# Serre spectral sequence of $\mathbb{Z}_2\rightarrow E\mathbb{Z}_2\rightarrow B\mathbb{Z}_2=\mathbb{R}P^\infty$

As the title shows, we have a fibration of $\mathbb{Z}_2\rightarrow E\mathbb{Z}_2\rightarrow B\mathbb{Z}_2\sim\mathbb{R}P^\infty$. I am trying to check my understanding of Serre spectral sequence with this simple example. First note that $H_i(\mathbb{Z}_2,\mathbb{Z})=\mathbb{Z}\oplus\mathbb{Z}$ for $i=0$ (since this fibre has two disconnected components) and otherwise $0$. By universal coefficient theorem , we get $H^i(\mathbb{Z}_2,\mathbb{Z})=\mathbb{Z}\oplus\mathbb{Z}$. We also know that $H^0(\mathbb{R}P^\infty,\mathbb{Z})=\mathbb{Z}$, $H^i(\mathbb{R}P^\infty,\mathbb{Z})=\mathbb{Z}_2$ with $i$ even and positive, and otherwise $0$. Therefore, the spectral sequence reads

\begin{array}{|c c c c c c} 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ 0& 0 &0 & 0 & 0 & 0 & 0 & 0 \\ \mathbb{Z}\oplus\mathbb{Z}& 0 &\mathbb{Z}_2\oplus\mathbb{Z}_2 & 0 & \mathbb{Z}_2\oplus\mathbb{Z}_2 & 0 & \mathbb{Z}_2\oplus\mathbb{Z}_2 & 0 \\ \hline \end{array}

The differentials all seems to be trivial. It determines the cohomology groups of $E\mathbb{Z}_2$. However, the cohomology ring of $E\mathbb{Z}_2$ has to be trivial since it should be a (weakly) contractible space. There is a contradiction here. What's wrong with my understanding? Perhaps, Serre spectral sequence does not work for fibre disconnected or base space not simply-connected? Thanks in advance!

• When the base has nontrivial fundamental group, $H^*(F)$ carries an action by the fundamental group of the base. You may think of this as a local coefficient system on the base, and the $E^2$ page is the cohomology of $B$ with respect to the local coefficient system $H^*(F)$. – user98602 Sep 19 '18 at 20:11
• Identifying the group cohomology of the $G$-module $A$ with $H^*(BG;A)$ (thinking of the action as defining a local system), the $E^2$ page of your spectral sequence (for $EG \to BG$) becomes $H^*(G;\Bbb Z[G])$. That is classically known to be a single copy of $\Bbb Z$ concentrated in degree 0. – user98602 Sep 19 '18 at 20:13
• (1) $\Bbb Z[G] = H^0(G;\Bbb Z) = H^*(G;\Bbb Z)$. So $q = 0$ in that term. (2) You push around the fiber along a loop, successively trivializing the fibration over little bits of the loop, to get a homotopy equivalence $\gamma_*: F \to F$, well-defined up to homotopy. In the case of a covering space this is the Galois action on fibers. (3) Yes. (4) No, that is not what I said, I said it is a single copy of $\Bbb Z$. If it was $\Bbb Z^2$, then $E(\Bbb Z/2)$ will not be connected. $\Bbb Z[G]$ here is not endowed with the trivial $G$-action, but rather $G$ acts by translation on basis elements. – user98602 Sep 19 '18 at 21:00
• That $H^*(G;\Bbb Z[G]) = \Bbb Z$ in degree 0 and 0 otherwise (to be clear, here the notation refers to group cohomology) is proved in any book on group cohomology, though I do not have a reference on hand. There is an explicit deformation retraction of the chain complex defining group cohomology onto a degree 0 generator, which is not hard to write down by hand. That $H^*(BG;A) \cong H^*(G;A)$, where $A$ is a $G$-module and considered over $BG$ as a local system (aka $\pi_1(BG)$-module) follows from the specific model of $BG$ constructed in eg Hatcher for a finite group: its cellular chains... – user98602 Sep 19 '18 at 21:03
• ...give the chain complex defining group cohomology. – user98602 Sep 19 '18 at 21:03