Edit: I would very much like someone to comment on this. I have worked through Vakil's notes Spectral sequences: friend or foe? and I believe my description is correct. But there are so many indices and signs that it is easy to make mistakes. Also relevant are probably the pages 162 and following in the book Differential Forms in Algebraic Topology by Bott and Tu. The main difference is, that they state that a zigzag as I describe it always exists. They do not state that I can just start to push the element around until I am at the right position. End of the edit.
(The question I have is at the very end.)
For every bounded double complex $M$ there is an associated spectral sequence $E_*^{*,*}$ which converges to the homology of the total complex of $M$. It is normally constructed using exact couples, or more classical the spectral sequence associated to a filtered complex. Unfortunately, the proof is always a mess, and it is hard to see what the differentials actually do. I like to have a more direct describtion of the differentials which makes the connections to diagram chasing clear. There is an exercise in Ravi Vakil's FOAG which asks to construct the differentials on page $E_2^{*,*}$ in a way reminiscent of the connecting morphism in the snake lemma. If I generalize the construction, then I get the following description. I use $u$ and $r$ to denote the differentials of the double complex. $u$ for up and $r$ for right. When you read the quote below, you should picture me trying to explain what I have learned about spectral sequences to a friend.
We define the spectral sequence recursively. The objects on the page $E_{r+1}^{*,*}$ are the homology groups of the chain complexes $(E_r^{*,*},d_r)$ from the previous page, as they should be. The objects on the first page are the objects of $M$. The differentials on the first page are the maps $u$. Hence $E_0^{*,*}$ is the double complex $M$ without the differential $r$. The differentials on the second page are the maps induced by $r$ between the homology groups. Here is another description of the differential $d_1^{p,q}$. Take an element $h$ in $E_1^{p,q}$. It is an equivalence class of elements in $E_0^{p,q} = M^{p,q}$. In other words there is a subobject of $M^{p,q}$ which surjects onto $E_1^{p,q}$. Choose an element $x$ on the first page such that $[x] = h$. Apply $r$ to $x$. Check that $rx$ is in the kernel of $u$. Push $rx$ back to the page $E_1^{p,q}$. The equivalence class $[rx]$ is $d_1^{p,q}h$. Note that the recipe involves one choice at the beginning, but because we divide by the image of $u$ at the end, the resulting class does not depend on that choice. The procedure generalizes. The objects $E_2^{p,q}$ is a homology of homology. So its elements are classes of classes. It is not helpful to think like this. Rather you should remember that there is surjection from (a subobject of) the previous page onto $E_2^{p,q}$ and hence a way to lift elements to pages with smaller numbers. Take some $h \in E_2^{p,q}$. Choose $x$ on page 1 such that $[x] = h$ and then choose $y$ on page 0 such that $[y] = x$. Now we are on page zero, so we can use $r$ and $u$ to walk on the page. Apply $r$ to $y$ to go one step to the right. Go one step down: choose some $z$ such that $uz = y$. Apply $r$ to $z$. Check that $rz$ is in the kernel of $u$. It is, so you are allowed to push $rz$ to page $1$. Show that $[rz]$ is in the kernel of $d_1$. Define $d_2h := [[rz]]$. It is a messy diagram chase to show that $d_2$ is well defined. In general, if you are on the $r$th page, have some class $h\in E_r^{p,q}$ and you want to define what $d_rh$ is, lift $h$ all the way up to page $0$, use $r$ and $u$ to walk in a zigzag to the position to which $d_r$ points and then push the result down to page $r$ where you came from. The picture below shows a cartoon of the procedure for the first few steps.
My question is: Is the description I have given above correct? Is this how the differentials act in a category of modules (or on generalized elements)? Is there a reference?
