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Following the first chapter of Hatcher's great book "Spectral Sequences in Algebraic Topology", I got into problems with spectral sequences of cohomological type. Fix a ring $R$ once and for all. Please let me first recapitulate the homological situation.

An exact couple consists of bigraded $R$-modules $A$ and $E$ and bigraded $R$-module homomorphisms $i$, $j$, and $k$, such that

$$\begin{array}{rcl}A&\xrightarrow{i}&A\newline {\scriptsize k}\nwarrow&&\nearrow{\scriptsize &&\swarrow{\scriptsize j}\newline&E&\end{array}$$

is exact at every corner. Its derived pair with $A'=i(A)$ and $E'=H(E)$ with respect to $d=j\circ k$ is again exact. Before you wonder about the indices in what comes next, please continue reading up to the canonical example. This will explain the degrees, if I have not made a mistake. Let $a\in \mathbb{N}$ (in the example below we have $a=1$). Set $r=0$ for the moment. An exact couple with bidegrees

$$\begin{array}{rcl}A^{(r)}&\xrightarrow{(1,-1)}&A^{(r)}\newline {\scriptsize (-1,0)}\nwarrow&&\nearrow{\scriptsize &&\swarrow{\scriptsize (-(a-1+r),(a-1+r))}\newline&E^{(r)}&\end{array}$$

induces a spectral sequence $E_{p,q}^{r+a}=E_{p,q}^{(r)}$ of homological type with differentials $d^{r+a}=j^{(r)}\circ k^{(r)}$. Here ${\scriptsize something}^{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram above.

Here is the canonical example. Let $0=C_{-1}\subseteq C_0\subseteq\ldots C_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take for example the singular complex of a topological space $X$ filtered by a filtration of $X$. We have many long exact sequences in homology, here are three of them: $$\begin{array}{lcccccr} \to H_{p+q}(C_p,C_{p-1})&\xrightarrow{k}&H_{p+q-1}(C_{p-1})&\xrightarrow{i}&H_{p+q-1}(C_p)&\xrightarrow{j}&H_{p+q-1}(C_p,C_{p-1})\to\newline \to H_{p+q}(C_{p-1},C_{p-2})&\xrightarrow{k}&H_{p+q-1}(C_{p-2})&\xrightarrow{i}&H_{p+q-1}(C_{p-1})&\xrightarrow{j}&H_{p+q-1}(C_{p-1},C_{p-2})\to\newline \to H_{p+q}(C_{p-2},C_{p-3})&\xrightarrow{k}&H_{p+q-1}(C_{p-3})&\xrightarrow{i}&H_{p+q-1}(C_{p-2})&\xrightarrow{j}&H_{p+q-1}(C_{p-2},C_{p-3})\to \end{array}$$

There is an exact couple as above with $a=1$ and $E_{p,q}=H_{p+q}(C_p,C_{p-1})$ and $A_{p,q}=H_{p+q}(C_p)$. Please note, that all the bigrades are correct. To follow $d^1:E_{p,q}^1\to E_{p-1,q}^1$, you start at the upper left corner, apply $k$, go one row down (the entries are equal here if you move one right) and apply $j$. One can also follow $d^2$ but this is not quite correct since you have to deal with representatives in $E^1$. Here you start at the upper left corner and get to the lower right.

Here is Hatcher's illuminating argument for convergence. I have never seen this so clearly presented.

The spectral sequence $E^1(C_\bullet)$ of homological type converges to $H_{p+q}(C)$ if it is bounded (=only finitely many non-zero entries on every fixed diagonal $p+q$). One has $$E^\infty_{p,q}=i(H_{p+q}(C_p))/i(H_{p+q}(C_{p-1})).$$ ($i$ denotes the image in the colimit $H_{p+q}(C)$.)

Proof. Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple

$$\begin{array}{c} \to E^{r}_{p+r,q-r+1}\xrightarrow{k^{r}} A^{r}_{p+r-1,q-r+1}\xrightarrow{i^{r}}A^{r}_{p+r,q-r}\newline \xrightarrow{j^{r}}E^{r}_{p,q}\xrightarrow{k^{r}}\newline A^{r}_{p-1,q}\xrightarrow{i^{r}}A^{r}_{p,q-1}\xrightarrow{j^{r}}E^{r}_{p-r,q-1+r}\to. \end{array}$$

The first and the last term are $0$ because of the bounding assumption. We have $A_{p,q}^{r}=i(A^1_{p-r,q+r})$ and since $C_n$ is zero for $n<0$, the last three terms are $0$. Exactness implies the result.

