i-1}^B), \end{align*} two boundary traces appear on the right: $\operatorname{tr}(f_i^B)$ at the current index $i$, and $\operatorname{tr}(f_{i-1}^B)$ at the shifted index $i - 1$. When we form the alternating sum $\sum_i (-1)^i \operatorname{tr}(f_i^H)$, each boundary trace $\operatorname{tr}(f_m^B)$ appears twice: once as $\operatorname{tr}(f_i^B)$ with $i = m$ (sign $(-1)^m$, coefficient $-1$ from the identity), and once as $\operatorname{tr}(f_{i-1}^B)$ with $i = m + 1$ (sign $(-1)^{m+1}$, coefficient $-1$ from the identity). The two contributions to $\operatorname{tr}(f_m^B)$ in the sum are therefore $(-1)^m \cdot (-1) = -(-1)^m$ and $(-1)^{m+1} \cdot (-1) = (-1)^m$, which add to zero. This is a standard telescoping pattern: whenever an identity involves the same quantity at two consecutive indices, alternating summation causes the quantity to cancel out. The finiteness of $K$ ensures that only finitely many terms are nonzero, so there are no boundary terms "at infinity" — and the convention $B_{-1} = 0$ handles the bottom of the complex. Note what the cancellation does **not** do: it does not identify $\operatorname{tr}(f_i^C)$ with $\operatorname{tr}(f_i^H)$ term by term. The equality holds only in the alternating sum. Individual chain-level traces $\operatorname{tr}(f_i^C)$ have no direct homological meaning; what is invariant is the Euler-characteristic-weighted sum. [/guided] [/step]