De Rham cohomology of mobius band

Question

I need to calculate De Rham cohomology of mobius band without the boundary. Here I have seen the calculation of mobius strip cohomology which includes boundary. But when it's not included then I guess the choice of open sets $U$ and $V$ would be diffeomorphic to $\mathbb{R}^2$ and $U\cap V$ is also diffeomorphic to two copies of $\mathbb{R}^2$ then all the nonzero cohomology would be zero. But I am not sure whether my guess is correct or not as pictorially I could not draw the open sets $U$ and $V$. Any insight would be very helpful.

Answer 1

Let $M = [0,1]\times \mathbb R /\sim$, where $(0,x) \sim (1,-x)$ are identified. This gives a Möbius strip without boundary, as you can hopefully see. Now define a homotopy $ H : M \times [0,1] \to M$ by $H( (s,r),t)= (s, (1-t)r)$, that is, we shrink the radial component of the Möbius strip as $t$ increases. We see that $H(\cdot, 0)$ is the identity on $M$ and $H(\cdot, 1)$ is the radial projection onto $S^1\cong S^1\times \{0\} \subseteq M$. This is the desired homotopy.

Now by the invariance of de Rham Cohomology under homotopies, to compute $H^k_{dR}(M)$ it suffices to compute $H^k_{dR}(S^1)$. This is almost always the first example one computes in a course on differential geometry, giving us $$H_{dR}^k(M) \cong H^k_{dR}(S^1) \cong \begin{cases} \mathbb R, & k = 0,1\\ (0) & \text{else.}\end{cases}$$