Now the cohomological situation.

Let $0=C_{-1}\subseteq C_0\subseteq\ldots C_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take again for example the singular complex of a topological space $X$ filtered by a filtration of $X$. Understand the cohomology $H^n(C)$ of $C$ as $H_n(\hom(C,\mathbb{Z}))$, thehomology of the dualized complex as one does in topology.

Again we have many long exact sequences: $$\begin{array}{lcccccr} \to H^{p+q}(C_p,C_{p-1})&\xrightarrow{k}&H^{p+q}(C_p)&\xrightarrow{i}&H^{p+q}(C_{p-1})&\xrightarrow{j}&H^{p+q+1}(C_p,C_{p-1})\to\newline \to H^{p+q}(C_{p+1},C_p)&\xrightarrow{k}&H^{p+q}(C_{p+1})&\xrightarrow{i}&H^{p+q}(C_{p})&\xrightarrow{j}&H^{p+q+1}(C_{p+1},C_{p})\to\newline \to H^{p+q}(C_{p+2},C_{p+1})&\xrightarrow{k}&H^{p+q}(C_{p+2})&\xrightarrow{i}&H^{p+q}(C_{p+1})&\xrightarrow{j}&H^{p+q+1}(C_{p+2},C_{p+1})\to \end{array}$$

One can again follow $d^1:E_{p,q}^1\to E_{p+1,q}^1$, etc. What is an exact couple of cohomological type? The only thing which fits into the picture is this: Set $A^{p,q}=H^{p+q}(C_p)$ and $E^{p,q}=H^{p+q}(C_p,C_{p-1})$ and $a=1$.

An exact couple with bidegrees

$$\begin{array}{rcl}A_{(r)}&\xrightarrow{(-1,1)}&A_{(r)}\newline {\scriptsize (0,0)}\nwarrow&&\nearrow{\scriptsize &&\swarrow{\scriptsize (a+r,-(a-1+r))}\newline&E_{(r)}&\end{array}$$

induces a spectral sequence $E_{r+a}=E_{(r)}$ of cohomological type with differentials $d_{r+a}=j_{(r)}\circ k_{(r)}$. Here ${\scriptsize something}_{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram.

Now I would like to establish the following result. This is done nowhere since everything is said to be "dual" to the homological case. But if you look at the indices...hmm:

The spectral sequence $E_1(C_\bullet)$ of cohomological type converges to $H^{p+q}(C)$ if it is bounded. One has $$E_\infty^{p,q}=ker(H^{p+q}(C)\to H^{p+q}(C_{p-1}))/ker(H^{p+q}(C)\to H^{p+q}(C_{p})).$$

Proof? Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple

$$\begin{array}{c} \to E_{r}^{p-r,q+r-1}\xrightarrow{k_{r}} A_{r}^{p-r,q+r-1}\xrightarrow{i_{r}}A_{r}^{p-r-1,q+r}\newline \xrightarrow{j_{r}}E_{r}^{p,q}\xrightarrow{k_{r}}\newline A_{r}^{p,q}\xrightarrow{i_{r}}A_{r}^{p-1,q+1}\xrightarrow{j_{r}}E_{r}^{p+r,q-r+1}\to. \end{array}$$

The last term is $0$ because of the bounding assumption. We have $A_{r}^{p,q}=i(A^{p+r,q-r}_1)$ and since $C_n$ is zero for $n<0$, the second and the third term are $0$. Exactness implies that $$E_r^{p,q}=ker(i(H^{p+q}(C_{p+r}))\to i(H^{p+q}(C_{p+r-1}))).$$ I cannot see how the result follows from this. Perhaps it is "only" a limit trick but I can not see it. So the question is:

How does Lemma 1.2. of Hatcher's text works in the cohomological case?

3 deleted 26 characters in body

Following the first chapter of Hatcher's great book "Spectral Sequences in Algebraic Topology", I got into problems with spectral sequences of cohomological type. Fix a ring $R$ once and for all. Please let me first recapitulate the homological situation.

