Editing Spectral sequence (section)

==== Long exact sequences ====

Using the spectral sequence of a filtered complex, we can derive the existence of [[long exact sequence]]s. Choose a short exact sequence of cochain complexes 0 → {{var|A}}{{i sup|•}} → {{var|B}}{{i sup|•}} → {{var|C}}{{i sup|•}} → 0, and call the first map {{var|f}}{{i sup|•}} : {{var|A}}{{i sup|•}} → {{var|B}}{{i sup|•}}. We get natural maps of homology objects ''H<sup>n</sup>''({{var|A}}{{i sup|•}}) → ''H<sup>n</sup>''({{var|B}}{{i sup|•}}) → ''H<sup>n</sup>''({{var|C}}{{i sup|•}}), and we know that this is exact in the middle. We will use the spectral sequence of a filtered complex to find the connecting homomorphism and to prove that the resulting sequence is exact.To start, we filter {{mvar|B}}{{i sup|•}}:

:<math>F^0 B^n = B^n</math>
:<math>F^1 B^n = A^n</math>
:<math>F^2 B^n = 0</math>

This gives:

:<math>E^{p,q}_0
= \frac{F^p B^{p+q}}{F^{p+1} B^{p+q}} = \begin{cases}
0 & \text{if } p < 0 \text{ or } p > 1 \\
C^q & \text{if } p = 0 \\
A^{q+1} & \text{if } p = 1 \end{cases}</math>
:<math>E^{p,q}_1
= \begin{cases}
0 & \text{if } p < 0 \text{ or } p > 1 \\
H^q(C^\bull) & \text{if } p = 0 \\
H^{q+1}(A^\bull) & \text{if } p = 1 \end{cases}</math>

The differential has bidegree (1, 0), so ''d''<sub>0,''q''</sub> : ''H<sup>q</sup>''({{var|C}}{{i sup|•}}) → ''H''<sup>''q''+1</sup>({{var|A}}{{i sup|•}}). These are the connecting homomorphisms from the [[snake lemma]], and together with the maps {{mvar|A}}{{i sup|•}} → {{mvar|B}}{{i sup|•}} → {{mvar|C}}{{i sup|•}}, they give a sequence:

:<math>\cdots\rightarrow H^q(B^\bull) \rightarrow H^q(C^\bull) \rightarrow H^{q+1}(A^\bull) \rightarrow H^{q+1}(B^\bull) \rightarrow\cdots</math>

It remains to show that this sequence is exact at the ''A'' and ''C'' spots.  Notice that this spectral sequence degenerates at the ''E''<sub>2</sub> term because the differentials have bidegree (2, &minus;1). Consequently, the ''E''<sub>2</sub> term is the same as the ''E''<sub>∞</sub> term:

:<math>E^{p,q}_2
\cong \text{gr}_p H^{p+q}(B^\bull)
= \begin{cases}
0 & \text{if } p < 0 \text{ or } p > 1 \\
H^q(B^\bull)/H^q(A^\bull) & \text{if } p = 0 \\
\text{im } H^{q+1}f^\bull : H^{q+1}(A^\bull) \rightarrow H^{q+1}(B^\bull) &\text{if } p = 1 \end{cases}</math>

But we also have a direct description of the ''E''<sub>2</sub> term as the homology of the ''E''<sub>1</sub> term. These two descriptions must be isomorphic:

:<math> H^q(B^\bull)/H^q(A^\bull) \cong \ker d^1_{0,q} : H^q(C^\bull) \rightarrow H^{q+1}(A^\bull)</math>
:<math> \text{im } H^{q+1}f^\bull : H^{q+1}(A^\bull) \rightarrow H^{q+1}(B^\bull) \cong H^{q+1}(A^\bull) / (\mbox{im } d^1_{0,q} : H^q(C^\bull) \rightarrow H^{q+1}(A^\bull))</math>

The former gives exactness at the ''C'' spot, and the latter gives exactness at the ''A'' spot.