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Using Mayer-Vietoris to calculate the de Rham cohomology of the Möbius band $M$, what is the choice of separation?
i.e. $M=U\cup V$, which $U$ and $V$ well chosen for calculation?

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    Yes. But the cilinder is also homotopy equivalent to the circle. So you already know the cohomology. You can also write it as the union of two contractible sets. The intersection contains two elements.2012-01-12

1 Answers 1

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You can take these two open sets (in blue and red) :
Möbius band covered by two open sets

The long exact Mayer-Vietoris sequence is $0\rightarrow H^0(M)\rightarrow H^0(U)\oplus H^0(V)\rightarrow H^0(U\cap V)\rightarrow H^1(M)\rightarrow H^1(U)\oplus H^1(V)\rightarrow$

But the Möbius strip is connected, so $H^0(M)=\mathbb R$.
The open sets are connected, so $H^0(U)=H^0(V)=\mathbb R$.
The intersection has two connected components so $H^0(U\cap V)=\mathbb R^2$.
The opens sets are contractible so $H^1(U)=H^1(V)=0$.

So you have the exact sequence $0\rightarrow\mathbb R\rightarrow \mathbb R^2\rightarrow\mathbb R^2\rightarrow H^1(M)\rightarrow0$

So you have $1-2+2-\dim H^1(M)=0\Rightarrow \dim H^1(M)=1$.

Next, $H^1(U)\oplus H^1(V)=0\rightarrow H^1(U\cap V)=0\rightarrow H^2(M)\rightarrow H^2(U)\oplus H^2(V)=0$ gives you $H^2(M)=0$.

Conclusion: $H^k(M)=\left\{\begin{array}{ll}\mathbb R & \text{ if }k=0,1 \\ 0 & \text{ else}\end{array}\right.$

Remark: without the MV sequence, you could notice that the median circle $\mathbb S^1$ is a deformation retract of the Möbius strip and use https://math.stackexchange.com/a/162378/33615.

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    @WernerGermánBusch The first one is dimensional formula for exact sequence. You can refer to http://math.stackexchange.com/questions/255384/dimensions-of-vector-spaces-in-an-exact-sequence.2016-05-31