An exact couple consists of bigraded $R$-modules $A$ and $E$ and bigraded $R$-module homomorphisms $i$, $j$, and $k$, such that

$$\begin{array}{rcl}A&\xrightarrow{i}&A\newline {\scriptsize k}\nwarrow&&\nearrow{\scriptsize j}\newline&E&\end{array}$$

is exact at every corner. Its derived pair with $A'=i(A)$ and $E'=H(E)$ with respect to $d=j\circ k$ is again exact. Before you wonder about the indices in what comes next, please continue reading up to the canonical example. This will explain the degrees, if I have not made a mistake. Let $a\in \mathbb{N}$ (in the example below we have $a=1$). Set $r=0$ for the moment. An exact couple with bidegrees

$$\begin{array}{rcl}A^{(r)}&\xrightarrow{(1,-1)}&A^{(r)}\newline {\scriptsize (-1,0)}\nwarrow&&\nearrow{\scriptsize (-(a-1+r),(a-1+r))}\newline&E^{(r)}&\end{array}$$

induces a spectral sequence $E_{p,q}^{r+a}=E_{p,q}^{(r)}$ of homological type with differentials $d^{r+a}=j^{(r)}\circ k^{(r)}$. Here ${\scriptsize something}^{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram above.

Here is the canonical example. Let $0=C_{-1}\subseteq C_0\subseteq\ldots C_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take for example the singular complex of a topological space $X$ filtered by a filtration of $X$. We have many long exact sequences in homology, here are three of them: $$\begin{array}{lcccccr} \to H_{p+q}(C_p,C_{p-1})&\xrightarrow{k}&H_{p+q-1}(C_{p-1})&\xrightarrow{i}&H_{p+q-1}(C_p)&\xrightarrow{j}&H_{p+q-1}(C_p,C_{p-1})\to\newline \to H_{p+q}(C_{p-1},C_{p-2})&\xrightarrow{k}&H_{p+q-1}(C_{p-2})&\xrightarrow{i}&H_{p+q-1}(C_{p-1})&\xrightarrow{j}&H_{p+q-1}(C_{p-1},C_{p-2})\to\newline \to H_{p+q}(C_{p-2},C_{p-3})&\xrightarrow{k}&H_{p+q-1}(C_{p-3})&\xrightarrow{i}&H_{p+q-1}(C_{p-2})&\xrightarrow{j}&H_{p+q-1}(C_{p-2},C_{p-3})\to \end{array}$$

There is an exact couple as above with $a=1$ and $E_{p,q}=H_{p+q}(C_p,C_{p-1})$ and $A_{p,q}=H_{p+q}(C_p)$. Please note, that all the bigrades are correct. To follow $d^1:E_{p,q}^1\to E_{p-1,q}^1$, you start at the upper left corner, apply $k$, go one row down (the entries are equal here if you move one right) and apply $j$. One can also follow $d^2$ but this is not quite correct since you have to deal with representatives in $E^1$. Here you start at the upper left corner and get to the lower right.

Here is Hatcher's illuminating argument for convergence. I have never seen this so clearly presented.

The spectral sequence $E^1(C_\bullet)$ of homological type converges to $H_{p+q}(C)$ if it is bounded (=only finitely many non-zero entries on every fixed diagonal $p+q$). One has $$E^\infty_{p,q}=i(H_{p+q}(C_p))/i(H_{p+q}(C_{p-1})).$$ ($i$ denotes the image in the colimit $H_{p+q}(C)$.)

Proof. Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple

$$\begin{array}{c} \scriptsize \to E^{r}_{p+r,q-r+1}\xrightarrow{k^{r}} A^{r}_{p+r-1,q-r+1}\xrightarrow{i^{r}}A^{r}_{p+r,q-r}\newline \xrightarrow{j^{r}}E^{r}_{p,q}\xrightarrow{k^{r}}\newline A^{r}_{p-1,q}\xrightarrow{i^{r}}A^{r}_{p,q-1}\xrightarrow{j^{r}}E^{r}_{p-r,q-1+r}\to. \end{array}$$

The first and the last term are $0$ because of the bounding assumption. We have $A_{p,q}^{r}=i(A^1_{p-r,q+r})$ and since $C_n$ is zero for $n<0$, the last three terms are $0$. Exactness implies the result.

Now the cohomological situation.

Let $0=C_{-1}\subseteq C_0\subseteq\ldots C_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take again for example the singular complex of a topological space $X$ filtered by a filtration of $X$. Understand the cohomology $H^n(C)$ of $C$ as $H_n(\hom(C,\mathbb{Z}))$, thehomology of the dualized complex as one does in topology.

