Eliminating F-points gives the exact coarse time operator

For any right-hand side, solve the fine equations in increasing index: the first unknown is determined by its identity diagonal block, and each later unknown is uniquely $u_i=r_i+{\mathrm{\Phi }}_i{u}_{i-1}$. This proves $A_t$ is invertible. Holding all C-values fixed, the F-equations in each interval between consecutive C-points have the same forward-substitution form with identity diagonal blocks, so they uniquely determine the F-values; hence $A_{FF}$ is invertible.

In the reordered block system, the F rows give $u_F=A_{FF}^{-1}(b_F-A_{FC}{u}_C)$. Substituting this expression in the C rows gives $(A_{CC}-A_{CF}{A}_{FF}^{-1}{A}_{FC})u_C=b_C-A_{CF}{A}_{FF}^{-1}{b}_F$. Conversely, any solution $u_C$ of this coarse system, together with the displayed formula for $u_F$, satisfies both the F and C rows. Thus elimination preserves exactly the solution set.

Repeated substitution in $u_i={\mathrm{\Phi }}_i{u}_{i-1}+g_i$, for $i=(j-1)m+1,\dots ,jm$, gives $u_{jm}=({\mathrm{\Phi }}_{jm}\cdots {\mathrm{\Phi }}_{(j-1)m+1})u_{(j-1)m}+{\sum }_{s=1}^m({\mathrm{\Phi }}_{jm}\cdots {\mathrm{\Phi }}_{(j-1)m+s+1})g_{(j-1)m+s}$. For $s=m$ the product is empty and equals $I$. This is exactly the coarse recurrence in the Statement; the coefficient product reduces to ${\mathrm{\Phi }}^m$ when every ${\mathrm{\Phi }}_i=\mathrm{\Phi }$.