Let $F\stackrel i \to E\stackrel \pi\to S^1$ a fiber bundle over the circle $S^1$. There is a long exact sequence sequence in cohomology, called Wang: $$\dots\to H^k(E)\stackrel {i^*}\to H^k(F)\stackrel {f^*-I}\to H^k(F)\stackrel \delta\to H^{k+1}(E)\to \dots ,$$ 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: $$\dots\to H_q(E)\to E^1_{1,q-1}\stackrel d\to E^1_{0,q-1}\to H_{q-1}(E)\to\dots~.$$ 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.

  • 1
    $\begingroup$ On behalf of @Gary Kennedy: An elementary argument is outlined in Lemma 8.4 of Milnor's book Singular Points of Complex Hypersurfaces. $\endgroup$
    – dantopa
    Apr 17, 2019 at 20:58

1 Answer 1


Here is the Mayer-Vietoris argument :

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

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 : $$...\rightarrow H^i(E)\rightarrow H^i(F\times\{N\})\oplus H^i(F\times\{S\})\rightarrow H^i(F\times\{N\})\oplus H^i(F\times\{S\})\rightarrow H^{i+1}(E)\rightarrow ...$$ But what are the maps between these four copies of $H^i(F)$ ? So this enough to give the maps $H^i(F\times\{N\})\to H^i(F\times\{N\})\oplus H^i(F\times\{S\})$ and $H^i(F\times\{S\})\to H^i(F\times\{N\})\oplus H^i(F\times\{S\})$.

For the first one, remember that this map comes from the restriction $H^i(\pi^{-1}(U))\to H^i(\pi^{-1}(U\cap V))$. So this the identity on $H^i(F\times\{N\})$ and a map $u:H^i(F\times\{N\})\to H^i(F\times\{S\})$ which intuitively means "cohomology class lying on $F\times\{N\}$ that have been moved to $F\times\{S\}$ along the path from the north pole to the south avoiding the east (because $U=S^1\setminus\{E\}$). 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 $H^i(F\times\{S\})\to H^i(F\times\{N\})\oplus H^i(F\times\{S\})$ is the identity on the second component and a map $v:H^i(F\times\{S\})\to H^i(F\times\{N\})$ which intuitively moves the cohomology classes on $F\times\{S\}$ to $F\times\{N\}$ along the path from the south pole to the north which avoids the west.

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

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 $\operatorname{coker}\begin{pmatrix}1&v\\u&a\end{pmatrix}\simeq \operatorname{coker}(f^*-I)$.

Thus, you can replace the map by $f^*-I:H^i(F\times\{N\})\to H^i(F\times\{N\})$. Of course $H^i(F\times\{N\})$ 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.

  • $\begingroup$ Thank you for your detailled and clear answer! $\endgroup$
    – klirk
    Nov 17, 2018 at 17:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .