Cfrgtkky

Let $Fstackrel i to Estackrel pito S^1$ a fiber bundle over the circle $S^1$. There is a long exact sequence sequence in cohomology, called Wang:
$$dotsto H^k(E)stackrel {i^*}to H^k(F)stackrel {f^*-I}to H^k(F)stackrel deltato H^{k+1}(E)to dots ,$$
where $I$ is the identity and $f^*$ is induced by the monodromy of the bundle.

In the case of a bundle over $S^n$, $n>1$, the long exact sequence above follows from the Serre spectral sequence (see Wikipedia). However, using the presented argument, one obtains:
$$dotsto H_q(E)to E^1_{1,q-1}stackrel dto E^1_{0,q-1}to H_{q-1}(E)todots~.$$
Thus in order to conclude the Wang sequence one must identify the first page and the boundary map of the Serre spectral sequence. I hoped to avoid this.

Another way of proving the result was mentioned in this mathoverflow answer.

The strategy is to apply Mayer-Vietoris to the bundle. As I didn't do an argument like this before, I did not understand how to use monodromy.

I would appreciate if anyone could point me in the right direction in either of the two arguments.

Here is the Mayer-Vietoris argument :

Let $N,W,S,E$ be the north, west, south and east pole on the circle. Let $U=S^1setminus{E}$ and $V=S^1setminus {W}$. This is an open covering of $S^1$ such that $U$ and $V$ are contractible so that $pi^{-1}(U)$ is homeomorphic to $Ftimes U$ and even homotopicaly equivalent to $Ftimes{N}$. Similarly $pi^{-1}(V)simeq Ftimes{S}$ and $pi^{-1}(Ucap V)simeq Ftimes{N}cup Ftimes{S}$.

Now write the Mayer-Vietoris exact sequence associated to the covering of the total space $E$ by $pi^{-1}(U)$ and $pi^{-1}(V)$. This is :
$$...rightarrow H^i(E)rightarrow H^i(pi^{-1}(U))oplus H^i(pi^{-1}(V))rightarrow H^i(pi^{-1}(U cap V))rightarrow H^{i+1}(E)rightarrow ...$$
Using the above homotopy equivalence, this is :
$$...rightarrow H^i(E)rightarrow H^i(Ftimes{N})oplus H^i(Ftimes{S})rightarrow H^i(Ftimes{N})oplus H^i(Ftimes{S})rightarrow H^{i+1}(E)rightarrow ...$$
But what are the maps between these four copies of $H^i(F)$ ? So this enough to give the maps $H^i(Ftimes{N})to H^i(Ftimes{N})oplus H^i(Ftimes{S})$ and $H^i(Ftimes{S})to H^i(Ftimes{N})oplus H^i(Ftimes{S})$.

For the first one, remember that this map comes from the restriction $H^i(pi^{-1}(U))to H^i(pi^{-1}(Ucap V))$. So this the identity on $H^i(Ftimes{N})$ and a map $u:H^i(Ftimes{N})to H^i(Ftimes{S})$ which intuitively means "cohomology class lying on $Ftimes{N}$ that have been moved to $Ftimes{S}$ along the path from the north pole to the south avoiding the east (because $U=S^1setminus{E}$). You can prove this more rigorously by writing down all the diagrams with all the maps, including the homotopy equivalence. I will not do it here.

Similarly the second map $H^i(Ftimes{S})to H^i(Ftimes{N})oplus H^i(Ftimes{S})$ is the identity on the second component and a map $v:H^i(Ftimes{S})to H^i(Ftimes{N})$ which intuitively moves the cohomology classes on $Ftimes{S}$ to $Ftimes{N}$ along the path from the south pole to the north which avoids the west.

So the map in the long exact sequence looks like
$$H^i(Ftimes{N})oplus H^i(Ftimes{S})xrightarrow{begin{pmatrix}1&v\u&1 end{pmatrix}} H^i(Ftimes{N})oplus H^i(Ftimes{S})$$

The point is that there is an extra copy $H^i(F)$ on each side here and we would like to remove it. So let us have a look at its kernel and its cokernel :

• the kernel is the set of couple of classes $(x,y)$ such that $x+v(y)=0$ and $u(x)+y=0$. So this is the same as the classes $(x,-u(x))$ where $x=vu(x)$. But $vu$ is exactly the monodromy : we take a class and move it along a path going from the north to the south pole avoiding the east, and then from the south to the north avoiding the west, so a whole loop. Hence the kernel is $ker(f^*-I)$
  


• similarly, the cokernel is the set of couple $(x,y)$ modulo the relation $(x,y)=0$ iff $(x,y)=(a,u(a))+(v(b),b)$. But the map $(x,y)mapsto x-v(y)$ induces an isomorphism $operatorname{coker}begin{pmatrix}1&v\u&aend{pmatrix}simeq operatorname{coker}(f^*-I)$.

Thus, you can replace the map by $f^*-I:H^i(Ftimes{N})to H^i(Ftimes{N})$. Of course $H^i(Ftimes{N})$ is just $H^i(F)$ and you get the Wang sequence.

Note that to be more precise, one need to work at the level of complexes. But the argument is the same.