I'm trying to follow Hopkins' construction of the Serre Spectral Sequence, but some "obvious" things are not that obvious to me.
He starts with considering a double complex $C_{\bullet,\bullet}$ with $C_{p,q}$ to be a free $\mathbb{Z}$-module generated by the maps $\Delta[p]\times\Delta[q]\rightarrow E$ ($E$ is a total space of Serre fibration over $B$) which fit into the diagram
\begin{matrix}
\Delta[p]\times\Delta[q] & \to & E \\\
\downarrow & & \downarrow \\\
\Delta[p] & \to & B
\end{matrix}
with obvious differentials (coordinate by coordinate differentials as in normal singular complex).
There are two filtrations, he uses the first one (by rows) to determine the homology of the total complex, ane the second one to get $E^2_{p,q}=H_p(B,\underline{H_q}(F)$. I have a problem with the second part. He fixes $p$ and a map $c$ in the bottom row and interprets the diagram as a map $\Delta[q]\rightarrow F_c$ where $F_c$ is the image of $\Delta[p]\times\Delta[q]$ in $E$ such that the diagram commutes, which basically means that $C_{p,q}=\bigoplus_c \mathbb{Z}[F_c]$ (as a module, that's true). Now he calculates the homology of the column and says that it's $\bigoplus_c H_*(F_c)$. Why can we apply the vertical differential here? Do we need to check some compatibility condition?
Next, we want to use that $E^1_{p,q}$ is a module of singular $p$-chains with coefficients in a local system $\underline{H_q}(F)$ and say that the horizontal differential is just the regular differential to get the desired output. But how do we know that this differential works in a nice way?
Best Answer
The double complex has a horizontal differential $\partial'$ and a vertical differential $\partial''$ such that $\partial'\partial''=\partial''\partial'$. This gives rise to a total complex $TC_n=\bigoplus_{p+q=n}C_{pq}$ with differential $\partial|C_{pq}=\partial'+(-1)^p\partial''$. This can be filtered by $F_p=\bigoplus_{i\le p}C_{i,n-i}$ and so we get a spectral sequence $E^r$ converging to $H_\ast(TC)$, with $E^0_{pq}=C_{pq}$ and $d^0=\partial''$ (up to sign). Thus $E^1_{pq}=H_q(C_{p\ast})$ is the vertical homology of our double complex. Now the differential $d^1$ is induced by the chain map $\partial'$: any element of $E^1_{pq}$ is given by a $c\in C_{pq}$ with $\partial''c=0$, and that means $\partial c=\partial'c$. From here we see that $E^2$ is the horizontal homology of the vertical homology of our double complex.
Hope I didn't miss the bulk of your question. For more information I suggest Ken Brown's amazing textbook Cohomology of Groups.