# Wang Sequence for the circle $S^1$

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.

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

## 1 Answer

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.

• Thank you for your detailled and clear answer! – klirk Nov 17 '18 at 17:10