Step 1: Define $E_0$ and $d_0$. Set: $E_0^{p,q} = F_pC_{p+q} / F_{p-1}C_{p+q}$

The differential $d$ on $C$ induces $d_0$ on $E_0$ by passing to quotients. Since $d(F_pC) \subseteq F_pC$, this is well-defined.

Step 2: Compute $E_1$. By definition: $E_1^{p,q} = H_{p+q}(E_0^{p,\bullet}, d_0) = H_{p+q}(F_pC / F_{p-1}C)$

This is the homology of the $p$-th graded piece.

Step 3: Construct $d_1$. The differential $d_1: E_1^{p,q} \to E_1^{p-1,q}$ comes from the connecting homomorphism in the long exact sequence: $\cdots \to H_n(F_{p-1}C) \to H_n(F_pC) \to H_n(F_pC/F_{p-1}C) \xrightarrow{\partial} H_{n-1}(F_{p-1}C) \to \cdots$

Specifically, $d_1 = \partial: E_1^{p,q} \to E_1^{p-1,q}$ where we identify: $E_1^{p,q} = H_{p+q}(F_pC/F_{p-1}C)$

Step 4: Verify $d_1 \circ d_1 = 0$. This follows from the exactness of the long exact sequence: the composition of consecutive connecting homomorphisms is zero.

Step 5: Compute $E_2$. By definition: $E_2^{p,q} = \ker(d_1: E_1^{p,q} \to E_1^{p-1,q}) / \text{Im}(d_1: E_1^{p+1,q} \to E_1^{p,q})$

From the long exact sequence, this is related to the relative homology groups.

Step 6: Higher pages. For $r \geq 2$, define: $Z_r^{p,q} = \{z \in F_pC_{p+q} : dz \in F_{p-r}C_{p+q-1}\}$ $B_r^{p,q} = \{b \in F_pC_{p+q} : b = dz + b' \text{ where } z \in F_{p+r-1}C_{p+q+1}, b' \in F_{p-1}C_{p+q}\}$

Then: $E_r^{p,q} = (Z_r^{p,q} + F_{p-1}C_{p+q}) / (B_r^{p,q} + F_{p-1}C_{p+q})$

The differential $d_r: E_r^{p,q} \to E_r^{p-r,q+r-1}$ is induced by $d$ on $C$.

Step 7: Convergence. As $r \to \infty$: $E_\infty^{p,q} = (Z_\infty^{p,q} + F_{p-1}C_{p+q}) / (B_\infty^{p,q} + F_{p-1}C_{p+q})$

where: $Z_\infty^{p,q} = \{z \in F_pC_{p+q} : dz = 0\} = Z(F_pC)_{p+q} \cap F_pC_{p+q}$ $B_\infty^{p,q} = B(C)_{p+q} \cap F_pC_{p+q}$

Thus: $E_\infty^{p,q} \cong \frac{Z(F_pC)_{p+q}}{B(C)_{p+q} \cap F_pC_{p+q} + Z(F_{p-1}C)_{p+q}} = F_pH_{p+q}(C) / F_{p-1}H_{p+q}(C)$

This shows convergence to the associated graded of the filtered homology.