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The associated graded of filtered homology is a subquotient of chain level data

Statement

Put Zn=ker⁡dn, Dn=im⁡dn+1 and Pp=FpCn∩Zn. Then the induced homology filtration satisfies gr⁡pFHn(C)≅PpPp−1+(FpCn∩Dn). The isomorphism is induced by the cycle quotient Zn→Hn(C) and is natural for filtered chain maps.

Facts & Assumptions

Given: A filtered chain complex, an integer n, and an integer filtration index p.

[F1]

FpHn(C) is the image of the inclusion on homology (Induced filtration on homology).

[F2]

Coimage equals image, quotients descend, and the sum/intersection and modular identities hold (Spectral sequence subquotient and local lifting calculus).

Proof

technique · direct
1.1F1F2

The identity dndn+1=0 factors the boundary image Dn through Zn: composing with the epic map onto Dn gives zero, so cancellation gives the factorization. Cycles of FpC are Pp by the pullback property. Its boundaries map into Dn, so the image on homology in [F1] is precisely the image of Pp→Zn/Dn. By [F2] this is (Pp+Dn)/Dn.

2.1F2step 1.1

The kernel of Pp→(Pp+Dn)/(Pp−1+Dn) is Pp∩(Pp−1+Dn). Modularity [F2], using Pp−1⊆Pp, gives Pp−1+(Pp∩Dn)=Pp−1+(FpCn∩Dn). The map is epic since Dn is killed and Pp supplies the remaining summand.

3.1F2step 1.1step 2.1∎

Nested quotients [F2] identify (Pp+Dn)/(Pp−1+Dn) with FpHn(C)/Fp−1Hn(C). Coimage-to-image applied to the epic map in step 2.1 gives the stated formula. A filtered chain map carries each Pp and Dn into the corresponding target subobject; all arrows just used are uniquely induced by these restrictions, so their squares commute.

Source notes

Stacks §12.24, equations 12.24.5.1–12.24.5.2; the omitted intermediate quotient calculation is proved here.

Depends on

Dependency tree · 2 levels

Sources