bounded finite length complex euler identities

Set $Z_i=\ker (d_i)$ and $B_i=\mathrm{im}(d_{i+1})$. There are short exact sequences $0\to Z_i\to D_i\to B_{i-1}\to 0$ and $0\to B_i\to Z_i\to H_i(D)\to 0$, with maps induced by the differential and quotient. Since $D_i$ has finite length, all these modules do. Additivity gives ${\mathrm{\ell }}_R(D_i)={\mathrm{\ell }}_R(H_i(D))+{\mathrm{\ell }}_R(B_i)+{\mathrm{\ell }}_R(B_{i-1})$.

For any finite exact sequence $0\to E_0\to \cdots \to E_t\to 0$ of finite-length modules, set $J_j$ to be the image in $E_j$, so $J_0=J_{t+1}=0$ and $0\to J_j\to E_j\to J_{j+1}\to 0$ is exact. Hence ${\mathrm{\ell }}_R(E_j)={\mathrm{\ell }}_R(J_j)+{\mathrm{\ell }}_R(J_{j+1})$, and summation with alternating signs cancels every image length, yielding ${\sum }_j(-1{)}^j{\mathrm{\ell }}_R(E_j)=0$.

Choose integers $a\le b$ with $D_i=0$ outside $[a,b]$. Then $B_{a-1}=B_b=0$. In the alternating sum of the preceding equality, ${\mathrm{\ell }}_R(B_j)$ has coefficient $(-1{)}^j+(-1{)}^{j+1}=0$. Only ${\sum }_{i=a}^b(-1{)}^i{\mathrm{\ell }}_R(H_i(D))$ remains. This equals $\chi (D)$, including the zero complex and a complex with only one nonzero term.

The long exact sequence for $0\to A\to B\to C\to 0$ has successive blocks $H_i(A),H_i(B),H_i(C),H_{i-1}(A)$. Boundedness permits cutting it between zero endpoints, and all its terms have finite length by hypothesis. The alternating signs on each block can be taken as $(-1{)}^i,-(-1{)}^i,(-1{)}^i$: the next block starts with $(-1{)}^{i-1}$, the opposite of the previous block's last sign. The exact-sequence cancellation therefore gives $\chi (A)-\chi (B)+\chi (C)=0$.

The shift differential has the same kernels and images as $d_D$ in the corresponding degrees, so $H_i(D[1])=H_{i-1}(D)$. Reindexing the finite Euler sum gives $\chi (D[1])={\sum }_j(-1{)}^{j+1}{\mathrm{\ell }}_R(H_j(D))=-\chi (D)$. These establish all three assertions.