I am computing the cohomology of $T^2$ by Meyer-Vietoris sequence. $T^2$ can be seen as the union of two open sets U and V s.t. U and V are diffeomorphic to a cylinder respectively. Thus U$\cap$V is a disjoint union of two cylinders which is homotopic to a disjoint union of two circles. In order to apply Meyer-Vietoris sequence I need to figure out the cohomology of disjoint union of two circles, but I am stuck on it for a long time, can you give me some ideas? Thank you!
[Math] What’s the cohomology of disjoint union of two circles
algebraic-topologydifferential-topologymanifolds
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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.
Its easy to visualize when looking at the fundamental domain of the torus:
$U$ is a neighbourhood of $x$, $V$ is a neighbourhood of $y$. Removing $x$ and $y$ from the torus is homotopy equivalent to removing the whole neighbourhoods $U$ and $V$. Further we can choose $U,V$ so large, that they fill the entire triangle they lie in.
So the fundamental domain is homotopy equivalent to the union of the boundary of the square with the diagonal. But some parts of the boundary are identified with each other. Doing the identifications, we obtain the wedge sum of three circles (their common point is $A$). One circle corresponds to the left side = right side of the bundary, one circle corresponds to the top side = bottom side and one circle corresponds to the diagonal.
More precisely, first identifying top with bottom, we get two distinct edges connecting the bottom/top left vertex with the bottom/top right vertex (corresponding to the bottom/top edge and the diagonal) and two circles (corresponding to the left and right edges).
Then we identify the left and right edges. The two circles we already had get identified, and as the bottom/top left edge gets identified with the bottom/top right edge, the edges which previously connected these two vetices become circles.
So in the end we get three circles which are connected in their common point $A$, the bottom/top now left/right vertex.
I hope the detailled description did not make things more confusing. I probably should've also drawn this.
Best Answer
consider the Mayer-Vietoris long exact sequence of $A\cup B$, and you assume that co-homology of empty set is $0$ , this will give you a direct isomorphism $H^n(A\cup B)= H^n(A) \oplus H^n(B)$.
So your answer would be $\mathbb{Z\oplus Z}$.