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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?

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At the time this construction was first published, Mike Hopkins went to elementary school: A Dress, "Zur Spectralsequenz von Faserungen", Inventiones Math. 3 (1967). – Johannes Ebert Mar 30 '12 at 19:46
Yes, it's based on Dress' article, but it seems to be much easier and simplified proof. Maybe I should have written "Hopkins' article". – mathdonk Mar 30 '12 at 19:59
A good warm up would be Hopkins' definition of singular homology. I would then move on to Hopkins' definition of fibration. There must be a mistake in terminology: it should not be called, "Serre fibration." The correct terminology should be "Hopkin's fibration." – John Klein Mar 30 '12 at 21:29
@mathdonk: You got stuck at a point where Hopkins writes "From this one easily checks..." (if we are talking about the same text). Maybe you should have a look at Dress' paper, he gives full details. P.S.: why do you say that a proof is "much easier and simplified" if you fail to understand the main step? – Johannes Ebert Mar 30 '12 at 23:20
@Dan: I believe it is… – Johannes Ebert Mar 31 '12 at 9:23
up vote 4 down vote accepted

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.

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I know how the spectral sequence of the double complex works, I just don't get why one can calculate the homology (of rows and columns) after doing this identifications. You definitely have to check something, but I don't know what exactly... – mathdonk Mar 30 '12 at 19:20
The treatment of double complexes in Bott ant Tu "Differential forms in Algebraic Topology" is also very good. – Mark Grant May 22 '12 at 6:02

I wanted to post the following as a comment, but it's too long.

It might help to realize where the differential comes from. Let $p: E \to B$ be a Hurewicz fibration. Assume $B$ is a connected CW complex. Then $B$ has a cellular filtration $B_k \subset B_{k+1} \dots$. If we pull back $p$ along this filtration we obtain a filtration (in fact, by cofibrations!) of $E$ $$ E_0 \subset E_1 \subset \cdots $$ I claim that the homology spectral sequence of this filtration is then the Serre spectral sequence. The $d_1$-differential of the above filtration is given by the composition $$ E_k/E_{k-1} \overset{\delta}\to \Sigma E_{k-1} \to \Sigma E_{k-1}/E_{k-2} \qquad (\ast) $$ where the map $\delta$ is the Barratt-Puppe extension of the cofibration $E_{k-1} \to E_k$ and the second displayed map is given by collapsing $E_{k-2}$ to a point.

Furthermore, it's easy to check that $$ E_{k}/E_{k-1} \simeq F_+ \wedge B_k/B_{k-1} \qquad (\ast\ast) $$ So I guess that your question, in the end, amounts to the following: With respect to the equivalence $(\ast\ast)$, what does the map $E_k/E_{k-1} \to \Sigma E_{k-1}/E_{k-2}$ look like when considered as a map $$ F_+ \wedge B_k/B_{k-1} \to F_+ \wedge \Sigma B_{k-1}/B_{k-2} \quad ? $$ In other words, on the level of homology, is this map of the form $\text{id}\wedge c$, where $c:B_k/B_{k-1} \to \Sigma B_{k-1}/B_{k-2} $ is the map for the filtration $B_k \subset B_{k+1} \dots$ constructed like the one in $(\ast)$?

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And the answer to this question is that the map in question is $\alpha \wedge c$, where $\alpha$ is an admissible map for the fibration. An $H_*$-orientable fibration is one in which all admissible maps induce the same map, so it does induce $\id\wedge c$. – Jeff Strom Mar 31 '12 at 0:39

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