Let $R$ be a PID (such as $\mathbb{Z}$) and $R^m\overset{A}{\longrightarrow} R^n\overset{B}{\longrightarrow} R^o$ matrices with $BA=0$ and Smith normal forms $A=P\underbrace{\mathrm{diag}(a_1,\ldots,a_r,0,\ldots,0)}_DQ^{-1}$ and $B=P'\underbrace{\mathrm{diag}(b_1,\ldots,b_s,0,\ldots,0)}_{D'}Q'^{-1}$.
Topology for Computing by Zomorodian p.137-138 claims that $$\textstyle{\frac{\operatorname{Ker} B}{\operatorname{Im} A}\cong R^{n-r-s}\oplus\bigoplus_i\frac{R}{Ra_i}.}$$
How can one prove this claim? Any reference in the literature would also be welcome.
The free part is clearly true, because $\mathrm{rnk} \frac{\mathrm{Ker} B}{\mathrm{Im} A}=\mathrm{rnk}\,\mathrm{Ker} B - \mathrm{rnk}\,\mathrm{Im} A=$ $\mathrm{rnk}\,\mathrm{Ker} D' - \mathrm{rnk}\,\mathrm{Im} D=$ $n-s-r$. But I'm not sure about the torsion part...
Motivation: Basically, my question is, if we have a finite chain complex of free modules, given by matrices $\partial_1,\ldots,\partial_m$, can be compute the Smith normal form of each matrix separately $\partial_i=P_iD_iQ_i^{-1}$ to obtain homologies, or must this process be done simultaneously so that $Q_i=P_{i+1}$? The software packages that I know only compute SNF separately for each matrix.
Edit: Hmm, since $0=BA=P'D'Q'^{-1}PDQ^{-1}$ implies $0=D'Q'^{-1}PD$, the automorphism $Q'^{-1}P:R^n \to R^n$ must map $\langle e_1,\ldots,e_r\rangle$ into $\langle e_{s+1},\ldots,e_n\rangle$.
How can I make this question get more attention?