For the first difficulty, let $E^r$ be the rth page of a first quadrant spectral sequence with elements $E^r_{p,q}$ , where $p$ is the filtering degree.

Difficulty 1: On bullet 7 of Mosher and Tangora of page 76, he makes the following assertion - No element of $E^r_{n,0}$ can be a boundary. I don't understand why the boundary map $d^1_{n+1,0}:E^1_{n+1,0} \to E^1_{n,0}$ needs to be 0. If it isn't 0 then elements of $E^0_{n,0}$ will be boundaries.

Specializing to when $E^r$ is the $rth$ page of the serre spectral sequence of a fibration $(E,p,B)$... Difficulty 2: Mosher and Tangora say that assertion implies that there is a well defined map from $E^r_{n,0} \to E^2_{n,0}$. I put this as a separate difficulty because I understand the assertion in difficulty 1 when $r\geq 1$ and hence when $r\geq 2$. What is this map?


1) - You are right. $d^1$ is not an invariant of the fibration, and need not be 0 on $E^1_{p,0}$. Fix a $CW$ decomposition of $B$. Then $E^1_{p,0}=C_p(B)$ with the differential given by the cellular differential. So unless the cellular chain complex of the base is the homology, for your choice of the CW decomposition, $d_1$ is not zero here.

2)Because the $d_{r-1}$'s mapping into $E^{r-1}_{*,0}$ are mapping the image of $0$(convergence criterion #2 in mosher and tangora implies that this is a first quadrant SS), $E^r_{*,0}=ker d_{r-1} \subset E^{r-1}_{*,0}$, whenever $r>2$. Repeat to get that $E^r_{*,0}=ker d_{r-1}\cap... ker d_{2} \subset E^2$. Thus this is an inclusion map - in sum there is no quotienting happening.


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