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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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What choices have you tried? – Mariano Suárez-Alvarez Jan 10 '12 at 21:22
To summon the muses, you might consider computing the cohomology of a cylinder first. – Mariano Suárez-Alvarez Jan 10 '12 at 21:28
Can you write the Möbius band as a union of two open disks, with intersection having two connected components? – Grumpy Parsnip Jan 10 '12 at 22:20
It may also help to draw the band as a square with identifications made, i.e. to break up the square in a way that descends to the quotient. That way you don't have to picture a mobius band in your head. – Dylan Wilson Jan 11 '12 at 16:01
(Of course, if the exercise doesn't require you to use Mayer-Vietoris, then there is a much easier way to get the answer...) – Dylan Wilson Jan 12 '12 at 1:25

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

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Why is 1-2+2-dim(H^1) = 0 ? And why is it H^1 (U int. V) = 0 ? – Werner Germán Busch Jul 1 '15 at 4:05

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