Exact couple

Remarks

The map $d = j \circ k \colon E \to E$ satisfies $d \circ d = j (k j) k = 0$, by clause 2. The derived couple has $D' = i(D)$ and $E' = \ker d / \im d$, with $i'$ the restriction of $i$, $j'(i(x)) = [j(x)]$ and $k'([e]) = k(e)$; it is again an exact couple. Iterating gives groups $E^r$ with differentials $d^r$ such that $E^{r+1}$ is the homology of $(E^r, d^r)$, that is, a spectral sequence. In practice $D$ and $E$ are bigraded, and the bidegrees of $i, j, k$ fix the bidegree of each $d^r$.

Example: let $C$ be a chain complex of free abelian groups and $p$ a prime. The short exact sequence $0 \to C \xrightarrow{p} C \to C / pC \to 0$ gives a long exact sequence in homology. It is an exact couple with $D = H_*(C)$, $E = H_*(C/pC)$, $i$ multiplication by $p$, $j$ reduction mod $p$ and $k$ the connecting map.

Non-example: take $D = E = \Z$ with $i = 0$, $j = \id$ and $k = 0$. Clause 1 holds, since $\ker j = 0 = \im i$, and clause 2 holds, since $\ker k = \Z = \im j$. Clause 3 fails, since $\ker i = \Z$ while $\im k = 0$.

Exact couples were introduced by William S. Massey.