Again we have many long exact sequences: $$\begin{array}{lcccccr} \to H^{p+q}(C_p,C_{p-1})&\xrightarrow{k}&H^{p+q}(C_p)&\xrightarrow{i}&H^{p+q}(C_{p-1})&\xrightarrow{j}&H^{p+q+1}(C_p,C_{p-1})\to\newline \to H^{p+q}(C_{p+1},C_p)&\xrightarrow{k}&H^{p+q}(C_{p+1})&\xrightarrow{i}&H^{p+q}(C_{p})&\xrightarrow{j}&H^{p+q+1}(C_{p+1},C_{p})\to\newline \to H^{p+q}(C_{p+2},C_{p+1})&\xrightarrow{k}&H^{p+q}(C_{p+2})&\xrightarrow{i}&H^{p+q}(C_{p+1})&\xrightarrow{j}&H^{p+q+1}(C_{p+2},C_{p+1})\to \end{array}$$

One can again follow $d^1:E_{p,q}^1\to E_{p+1,q}^1$, etc. What is an exact couple of cohomological type? The only thing which fits into the picture is this: Set $A^{p,q}=H^{p+q}(C_p)$ and $E^{p,q}=H^{p+q}(C_p,C_{p-1})$ and $a=1$.

An exact couple with bidegrees

$$\begin{array}{rcl}A_{(r)}&\xrightarrow{(-1,1)}&A_{(r)}\newline {\scriptsize (0,0)}\nwarrow&&\nearrow{\scriptsize (a+r,-(a-1+r))}\newline&E_{(r)}&\end{array}$$

induces a spectral sequence $E_{r+a}=E_{(r)}$ of cohomological type with differentials $d_{r+a}=j_{(r)}\circ k_{(r)}$. Here ${\scriptsize something}_{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram.

Now I would like to establish the following result. This is done nowhere since everything is said to be "dual" to the homological case. But if you look at the indices...hmm:

The spectral sequence $E_1(C_\bullet)$ of cohomological type converges to $H^{p+q}(C)$ if it is bounded. One has $$E_\infty^{p,q}=ker(H^{p+q}(C)\to H^{p+q}(C_{p-1}))/ker(H^{p+q}(C)\to H^{p+q}(C_{p})).$$

Proof? Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple

$$\begin{array}{c} \scriptsize \to E_{r}^{p-r,q+r-1}\xrightarrow{k_{r}} A_{r}^{p-r,q+r-1}\xrightarrow{i_{r}}A_{r}^{p-r-1,q+r}\newline \xrightarrow{j_{r}}E_{r}^{p,q}\xrightarrow{k_{r}}\newline A_{r}^{p,q}\xrightarrow{i_{r}}A_{r}^{p-1,q+1}\xrightarrow{j_{r}}E_{r}^{p+r,q-r+1}\to. \end{array}$$

The last term is $0$ because of the bounding assumption. We have $A_{r}^{p,q}=i(A^{p+r,q-r}_1)$ and since $C_n$ is zero for $n<0$, the second and the third term are $0$. Exactness implies that $$E_r^{p,q}=ker(i(H^{p+q}(C_{p+r}))\to i(H^{p+q}(C_{p+r-1}))).$$ I cannot see how the result follows from this. Perhaps it is "only" a limit trick but I can not see it. So the question is:

How does Lemma 1.2. of Hatcher's text works in the cohomological case?

2 fixed indexing
\to E^{r}{p+r,q-r+1}\xrightarrow{k^{r}} A^{rE^{r}_{p+r,q-r+1}\xrightarrow{k^{r}} {p+r-1,q-r+1}\xrightarrow{i^{r}}A^{r}{p+r,q-r}\newline\xrightarrow{j^{r}}E^{r}{p,q}\xrightarrow{k^{r}}\newlineA^{r}{p-1,q}\xrightarrow{i^{r}}A^{r}{p,q-1}\xrightarrow{j^{r}}E^{r}_{p-r,q-1+r}\to.\to E_{rE_{r}^{p-r,q+r-1}\xrightarrow{k_{r}} {p-r,q+r-1}\xrightarrow{k{r}} A_{r}^{p-r,q+r-1}\xrightarrow{i_{r}}A_{r}^{p-r-1,q+r}\newline
1