I am currently reading spectral sequences. I got two approaches one is the classical approach and another is the exact couple approach. Before asks the question I want to write my notation and conventions.
$\textbf{Definition :---}$ A homological spectral sequence is
- A collections of bigraded modules $\big\{E_{p,q}^r|(p,q)\in \Bbb Z\times \Bbb Z\big\}$ for each $r=1,2,...$
- A collection of homomorphisms $\big\{d^r_{p,q}:E_{p,q}^r\to E_{p-r,q+r-1}^r|(p,q)\in \Bbb Z\times \Bbb Z\big\}$ for each $r=1,2,...$ such that $d_{p,q}^r\circ d_{p-r,p+r-1}^r=0$
- $$E^{r+1}_{p,q}\simeq H^r_{p,q}:=\frac{\text{ker}\big(d_{p,q}^r:E_{p,q}^r\to E_{p-r,q+r-1}^r\big)}{\text{Im}\big(d_{p+r,q-r+1}^r:E_{p+r,q-r+1}^r\to E_{p,q}^r\big)}.$$
$\textbf{Definition :---}$ Let $\big\{C_n,\partial _{n}:C_{n}\to C_{n-1}|\partial_{n-1}\circ\partial_{n}=0\big\}_{n\in \Bbb Z}$ be a chain complex. We say it is filtered if
- There is a collection of submodules $$\cdots\subseteq F_{-1}C_n\subseteq F_0C_n\subseteq F_1C_n\subseteq\cdots\subseteq C_n\text{ for each }n\in\Bbb Z.$$
- $$\partial_n\big(F_pC_n\big)\subseteq F_pC_{n-1}\text{ for each }p\in\Bbb Z\text{ and for each }n\in \Bbb Z.$$
$\textbf{1. Theorem :---}$ A filtered chain complex $\{C_\bullet,\partial_\bullet\}$ determines a spectral sequence $\big\{E_{\bullet\bullet}^r,d^r\big\}_{r\in\Bbb N}$.
$\textbf{Outline of proof:---}$
For arbitrary $r$ define $Z^r_{s,t}:=\big\{c\in F_sC_{s+t}\big|\partial c\in F_{s-r}C_{s+t-1}\big\}$ and $Z^\infty_{s,t}:=\big\{c\in F_sC_{s+t}\big|\partial c=0\big\}$ for $(s,t)\in \Bbb Z\times\Bbb Z$.
Define $$E^r_{s,t}:=\frac{Z^r_{s,t}}{Z^{r-1}_{s-1,t+1}+\partial Z^{r-1}_{s+r-1,t-r+2}}.$$ The map $\partial$ sends $Z^r_{s,t}$ to $Z^r_{s-r,t+r-1}$ and $Z^{r-1}_{s-1,t+1}+\partial Z^{r-1}_{s+r-1,t-r+2}$ to $\partial Z^{r-1}_{s-1,t+1}$. Therefore it induces a homomorphism $d^r_{s,t}:E^r_{s,t}\to E^r_{s-r,t+r-1}$. So $E^r:=\big\{E_{s,t}^r:(s,t)\in \Bbb Z\times\Bbb Z\big\}$ is a bigraded module with $d^r$ is a differential of bidegree $(-r,r-1)$. $\blacksquare$
$\textbf{Definition:---}$ An exact couple is a $5$-tuple $(D,E,\alpha,\beta,\gamma)$ where $D,E$ are bigraded modules and $\alpha:D\to D,\beta:D\to E,\gamma:E\to D$ are bigraded maps such that $\text{ker}\ \alpha=\text{im}\ \gamma$, $\text{im}\ \alpha=\text{ker}\ \beta$, $\text{im}\ \beta=\text{ ker}\ \gamma$. More explicitly for $(p,q)\in \Bbb Z\times \Bbb Z$ we have,$$\text{im}\big(E_{p,q}\xrightarrow{\gamma_{p,q}}D_{p+a'',q+b''}\big)=\text{ker}\big(D_{p+a'',q+b''}\xrightarrow{\alpha_{p+a'',q+b''}}D_{p+a''+a,q+b''+b}\big),$$$$\text{im}\big(D_{p,q}\xrightarrow{\alpha_{p,q}}D_{p+a,q+b}\big)=\text{ker}\big(D_{p+a,q+b}\xrightarrow{\beta_{p+a,q+b}}D_{p+a+a',q+b+b'}\big),$$$$\text{im}\big(D_{p,q}\xrightarrow{\beta_{p,q}}E_{p+a',q+b'}\big)=\text{ker}\big(E_{p+a',q+b'}\xrightarrow{\gamma_{p+a',q+b'}}D_{p+a'+a'',q+b'+b''}\big).$$ Here bidegree of $\alpha,\beta,\gamma$ are $(a,b),(a',b'),(a'',b'')$ respectively.
My question is how do I construct an exact couple when we have the data as in the $\textbf{1. Theorem }$. What I have tried is the following.
Let $r\geq 2$. Define $$D_{s,t}:=\frac{Z^r_{s-r+1,t+r-1}}{\partial Z^{r-1}_{s,t+1}}\text{ for each }(s,t)\in \Bbb Z\times\Bbb Z.$$
Consider the inclusion induced maps $$\frac{Z^r_{s-r+1,t+r-1}}{\partial Z^{r-1}_{s,t+1}}=D^r_{s,t}\xrightarrow{\alpha^r_{s,t}} D^r_{r+1,t-1}=\frac{Z^r_{s-r+2,t+r-2}}{\partial Z^{r-1}_{s+1,t}},$$$$\text{ and }$$$$\frac{Z^r_{s-r+1,t+r-1}}{\partial Z^{r-1}_{s,t+1}}=D^r_{s,t}\xrightarrow{\beta^r_{s,t}}E^r_{s,t}=\frac{Z^r_{s,t}}{Z^{r-1}_{s-1,t+1}+\partial Z^{r-1}_{s+r-1,t-r+2}}.$$ Also consider the map induced from the boundary map the complex $ C_\bullet$, written by $$\frac{Z^r_{s,t}}{Z^{r-1}_{s-1,t+1}+\partial Z^{r-1}_{s+r-1,t-r+2}}=E^r_{s,t}\xrightarrow{\gamma^r_{s,t}}D^r_{s-1,t}=\frac{Z^r_{s-r,t+r-1}}{\partial Z^{r-1}_{s-1,t+1}}$$
Is there any wrong with my approach. Actually I have a problem with proving exactness. Any help will be appreciated. Thanks.