Bott Tu, Chapter 14, Filtered complexes I'm reading Differential Forms in Algebraic Topology from Bott,Tu and i confused in section of spectral sequence of filtered complexes. In this section book consider ungraded complexes They're just forgetting about graduation in a complex. Given $K$ with differencial $D$ and a filtration $K=K_0\supset K_1\supset K_2\supset K_3\supset...$  we define $A=\bigoplus_{p\geq0}K_p$ which is a differential complex consider $D$ acting in each coordinate (at least that's what i understand). Then we define $i:A\to A$ as the sum of inclusions $K_{p+1}\hookrightarrow K_p$ and obtain exact sequence
$$0\to\bigoplus_{p\geq0}K_p\overset{i}{\to}\bigoplus_{p\geq0}K_p\to \bigoplus_{p\geq0}K_p/K_{p+1}$$
Now, in the grading case the previous sequence gives an exact sequence of differential complexes and then we have a long exact sequence between cohomology groups
$$...\to H^k\left(\bigoplus_{p\geq0}K_p\right)\to H^k\left(\bigoplus_{p\geq0}K_p\right)\to H^k\left(\bigoplus_{p\geq0}K_p/K_{p+1}\right)\to H^{k+1}\left(\bigoplus_{p\geq0}K_p\right)\to...$$
and then sum all over cohomology groups we obtain an exact couple. I would appreciate it if someone could confirm if what I am saying is correct and if they could explain the following to me. Following the above (page 158) the book says " It is not difficult to see that the same
diagram exists in the ungraded case". What would the sequence in cohomology look like in the ungraded case? And then the book does an example with a finite filtration
$$K=K_0\supset K_1\supset K_2\supset K_3\supset\{0\}$$
but i don´t understand how obtain the exact couple in this example. Please someone explain it to me. Thanks.
 A: First the homology group $H(A)$ is defined to be $\ker D/\mathrm{im} D=\bigoplus\limits_{p\in\mathbb{Z}}H(K_p)$, where $D:A\to A$ is a group homomorphism satisfying $D^2=0$. Since $B=\bigoplus\limits_{p\in\mathbb{Z}}K_p/K_{p+1}$ is the quotient group, we can naturally descend $D$ to $B$. That is to say, $D:K_p/K_{p+1}\to K_p/K_{p+1},\; [b]\mapsto [Db]$ is induced by $D:K_p\to K_p$, $D:K_{p+1}\to K_{p+1}$. It's direct to check this definition is independent of the choice of the representative $b$. Then Bott and Tu asserted that it's not difficult to get an exact couple
The map $i_1,j_1$ are naturally induced from $i,j$. For example,
$$i_1:H(A)\to H(A),\;[a]\mapsto [ia],$$ where $i:A\to A$ is the direct sum of all inclusions $K_{P+1}\hookrightarrow K_p$. It should be noted that although we have a sequence of inclusion
$$\cdots=K=K\supset K_1\supset K_2\supset K_3\supset \cdots,$$ it's not automatically true that
$$\cdots \stackrel{\cong}{\leftarrow} H(K) \stackrel{\cong}{\leftarrow} H(K) \supset H(K_1)\supset H(K_2)\supset H(K_3)\supset 0,$$ because $H(K_3)$ and $H(K_2)$ can just be two different groups without inclusion relation. Instead, we have
$$\cdots \stackrel{\cong}{\leftarrow} H(K) \stackrel{\cong}{\leftarrow} H(K) \stackrel{i}{\leftarrow} H(K_1)\stackrel{i}{\leftarrow} H(K_2)\stackrel{i}{\leftarrow} H(K_3)\leftarrow 0,$$ where $i$ is an Abelian group homomorphism. Direct summing all the Abelian homomorphisms $i:H(K_p)\to H(K_{p+1})$, we finally get an explicit expression of $$i_i:H(A)=\bigoplus\limits_{p\in\mathbb{Z}}H(H_p)\to H(A)=\bigoplus\limits_{p\in\mathbb{Z}}H(H_p),$$ and of course the map $i_1$ is not an inclusion map! The same argument can be applied to understand $j_1$. Finally we should explain what is $k_1$. Roughly speaking, it's just an analogous of the Zig-Zag lemma to determine the map $k_1$. Note that we have the following commutative diagram

The commutativity comes from the definition of $D$ and $i$. That is, if we choose an element $a\in K_{p+1}\subset K$, then $ia=a\in K_p\subset K$, thus $Dia=Da\in K_p\subset K$. On the other hand, if we first apply $D$ to $a$, we get $Da\in K_{p+1}\subset K$, then $iDa=Da\in K_{p}\subset K$. When acting on an element of $H(A)$ it's just a matter of moving the position of $Da\in K$ at first or later which obviously does not change the result.
To determine $k_1[b]$, choose $[b]\in H(B)$. Note that $j$ is surjective, so there exists a $a\in A$ such that $ja=b$. By the commutativity of the diagram, $jDa=Dja=Db=0$, thus by exactness, there exists an $\tilde{a}$ such that $i(\tilde{a})=Da$. Define $k_1([b])=[\tilde{a}]\in H(A)$. It's an easy exercise to check that
(1) $\tilde{a}$ is closed, so that $[\tilde{a}]$ really lies in $H(A)$;
(2) $[\tilde{a}]$ does not depend on the choice of $b$ and $a$.
