## 12.21 Spectral sequences: exact couples

Definition 12.21.1. Let $\mathcal{A}$ be an abelian category.

An *exact couple* is a datum $(A, E, \alpha , f, g)$ where $A$, $E$ are objects of $\mathcal{A}$ and $\alpha $, $f$, $g$ are morphisms as in the following diagram

\[ \xymatrix{ A \ar[rr]_{\alpha } & & A \ar[ld]^ g \\ & E \ar[lu]^ f & } \]

with the property that the kernel of each arrow is the image of its predecessor. So $\mathop{\mathrm{Ker}}(\alpha ) = \mathop{\mathrm{Im}}(f)$, $\mathop{\mathrm{Ker}}(f) = \mathop{\mathrm{Im}}(g)$, and $\mathop{\mathrm{Ker}}(g) = \mathop{\mathrm{Im}}(\alpha )$.

A *morphism of exact couples* $t : (A, E, \alpha , f, g) \to (A', E', \alpha ', f', g')$ is given by morphisms $t_ A : A \to A'$ and $t_ E : E \to E'$ such that $\alpha ' \circ t_ A = t_ A \circ \alpha $, $f' \circ t_ E = t_ A \circ f$, and $g' \circ t_ A = t_ E \circ g$.

Lemma 12.21.2. Let $(A, E, \alpha , f, g)$ be an exact couple in an abelian category $\mathcal{A}$. Set

$d = g \circ f : E \to E$ so that $d \circ d = 0$,

$E' = \mathop{\mathrm{Ker}}(d)/\mathop{\mathrm{Im}}(d)$,

$A' = \mathop{\mathrm{Im}}(\alpha )$,

$\alpha ' : A' \to A'$ induced by $\alpha $,

$f' : E' \to A'$ induced by $f$,

$g' : A' \to E'$ induced by “$g \circ \alpha ^{-1}$”.

Then we have

$\mathop{\mathrm{Ker}}(d) = f^{-1}(\mathop{\mathrm{Ker}}(g)) = f^{-1}(\mathop{\mathrm{Im}}(\alpha ))$,

$\mathop{\mathrm{Im}}(d) = g(\mathop{\mathrm{Im}}(f)) = g(\mathop{\mathrm{Ker}}(\alpha ))$,

$(A', E', \alpha ', f', g')$ is an exact couple.

**Proof.**
Omitted.
$\square$

Hence it is clear that given an exact couple $(A, E, \alpha , f, g)$ we get a spectral sequence by setting $E_1 = E$, $d_1 = d$, $E_2 = E'$, $d_2 = d' = g' \circ f'$, $E_3 = E''$, $d_3 = d'' = g'' \circ f''$, and so on.

Definition 12.21.3. Let $\mathcal{A}$ be an abelian category. Let $(A, E, \alpha , f, g)$ be an exact couple. The *spectral sequence associated to the exact couple* is the spectral sequence $(E_ r, d_ r)_{r \geq 1}$ with $E_1 = E$, $d_1 = d$, $E_2 = E'$, $d_2 = d' = g' \circ f'$, $E_3 = E''$, $d_3 = d'' = g'' \circ f''$, and so on.

Lemma 12.21.4. Let $\mathcal{A}$ be an abelian category. Let $(A, E, \alpha , f, g)$ be an exact couple. Let $(E_ r, d_ r)_{r \geq 1}$ be the spectral sequence associated to the exact couple. In this case we have

\[ 0 = B_1 \subset \ldots \subset B_{r + 1} = g(\mathop{\mathrm{Ker}}(\alpha ^ r)) \subset \ldots \subset Z_{r + 1} = f^{-1}(\mathop{\mathrm{Im}}(\alpha ^ r)) \subset \ldots \subset Z_1 = E \]

and the map $d_{r + 1} : E_{r + 1} \to E_{r + 1}$ is described by the following rule: For any (test) object $T$ of $\mathcal{A}$ and any elements $x : T \to Z_{r + 1}$ and $y : T \to A$ such that $f \circ x = \alpha ^ r \circ y$ we have

\[ d_{r + 1} \circ \overline{x} = \overline{g \circ y} \]

where $\overline{x} : T \to E_{r + 1}$ is the induced morphism.

**Proof.**
Omitted.
$\square$

Note that in the situation of the lemma we obviously have

\[ B_\infty = g\left(\bigcup \nolimits _ r \mathop{\mathrm{Ker}}(\alpha ^ r)\right) \subset Z_\infty = f^{-1}\left(\bigcap \nolimits _ r \mathop{\mathrm{Im}}(\alpha ^ r)\right) \]

provided $\bigcup \mathop{\mathrm{Ker}}(\alpha ^ r)$ and $\bigcap \mathop{\mathrm{Im}}(\alpha ^ r)$ exist. This produces as limit $E_\infty = Z_\infty / B_\infty $, see Definition 12.20.2.

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