Herbrand Quotient

Statement (b) can be proved by applying long homology sequence. For consider the complexes of abelian groups: \begin{align} &A:\ldots\to A\xrightarrow f A\xrightarrow g A\xrightarrow f\ldots\\ &B:\ldots\to B\xrightarrow f B\xrightarrow g B\xrightarrow f\ldots\\ &A/B:\ldots\to A/B\xrightarrow f A/B\xrightarrow g A/B\xrightarrow f\ldots\\ \end{align} Then we get an exact sequence of complexes $$\{0\}\to B\to A\to A/B\to\{0\}$$ giving rise to the long homology sequence $$H_1(A/B)\xrightarrow{\delta_1}H_0(B)\xrightarrow{\kappa_0}H_0(A)\xrightarrow{\pi_0}H_0(A/B)\xrightarrow{\delta_0}H_1(B)\xrightarrow{\kappa_1}H_1(A)\xrightarrow{\pi_1}H_1(A/B)\tag 1$$ where we noted that $$H_n(A)=\begin{cases}A_g/A^f&n\equiv 0\pmod 2\\A_f/A^g&n\equiv 1\pmod 2\end{cases}$$ so that $q(A)=|H_1(A)|/|H_0(A)|$. The equation $q(A)=q(B)q(A/B)$ then follows from the exactness of $(1)$ for:$\newcommand\Ker{\operatorname{Ker}}\renewcommand\Im{\operatorname{Im}}$ \begin{align} q(B)(A/B) &=\frac{|H_1(B)|}{|H_0(B)|}\frac{|H_1(A/B)|}{|H_0(A/B)|}\\ &=\frac{|\Ker\kappa_1||\Im\kappa_1|}{|\Ker\kappa_0||\Im\kappa_0|} \frac{|\Ker\delta_1||\Im\delta_1|}{|\Ker\delta_0||\Im\delta_0|}\\ &=\frac{|\Im\kappa_1|}{|\Im\kappa_0|} \frac{|\Ker\delta_1|}{|\Ker\delta_0|}\\ &=\frac{|\Ker\pi_1|}{|\Ker\pi_0|} \frac{|\Im\pi_1|}{|\Im\pi_0|}\\ &=\frac{|H_1(A)|}{|H_0(A)|}\\ &=q(A) \end{align}