Calculating the cohomology with compact support of the open Möbius strip

Solution 1:

Jim's comment is right, the glueing information is hidden in the map $\mathbb R^2 \to \mathbb R^2$ corresponding to the map $\phi : H^2_c(U\cap V) \to H^2_c(U) \oplus H^2_c(V)$.

Recall that if $U$ and $V$ are opens of $\mathbb R^2$, if we have a map $\iota : \mathbb U \to \mathbb V$, it induces a map $\iota^* : \Omega^2_c(\mathbb U) \to \Omega^2_c(\mathbb V)$, such that $\iota^* (f dxdy) = g dxdy$ where $g$ satisfies $f = (g \circ \iota) * J$ where $J$ is the jacobian of $\iota$, and $g=0$ outside the image of $\iota$. Consequently, depending on the (non-changing) sign of $J$, we have $\int_U \omega = \pm \int_V (\iota^*(\omega))$ forall $\omega \in \Omega^2_c(U)$ (the change of variable formula is exactly what we have, but with an absolute value on the jacobian). Therefore you have to keep track wether all your inclusion maps preserve orientation or not.

Write $U \cap V = W = W_1 \cup W_2$, so that $H^2_c(U\cap V) = H^2_c(W_1) \oplus H^2_c(W_2)$. We know that $U,V,W_1,W_2$ are diffeomorphic to $\mathbb R^2$, so the isomorphisms are induced by $\alpha : \omega \in \Omega^2_c(U) \mapsto \int_U \omega$

$\phi$ is given by $\phi(\omega_1 \oplus \omega_2) = (\iota_{W_1 \to U}^*(\omega_1) - \iota_{W_2 \to U}^*(\omega_2), \iota_{W_1 \to V}^*(\omega_1) - \iota_{W_2 \to V}^*(\omega_2))$. So $\alpha \circ \phi (\omega_1 \oplus \omega_2) = (\int_U \iota_{W_1 \to U}^*(\omega_1) - \int_U \iota_{W_2 \to U}^*(\omega_2), \int_V \iota_{W_1 \to V}^*(\omega_1) - \int_V \iota_{W_2 \to V}^*(\omega_2))$.
In the Möbius case, we usually pick maps such that $\iota_{W_1 \to U},\iota_{W_2 \to U},\iota_{W_1 \to V}$ are orientation preserving, and $\iota_{W_2 \to V}$ is orientation-reversing, so we obtain : $\alpha \circ \phi (\omega_1 \oplus \omega_2) = (\int_{W_1} \omega_1 + \int_{W_2} \omega_2, \int_{W_1} \omega_1 - \int_{W_2} \omega_2)$ thus the corresponding map $\tilde{\phi} : \mathbb R^2 \to \mathbb R^2$ is $(x,y) \mapsto (x+y,x-y)$, which is an isomorphism. Therefore, its kernel and cokernel, $H^1_c(M)$ and $H^2_c(M)$, are both zero.

Solution 2:

This is one of the execrises from Bott-Tu. Here I try to write a more friendly version of mercio's answer. Note that the Mayer-Vietoris sequence of forms with compact support is induced by the short exact sequence $$ 0\leftarrow \Omega^{*}_{c}(M)\leftarrow \Omega_{c}^{*}(U)\oplus \Omega_{c}^{*}(V)\leftarrow \Omega_{c}^{*}(U\cap V)\leftarrow 0 $$ where the first map is $(u,v)\rightarrow u+v$, second map $(\omega)\rightarrow (-j_{*}\omega, j_{*}\omega)$, where $j: \Omega_{c}^*(U)\rightarrow \Omega_{c}^{*}(M)$ is the inclusion map.

Now back to the setting of Mobius band, we have two pieces $U,V$ each differeomorphic to $\mathbb{R}^{2}$. The induced map is given by $$ H^{0}_{c}(U\cap V)\rightarrow H^{0}_{c}(U)\oplus H^{0}_{c}(V)\rightarrow H^{0}_{c}(M)\rightarrow H^{1}_{c}(U\cap V)\rightarrow H^{1}_{c}(U)\oplus H^{1}_{c}(V)\rightarrow H^{1}_{c}(M)\rightarrow H^{2}_{c}(U\cap V)\rightarrow H^{2}_{c}(U)\oplus H^{2}_{c}(V)\rightarrow H^{2}_{c}(M) $$ By Poincare lemma we know that $H^{0}_{c}(U\cap V)=H^{0}_{c}(U)=H^{0}_{c}(V)=0$. Thus $H^{0}_{c}(M)=0$. Now after truncation we have

$$ 0\rightarrow H^{1}_{c}(M)\rightarrow H^{2}_{c}(U\cap V)\rightarrow H^{2}_{c}(U)\oplus H^{2}_{c}(V)\rightarrow H^{2}_{c}(M)\rightarrow 0 $$ where the middle map is given by $\omega=(\omega_1, \omega_2)\rightarrow (-(j_{U}(\omega), (j_{V})(\omega))$. We use the identification of the Mobius band to let $$ j_{U}(\omega_1)=\omega_1, j_{U}(\omega_2)=\omega_2, j_{V}(\omega_{1})=\omega_{1}, j_{V}(\omega_{2})=-\omega_{2} $$ Then the image is given by $(-\omega_1-\omega_2, \omega_1-\omega_2)$ is isomorphic to $\mathbb{R}\oplus \mathbb{R}$. As a result both the kernel and cokernel are zero. Thus $H^{*}_{c}(M)=0$.